Using diodes to reduce voltage by 1-2 V under a very high load (e.g., 20A)

[SharpEars]

Would it be possible to use diodes to reduce voltage by 1-2 V for a high load circuit. Let's say I have a supply that puts out 9 V 20 A (max) which feeds several 5 V linear voltage regulators. I want to reduce the input voltage to the regulators as much as possible, because that excess voltage above the minimum needed for a 5 V output will only result in heat within the regulator.

My idea is to connect one or more diodes in between the power supply output and the regulators' V_i to reduce the incoming voltage from 9 V to say 7 V (or maybe even less, depending on the regulator dropout). I was thinking of doing this using diodes, taking advantage of the approximate 0.7 V drop when they are forward biased.

Accuracy (whether it's 0.7V or 0.9V across the diode depending on temperature and current) is not an issue here, but overheating/exploding diodes when a continuous 20A load is applied across them is an issue.

Is this a feasible solution for a load this high? I would imagine that if the voltage across the diode is almost 1V and the current is 20A, the diode has to dissipate a full 20 watts without the junction temperature going over its rated limit. Are such diodes even made and will they accomplish the task of lowering the voltage by 0.7-0.9 volts?

[wraper]

I'd say you shouldn't use linear regulators at all unless you have a very good reason for this. Also you could use power resistors, just calculate the resistance for needed voltage drop under max current. If the current will be smaller, hence less voltage drop on the resistors, then voltage regulators won't be at max stress anyway.

[SharpEars]

I'd say you shouldn't use linear regulators at all unless you have a very good reason for this. Also you could use power resistors, just calculate the resistance for needed voltage drop under max current. If the current will be smaller, hence less voltage drop on the resistors, then voltage regulators won't be at max stress anyway.

Let's go under the assumption that I will stick with linear regulators. This is more of a learning project for me than anything, dealing with high currents.

Your idea of using a power resistor is very interesting in that as you mentioned, the voltage across the resistor will increase as the resistance of the load decreases. I will have to give this some thought. The problem I see immediately is that for 5-10A loads, the power resistor that is in series with the regulators will do very little and the voltage regulators will wind up sinking a lot of current also. With a diode, there will be very little change in forward voltage drop as a function of the load. If the resistor could increase its resistance with the load decreasing, so as to keep the voltage going to the regulator relatively constant, that would be great, but now I am describing a voltage regulator .

On the other hand the voltage across the resistor will be highly variable, while with my diode(s) idea the range is very narrow as a function of current. So, I would still like to know whether the diode idea is viable for such high currents.

[PSR B1257]

Let's say I have a supply that puts out 9 V 20 A (max) which feeds several 5 V linear voltage regulators.

9V input for a 5V regulator (7805) is barely over it's minimum - no problems whatsoever 

What load current do the regulators have to supply?
And what's the overall purpose of this project?

[SharpEars]

Let's say I have a supply that puts out 9 V 20 A (max) which feeds several 5 V linear voltage regulators.

9V input for a 5V regulator (7805) is barely over it's minimum - no problems whatsoever 

What load current do the regulators have to supply?

The regulators have to supply a maximum 20 A load. I haven't decided on the number of regulators yet or which ones (so please don't ask about the exact dropout voltage).

The purpose of this project is to learn about linear regulators powering high loads (so a 7805 won't do here).

[madires]

Would it be possible to use diodes to reduce voltage by 1-2 V for a high load circuit. Let's say I have a supply that puts out 9 V 20 A (max) which feeds several 5 V linear voltage regulators. I want to reduce the input voltage to the regulators as much as possible, because that excess voltage above the minimum needed for a 5 V output will only result in heat within the regulator.

Either way you'll produce hot air You would just move some power (to dissipate) away from the linear voltage regulators to the diodes. If your circuits will run happily with a buck converter, that would be the best choice then.

[wraper]

You will need to provide sufficient cooling for the diodes. Also, to get even power distribution between the diodes (to reduce the stress on the single diode), you could use a few schottky diodes in series. However resistors would be much better as they can withstand much higher temperatures. I'd formalize this question how to move the heat dissipation out of semiconductors, not how to stress one semiconductor instead of another.

[SharpEars]

You will need to provide sufficient cooling for the diodes. Also, to get even power distribution between diodes, you could use a few schottky diodes in series. However resistors would be much better as they can withstand much higher temperatures. I'd formalize this question how to move the heat dissipation out of semiconductors, not how to stress one semiconductor instead of another.

Thank you and perhaps my question should be: If I were to use diodes to drop voltage by a volt or two, which diodes should I use so that they can (harmlessly) sink 20 A a piece.

[madires]

Thank you and perhaps my question should be: If I were to use diodes to drop voltage by a volt or two, which diodes should I use so that they can (harmlessly) sink 20 A a piece.

MBR4045 (both diodes can be connected in parallel for 40A).

[PSR B1257]

The regulators have to supply a maximum 20 A load. I haven't decided on the number of regulators yet or which ones (so please don't ask about the exact dropout voltage).

In order to dissipate 80W in a linar regulator, you should take at least tree power transistors and build a voltage reglator using opamp's (ideally one per transistor).

[SharpEars]

Thank you and perhaps my question should be: If I were to use diodes to drop voltage by a volt or two, which diodes should I use so that they can (harmlessly) sink 20 A a piece.

MBR4045 (both diodes can be connected in parallel for 40A).

Can that thing really handle 20A of continuous current (i.e., not instantaneous)?

Also, if I connect it's two diodes in parallel I will cut the voltage drop across the device in half - something I don't want to do (i.e., I want it to drop 0.7 V when forward biased).

[wraper]

Also, if I connect it's two diodes in parallel I will cut the voltage drop across the device in half - something I don't want to do (i.e., I want it to drop 0.7 V when forward biased).

How on the earth? you cannot connect them in series anyway because they have one common pin.

[PSR B1257]

if I connect it's two diodes in parallel I will cut the voltage drop across the device in half

Take a look at this regulator: http://cds.linear.com/docs/en/datasheet/1083ffe.pdf
With some additional resistors at the output you can parallel them up.

[SharpEars]

The regulators have to supply a maximum 20 A load. I haven't decided on the number of regulators yet or which ones (so please don't ask about the exact dropout voltage).

In order to dissipate 80W in a linar regulator, you should take at least tree power transistors and build a voltage reglator using opamp's (ideally one per transistor).

I am not trying to build a linear regulator from scratch here. I am trying to reduce the voltage coming into pre-made linear regulators to the minimum possible for them to produce 5 V and 20 A of output, so as to move the heat generated from this process from the regulators and onto something else, whether that's diodes or power resistors or .

[SharpEars]

if I connect it's two diodes in parallel I will cut the voltage drop across the device in half

Take a look at this regulator: http://cds.linear.com/docs/en/datasheet/1083ffe.pdf
With some additional resistors at the output you can parallel them up.

Greate! Let's get much more specific.

Let's assume I am going to use a pair of these regulators in parallel to produce 15 A at 5V. You see the V_in >= 6.5 V for this regulator. I would like to make my V_in 6.6 V, while using a 9 V supply. What can I do to accomplish this, short of adding another voltage regulator in front of this one.

Now this question just got a lot more specific!

[PSR B1257]

Fine. But it doesn't matter, where you drop the voltage (i.e. dissipate the heat).

Anyway, if you use several regulators, put one diode in series of each one.

I'm still not sure, what you need this for. If you need a 5V rail, use a 5V PSU 

[SharpEars]

Fine. But it doesn't matter, where you drop the voltage (i.e. dissipate the heat).

Anyway, if you use several regulators, put one diode in series of each one.

I'm still not sure, what you need this for. If you need a 5V rail, use a 5V PSU 

Hey this is a learning exercise and 5 V 20 A PSUs don't come cheap in any case  .

[madires]

Can that thing really handle 20A of continuous current (i.e., not instantaneous)?

Yes, please read the datasheet!

Also, if I connect it's two diodes in parallel I will cut the voltage drop across the device in half - something I don't want to do (i.e., I want it to drop 0.7 V when forward biased).

That would be true for a pair of resistors with the same resistance, but not for diodes. Vf depends on the current, but the difference between a 10A and 20A load would be around 100mV for the MBR4045. BTW, it's a Schottky diode.

[SharpEars]

Can that thing really handle 20A of continuous current (i.e., not instantaneous)?

Yes, please read the datasheet!

Also, if I connect it's two diodes in parallel I will cut the voltage drop across the device in half - something I don't want to do (i.e., I want it to drop 0.7 V when forward biased).

That would be true for a pair of resistors with the same resistance, but not for diodes. Vf depends on the current, but the difference between a 10A and 20A load would be around 100mV for the MBR4045. BTW, it's a Schottky diode.

OK, in that case it sounds good except for the Schottky part, I want it to drop 0.7 volts per device (not 0.45 V). I'd have to put five of these in series to bring 9 V down to say 6.5 V. Is there a non-Schottky part that does the same thing?

[edavid]

I'm still not sure, what you need this for. If you need a 5V rail, use a 5V PSU 

Hey this is a learning exercise and 5 V 20 A PSUs don't come cheap in any case  .

I think you are going to learn how not to do things.

You can get a 5V 20A power supply for under $20 on eBay, or use an old ATX power supply for $0.

[SharpEars]

I'm still not sure, what you need this for. If you need a 5V rail, use a 5V PSU 

Hey this is a learning exercise and 5 V 20 A PSUs don't come cheap in any case  .

I think you are going to learn how not to do things.

You can get a 5V 20A power supply for under $20 on eBay, or use an old ATX power supply for $0.

And you get what you pay for:

That 5V 20A power supply from eBay will give up the smoke in a week and have 1 V PtP ripple (not to mention that just beacuse they say 20A doesn't mean they can really sustain a 20A load for any significant period of time). Also, what exactly do I learn by buying the finished product?

The ATX power supply is a horrible idea. The voltage swing on the 5V rail is ridiculous - it has horrible load regulation and many assume a minimal load on the 12V rail to even regulate the 5V rail. Been there, done that, threw it out in despair.

[Cliff Matthews]

FWIW - Cheap 20-30amp TO-220 dual diodes can be plucked from discarded PC PSU's by the bucket load. It usually the output caps that fail..

[wraper]

That 5V 20A power supply from eBay will give up the smoke in a week and have 1 V PtP ripple (not to mention that just beacuse they say 20A doesn't mean they can really sustain a 20A load for any significant period of time). Also, what exactly do I learn by buying the finished product?

The ATX power supply is a horrible idea. The voltage swing on the 5V rail is ridiculous - it has horrible load regulation and many assume a minimal load on the 12V rail to even regulate the 5V rail. Been there, done that, threw it out in despair.

That $20 figure actually in not too far from the price for the quiet decent PSU, this is only 100W after all.
http://eu.mouser.com/ProductDetail/Mean-Well/SE-100-5/?qs=aVx8dZh3YZ50NLD1xOL7vg%3D%3D

[madires]

OK, in that case it sounds good except for the Schottky part, I want it to drop 0.7 volts per device (not 0.45 V). I'd have to put five of these in series to bring 9 V down to say 6.5 V. Is there a non-Schottky part that does the same thing?

Digikey offers a parametic search for example.

[SharpEars]

OK, in that case it sounds good except for the Schottky part, I want it to drop 0.7 volts per device (not 0.45 V). I'd have to put five of these in series to bring 9 V down to say 6.5 V. Is there a non-Schottky part that does the same thing?

Digikey offers a parametic search for example.

Thanks for that and I think I have found the perfect part:

http://www.vishay.com/docs/94053/hfa15tb6.pdf

Nice voltage drops from 1.1-1.6 V (forward biased), depending on current. It comes in a TO-220 case so it will be easy to heatsink and the price isn't too bad.

What do you guys think?

[mariush]

Linear regulators like 7805 need about 2v above the output voltage to work, not 4v.. so there's really no need for 9v at the input.

Using diodes seems like a bad idea to me, as you're just moving heat dissipation from a nicely package device that can be easily heatsinked (to220,to247 etc) to a small part that's (in the case of radial diodes) hard to cool and in addition, a part with a forward voltage drop that varies with temperature.

My advice would be to simply suck it up and move to linear regulators better than plain 7805, if you really insist on using linear regulators.

A good one would be LT1084, or LM1084 , or AZ1084... whatever 1084 ... it's a regulator that can do 5A with a voltage drop of about 1.3v and can be wired in parallel to increase the current output: http://www.digikey.com/product-search/en?FV=fff40027%2Cfff80182&k=1084&mnonly=0&newproducts=0&ColumnSort=1000011&page=1&stock=0&pbfree=0&rohs=0&quantity=10&ptm=0&fid=0&pageSize=25

So you could basically reconfigure your input voltage to 6.5-7.5v (I'd use 7.5v to account for voltage drop in the power supply cable at 20A, your regulators would probably get something around 7v) and parallel a bunch of these to get your 20A  (see page 15 in this datasheet and note you can replace the 2 feet of wire with ballast resistors as the note says)

But even if you're stuck with using 9v at the input, you could just connect two regulators in series, one configured to output 7-7.5v, then feed the second one with this and configure it to output 5v.  I'd say much safer than relying on diodes or resistors.

[SharpEars]

Linear regulators like 7805 need about 2v above the output voltage to work, not 4v.. so there's really no need for 9v at the input.

Using diodes seems like a bad idea to me, as you're just moving heat dissipation from a nicely package device that can be easily heatsinked (to220,to247 etc) to a small part that's (in the case of radial diodes) hard to cool and in addition, a part with a forward voltage drop that varies with temperature.

My advice would be to simply suck it up and move to linear regulators better than plain 7805, if you really insist on using linear regulators.

A good one would be LT1084, or LM1084 , or AZ1084... whatever 1084 ... it's a regulator that can do 5A with a voltage drop of about 1.3v and can be wired in parallel to increase the current output: http://www.digikey.com/product-search/en?FV=fff40027%2Cfff80182&k=1084&mnonly=0&newproducts=0&ColumnSort=1000011&page=1&stock=0&pbfree=0&rohs=0&quantity=10&ptm=0&fid=0&pageSize=25

So you could basically reconfigure your input voltage to 6.5-7.5v (I'd use 7.5v to account for voltage drop in the power supply cable at 20A, your regulators would probably get something around 7v) and parallel a bunch of these to get your 20A  (see page 15 in this datasheet and note you can replace the 2 feet of wire with ballast resistors as the note says)

But even if you're stuck with using 9v at the input, you could just connect two regulators in series, one configured to output 7-7.5v, then feed the second one with this and configure it to output 5v.  I'd say much safer than relying on diodes or resistors.

I just want to point a few things out and add some restrictions, now that I am more familiar with the issue:

First: The supply voltage is 9 V (+/- 10%). This is all I have, so this part cannot be changed. I can tweak it down to 8.1V, so that is my starting point.

Second: I don't plan on using a radially packaged diode (did you really mean axially?). I plan on using a TO-220 packaged one that I can easily attach a heatsink to.

Third: I don't want two levels of regulators. Because I plan on using multiple regulators as it is for this project - at least four of them, probably LT1084 (5 A) adjustables. I don't want to have to use eight, because they are not as cheap as TO-220 diodes.

So, the conclusion is to use a TO-220 packaged diode to reduce the 8.1 V input voltage to as close to 6.5 V as I can.

The issue:

All of the diodes I've looked at have a voltage drop that is dependent on junction temperature and current. This is OK, but I would like for the voltage drop to be nearly constant, at least at currents from 0-5 A and junction temps below 60 degrees Celsius.

If anyone knows of a part that has a voltage drop of a fairly constant 1-1.5 V for 0-5 A forward current, I would be much obliged.

[mariush]

First: The supply voltage is 9 V (+/- 10%). This is all I have, so this part cannot be changed. I can tweak it down to 8.1V, so that is my starting point.

9v +/- 10% means your input could already be 8.1-9.9v.  If you tweak it down to 8.1v, you're probably still going to have some variation, depending on power supply temperature, current load etc.

Are you sure it's sane to use 9v with 20A of current? For 20A, you need good quality power cable between the power supply and the input of the regulators, and proper connectors ... if you plan on powering your product through a regular round DC jack as seen on laptops and other small devices, 20A may be a problem.

Regarding power cable, with 20A there's going to be some voltage loss in your cables.. for example, with two AWG18 wires (common on ATX power supplies, to make it easier for you to compare sizes), with 1 meter of cable and 9v @20 A  at the end of the cable you'll have 8.16v (there's about 0.8v drop in the cable alone), see here.
That's why a 6pin pci-express cable meant for 75w (12v x 6.25A) is designed with 3 positive and 3 negative cables, for redundancy and minimal voltage drop.
 
It would make more sense to reduce the input current by using higher voltage, for example by using a 16-18v laptop adapter, then using your own dc-dc converter to get 5v at 20A. Or if you really want smooth dc voltage, to use a dc-dc converter to get about 6.5v @ 20-25A and then use linear regulators to cut the switching noise.

Diodes in the TO- packages look tempting, but all have voltage drop that varies with temperature and current. If you're going to put them on the same heatsink with the linear regulators, they're going to get hot and (most of the time) their forward voltage drop will lower a bit, which would only make the linear regulator dissipate more power as heat, so you're not gaining anything. Pretty much the same amount of heat is dissipated. 

By using two linear regulators, you're dissipating the same amount of heat as a diode+regulator combo, but the regulators themselves have circuitry inside which insures the amount of voltage drop on each regulator remains the same regardless of temperature and load... solving exactly the problem you want to solve.

One more note... pay attention to datasheets of linear regulators, in particular to power dissipation. Some linear regulators can only dissipate about 15w of power, some can go up to 25-30w ... these matter when choosing the heatsink and when calculating the maximum input voltage acceptable.

If your 1084 in only  guaranteed at up to 15w that means that at 5A current output, your input voltage - output voltage can be maximum 3V.

[SharpEars]

First: The supply voltage is 9 V (+/- 10%). This is all I have, so this part cannot be changed. I can tweak it down to 8.1V, so that is my starting point.

9v +/- 10% means your input could already be 8.1-9.9v.  If you tweak it down to 8.1v, you're probably still going to have some variation, depending on power supply temperature, current load etc.

Are you sure it's sane to use 9v with 20A of current? For 20A, you need good quality power cable between the power supply and the input of the regulators, and proper connectors ... if you plan on powering your product through a regular round DC jack as seen on laptops and other small devices, 20A may be a problem.

I've got good quality power cable up to and include 12 AWG, so check...

Regarding power cable, with 20A there's going to be some voltage loss in your cables.. for example, with two AWG18 wires (common on ATX power supplies, to make it easier for you to compare sizes), with 1 meter of cable and 9v @20 A  at the end of the cable you'll have 8.16v (there's about 0.8v drop in the cable alone), see here.
That's why a 6pin pci-express cable meant for 75w (12v x 6.25A) is designed with 3 positive and 3 negative cables, for redundancy and minimal voltage drop.

Insignificant voltage drop due to 12 AWG cables, check...

It would make more sense to reduce the input current by using higher voltage, for example by using a 16-18v laptop adapter, then using your own dc-dc converter to get 5v at 20A. Or if you really want smooth dc voltage, to use a dc-dc converter to get about 6.5v @ 20-25A and then use linear regulators to cut the switching noise.

This is an exercise in building, not buying (up to a point)...

Diodes in the TO- packages look tempting, but all have voltage drop that varies with temperature and current. If you're going to put them on the same heatsink with the linear regulators, they're going to get hot and (most of the time) their forward voltage drop will lower a bit, which would only make the linear regulator dissipate more power as heat, so you're not gaining anything. Pretty much the same amount of heat is dissipated. 

They will be on their own individual private heatsinks, because TO-220 heatsinks are cheap, so this is beneficial. I can spread the heat among multiple separate heatsinks which have at least a little distance between them.

By using two linear regulators, you're dissipating the same amount of heat as a diode+regulator combo, but the regulators themselves have circuitry inside which insures the amount of voltage drop on each regulator remains the same regardless of temperature and load... solving exactly the problem you want to solve.

This I agree with, but (LT1084 * 4) additional regulators cost more and require more supporting circuitry, which is why i am leaning towards the same number of simple cheap TO-220 rectifiers.

One more note... pay attention to datasheets of linear regulators, in particular to power dissipation. Some linear regulators can only dissipate about 15w of power, some can go up to 25-30w ... these matter when choosing the heatsink and when calculating the maximum input voltage acceptable.

If your 1084 in only  guaranteed at up to 15w that means that at 5A current output, your input voltage - output voltage can be maximum 3V.

Another good reason, to reduce the voltage to as close to 6.5 V as I can...

[SharpEars]

Another idea, what about a pre-regulator for each of the 5 A regulators that uses a zener and a pass transistor to lower the voltage from 8.1 V to 6.5 V ? The zener can control the pass transistor to strip exactly 1.6 V off of the incoming voltage and supply 6.5 V to the succeeding regulator.

Perhaps something along the lines of figure 3 on this page:

http://www.bristolwatch.com/ele/zener_power_supply.htm

[langwadt]

Another idea, what about a pre-regulator for each of the 5 A regulators that uses a zener and a pass transistor to lower the voltage from 8.1 V to 6.5 V ? The zener can control the pass transistor to strip exactly 1.6 V off of the incoming voltage and supply 6.5 V to the succeeding regulator.

Perhaps something along the lines of figure 3 on this page:

http://www.bristolwatch.com/ele/zener_power_supply.htm

or add a PNP "helper" to the regulators, like this: http://www.seekic.com/uploadfile/ic-circuit/2012812215950388.gif

[SharpEars]

Another idea, what about a pre-regulator for each of the 5 A regulators that uses a zener and a pass transistor to lower the voltage from 8.1 V to 6.5 V ? The zener can control the pass transistor to strip exactly 1.6 V off of the incoming voltage and supply 6.5 V to the succeeding regulator.

Perhaps something along the lines of figure 3 on this page:

http://www.bristolwatch.com/ele/zener_power_supply.htm

or add a PNP "helper" to the regulators, like this: http://www.seekic.com/uploadfile/ic-circuit/2012812215950388.gif

But this is a current boosting circuit, I am trying to limit input voltage...

[langwadt]

Another idea, what about a pre-regulator for each of the 5 A regulators that uses a zener and a pass transistor to lower the voltage from 8.1 V to 6.5 V ? The zener can control the pass transistor to strip exactly 1.6 V off of the incoming voltage and supply 6.5 V to the succeeding regulator.

Perhaps something along the lines of figure 3 on this page:

http://www.bristolwatch.com/ele/zener_power_supply.htm

or add a PNP "helper" to the regulators, like this: http://www.seekic.com/uploadfile/ic-circuit/2012812215950388.gif

But this is a current boosting circuit, I am trying to limit input voltage...

I'll assume lowering the input voltage is to reduce the power power dissipation in regulator, why else would you
do it?

The booster accomplishes that by letting the transistor do most of the work once the current goes above ~0.7V/R3

[SharpEars]

Another idea, what about a pre-regulator for each of the 5 A regulators that uses a zener and a pass transistor to lower the voltage from 8.1 V to 6.5 V ? The zener can control the pass transistor to strip exactly 1.6 V off of the incoming voltage and supply 6.5 V to the succeeding regulator.

Perhaps something along the lines of figure 3 on this page:

http://www.bristolwatch.com/ele/zener_power_supply.htm

or add a PNP "helper" to the regulators, like this: http://www.seekic.com/uploadfile/ic-circuit/2012812215950388.gif

But this is a current boosting circuit, I am trying to limit input voltage...

I'll assume lowering the input voltage is to reduce the power power dissipation in regulator, why else would you
do it?

The booster accomplishes that by letting the transistor do most of the work once the current goes above ~0.7V/R3

The problem with this scheme is that you lose the protection and noise rejection circuitry built into the regulator, among other things.

[langwadt]

Another idea, what about a pre-regulator for each of the 5 A regulators that uses a zener and a pass transistor to lower the voltage from 8.1 V to 6.5 V ? The zener can control the pass transistor to strip exactly 1.6 V off of the incoming voltage and supply 6.5 V to the succeeding regulator.

Perhaps something along the lines of figure 3 on this page:

http://www.bristolwatch.com/ele/zener_power_supply.htm

or add a PNP "helper" to the regulators, like this: http://www.seekic.com/uploadfile/ic-circuit/2012812215950388.gif

But this is a current boosting circuit, I am trying to limit input voltage...

I'll assume lowering the input voltage is to reduce the power power dissipation in regulator, why else would you
do it?

The booster accomplishes that by letting the transistor do most of the work once the current goes above ~0.7V/R3

The problem with this scheme is that you lose the protection and noise rejection circuitry built into the regulator, among other things.

not really, the regulator will still be regulating part of the current, that takes care of the noise
if the regular lowers the voltage or shut down to protect itself the transistor will follow

[techricky]

You could just use a chassis/heatsink mounting Bridge rectifier, 40A rating is common and they are quite rugged, parallel two of the diodes  both ~ to + , or use - to + for 2 paralleled double Diode drops..

[PSR B1257]

chassis/heatsink mounting Bridge rectifier, 40A rating is common and they are quite rugged

Jup, that's what he should do.

[oldway]

In high power electronics, use of diodes to cause a voltage drop is a method frequently used to adapt the battery voltage to the range of acceptable voltages by the consumer.

Different diode groups is used (with heatsink) wich are short circuited by relays or contactors when the battery is discharging.
The advantage of this method is that it is very reliable.
The diodes are the safest semiconductors, they accept large overloads safely.
In case of default, they go short circuit, which does not interrupt the consumer power

[PSR B1257]

The zener can control the pass transistor to strip exactly 1.6 V off of the incoming voltage and supply 6.5 V to the succeeding regulator.

That's not necessary at all.
At low currents the voltage drop over the diode is small, but that's fine since low current means low power dissipation in the regulator anyway.

[jeff.remus]

[SharpEars]

You could just use a chassis/heatsink mounting Bridge rectifier, 40A rating is common and they are quite rugged, parallel two of the diodes  both ~ to + , or use - to + for 2 paralleled double Diode drops..

Wow, interesting idea I hadn't thought off. Will connecting to two double diodes in parallel lower the voltage drop to the equivalent of one diode's drop?

That is

A ___/--->|---~--->|---\____ B
         \--->|---~--->|---/

What is the voltage drop from point A to point B (assuming forward bias, of course)?

[c4757p]

It will not, because diodes are nonlinear. It will lower the drop, but not by half. There's a bit of math here:

Diodes are described by the Ebers-Moll / Shockley equation:

I = Is (exp(V/Vt) - 1)
I ~= Is (exp(V/Vt))       # Because exp(V/Vt) is very large, so exp(V/Vt)-1 is only a very tiny bit smaller
I/Is = exp(V/Vt)
ln(I/Is) = V/Vt

Let V' be the new voltage drop after reducing the current by half:

ln(0.5I/Is) = V'/Vt
ln(0.5) + ln(I/Is) = V'/Vt
ln(I/Is) = V'/Vt - ln(0.5) = (V' - ln(0.5)Vt) / Vt
V/Vt = (V' - ln(0.5)Vt) / Vt
V = V' - ln(0.5)Vt
V' = V + ln(0.5)Vt = V - (0.693)(0.026volts) = V - 0.018volts

So the drop is reduced by about 0.018volts.

Notice that's a constant, regardless of the actual current!

So, you double the drop by putting them in series, and then reduce the drop by 18mV each (36mV total for the two in series).

[SharpEars]

It will not, because diodes are nonlinear. It will lower the drop, but not by half. There's a bit of math here:

Diodes are described by the Ebers-Moll / Shockley equation:

I = Is (exp(V/Vt) - 1)
I ~= Is (exp(V/Vt))       # Because exp(V/Vt) is very large, so exp(V/Vt)-1 is only a very tiny bit smaller
I/Is = exp(V/Vt)
ln(I/Is) = V/Vt

Let V' be the new voltage drop after reducing the current by half:

ln(0.5I/Is) = V'/Vt
ln(0.5) + ln(I/Is) = V'/Vt
ln(I/Is) = V'/Vt - ln(0.5) = (V' - ln(0.5)Vt) / Vt
V/Vt = (V' - ln(0.5)Vt) / Vt
V = V' - ln(0.5)Vt
V' = V + ln(0.5)Vt = V - (0.693)(0.026volts) = V - 0.018volts

So the drop is reduced by about 0.018volts.

Notice that's a constant, regardless of the actual current!

So, you double the drop by putting them in series, and then reduce the drop by 18mV each (36mV total for the two in series).

Thanks for that. The electronics math above is way beyond my level ¹.

1. My level: I can figure out the resistance of two resistors in parallel as long as they both have the same value  .

[TheElectricChicken]

You can do this very very easy, I have done it many many times it works just fine. Get a pack of 100 or even 1000 diodes from china on ebay. If they are 1A and you need 20A then you use 20 at a time in parallel to handle the current with comfort. Each bank of 20 drops .6Volts and you just add the banks in series to each other until you are close to your target voltage. a piece of cable can do the rest of the voltage drop if the current consumption is reasonably constant. like a roll of 20 amp cable can drop the rest, just measure the voltage across it and see if it is suitable.

The best place to get the BEST QUALITY power supplies is from old equipment. There is no equal to the quality and value. An old server can produce 12 and 5 volts at a total power of 1.2 KW and cost you $10. Here are a few power supplies I pulled out of a single large printer that I purchased for $1.

Two are 5V 10Amp and one is 12V 4amp. There are other power supplies in the printer but they are not on the desk in front of me. I brought these in to work on an arduino project. I think the arduino can work on 5v. 10A should be enough 8-) and the power supply cost like 5c when you divide up all the mechanical parts like stepper motors, switches, hookup cables, timing belts gears, LCD display and so much more. Try taking some old stuff apart. I used an audio amp as both the big power supply and the box for the power supply for my last project.
 

[PSR B1257]

need 20A then you use 20 at a time in parallel to handle the current with comfort.

At least until they die one at a time due to uneven current distribution (which of course leads to overheating).

Granted, it might work a while, especially if the diodes are of one batch. But sooner or later this setup will fail.

[madires]

need 20A then you use 20 at a time in parallel to handle the current with comfort.

At least until they die one at a time due to uneven current distribution (which of course leads to overheating).

Granted, it might work a while, especially if the diodes are of one batch. But sooner or later this setup will fail.

This is an interesting question. When the junction temperature rises, Vf decreases. But since Vf decreases, the current increases. So the hottest diode has to pass more current and will let out the magic smoke at some point.

[TheElectricChicken]

Granted, it might work a while, especially if the diodes are of one batch. But sooner or later this setup will fail.

This is an interesting question. When the junction temperature rises, Vf decreases. But since Vf decreases, the current increases. So the hottest diode has to pass more current and will let out the magic smoke at some point.

You're both wrong in the real world. I've done this many times. The technical explanation has to do with internal resistance temperature coefficients and the comparatively much much lower resistance of the cable that you scrap from an old machine to connect things up, in relation to the voltage drop due to internal resistance in the diodes, not just the junction drop. Don't use new cable because you cannot afford thick cable. Use scrapped cable so you can always overkill in everything that you do.

You should not be giving newbies bad advice, bad advice is anything that makes you sound like a prophet of DOOM, which you both do, don't tell people "don't bother", say, "try this". BETTER is to give newbies good advice they can always use. Here is a good diagram which works for any part even batteries.

[TheElectricChicken]

Granted, it might work a while, especially if the diodes are of one batch. But sooner or later this setup will fail.

batch has nothing to do with it. the internal resistance of the diode apart from junction drop evens things out. hot leads have higher resistance, in the real world it works fine. Also, you can totally destroy these silly pessimistic arguments by using 25 or 35! diodes in parallel.

Sit back and watch the end of the world not arrive when you switch it on. It just works and with hours of counseling we can move on with our lives.

[madires]

Granted, it might work a while, especially if the diodes are of one batch. But sooner or later this setup will fail.

This is an interesting question. When the junction temperature rises, Vf decreases. But since Vf decreases, the current increases. So the hottest diode has to pass more current and will let out the magic smoke at some point.

You're both wrong in the real world. I've done this many times. The technical explanation has to do with internal resistance temperature coefficients and the comparatively much much lower resistance of the cable that you scrap from an old machine to connect things up, in relation to the voltage drop due to internal resistance in the diodes, not just the junction drop. Don't use new cable because you cannot afford thick cable. Use scrapped cable so you can always overkill in everything that you do.

Yes, you can fix the problem by adding low value series resistors. But it's more helpful to explain what the problem is and then how to fix it. If a beginner just understands "yes, no problem", he will have no idea about the reason if something fails because of bad wiring. The point of your suggestion is the wire's resistance and the way of wiring. It doesn't tell why that helps. So the beginner doesn't know, that he has to select the wire, i.e. the diameter, very carefully. IMHO, I'd prefer to add dedicated series resistors to make it more fool proof.

[AF6LJ]

I'm still not sure, what you need this for. If you need a 5V rail, use a 5V PSU 

Hey this is a learning exercise and 5 V 20 A PSUs don't come cheap in any case  .

I think you are going to learn how not to do things.

You can get a 5V 20A power supply for under $20 on eBay, or use an old ATX power supply for $0.

I have been reading this thread, now re-reading it.. and I have to agree with this statement.
Nobody does anything by brute force unless they are trying to hack / modify existing gear.
Even still dropping 1.2 to 1.8V at 20 amps means you have 28 to 42W that has to be dissipated somewhere.
You cannot just make the heat disappear...

[TheElectricChicken]

Yes, you can fix the problem by adding low value series resistors. But it's more helpful to explain what the problem is and then how to fix it. If a beginner just understands "yes, no problem", he will have no idea about the reason if something fails because of bad wiring. The point of your suggestion is the wire's resistance and the way of wiring. It doesn't tell why that helps. So the beginner doesn't know, that he has to select the wire, i.e. the diameter, very carefully. IMHO, I'd prefer to add dedicated series resistors to make it more fool proof.

There are NO resistors to add. I've never said such a thing. In order to even out the current you simply follow the wiring diagram I drew. There is no need to understand or even know how much resistance the cable is providing, and no need to select. The method of wiring means that all resistances are reasonably equal and all currents are reasonably well distributed. SharpEars builds a powersupply, not an exploding space shuttle. I have built such a power supply, with diodes to drop the voltage.

If SharpEars follow me, then he has power supply same as I have a power supply. If he follow prophets of DOOM, he never start. Maybe you sell power supplies and not like someone build their own ? I dunno.

[TheElectricChicken]

Here is my power supply I built. I used some switches and a packet of 100 diodes which I already had. I also made a panel with square holes for the switches as well. No doubt people will say it's a mess and will cause the end of the world and burn down the city and dim the lights when switched on, and I hope so too, but so far it just works like a regular power supply. It just works like a regular 10-110v DC 2amp power supply. I flip switches on the front to short out the diodes. The diodes are in strings. Each string is half the size of the last string. The largest is I think 48 diodes from memory. It works well for it's purpose. Please feel free to say what YOU want a power supply for and claim that this wouldn't work. I'll be glad because I did not build it for naysayers, I build for myself. I think it is hilarious to use the old Marantz stereo I found, use the parts that are in it, and use the case of it as the case of the power supply and leave all the parts inside. Just add an alien face sucker to the front of the stereo (should show this to an audiophile who likes this one) and take control of the new Zombie stereo which now outputs voltage through the old speaker outputs. I think it's hilarious.

Here are some pictures how I amuse myself.

[TheElectricChicken]

IMHO, I'd prefer to add dedicated series resistors to make it more fool proof.

You appear to speak in contradiction here. resistors make it less foolproof because if the current is variable then using resistors will make the voltage drop variable. Diodes will clamp the voltage drop regardless of load.

[madires]

There are NO resistors to add. I've never said such a thing. In order to even out the current you simply follow the wiring diagram I drew. There is no need to understand or even know how much resistance the cable is providing, and no need to select. The method of wiring means that all resistances are reasonably equal and all currents are reasonably well distributed. SharpEars builds a powersupply, not an exploding space shuttle. I have built such a power supply, with diodes to drop the voltage.

And you wouldn't also add series resistors when paralleling BJTs?

Please see http://toshiba.semicon-storage.com/ap-en/design-support/faq/diode.html for "Is it OK to connect multiple diodes with the same part number in parallel?" for example.

[TheElectricChicken]

Please see http://toshiba.semicon-storage.com/ap-en/design-support/faq/diode.html for "Is it OK to connect multiple diodes with the same part number in parallel?" for example.

So not caring. this is like saying you require matched pairs of transistors or CPU's for a backyard job. You do not need oxygen free cable for this power supply either. Do you need oxygen free cable ? what do you think ? ridiculous. It's for people with lots of dollars and NO SENSE at all. same with quoting toshiba as if to suggest one diode will carry 4 percent more current than the next one. The amount of difference is proportional to the IQ of people who demand such precision for a DIY power supply. Hey, what do you think of my treating the stereo that way ? MwahHAhaHAhaHAHahAHAHhahaha

If you want to make ridiculous claims that  the diodes will have significant differences then quote the specification. It will state the tolerances of the diodes. A bag of diodes does not have enough variation to be measured at home.

Here is the circuit for above power supply.

[madires]

IMHO, I'd prefer to add dedicated series resistors to make it more fool proof.

You appear to speak in contradiction here. resistors make it less foolproof because if the current is variable then using resistors will make the voltage drop variable. Diodes will clamp the voltage drop regardless of load.

Sorry, Vf depends on the load current. Most diode datasheets provide a nice diagram showing the relation between Vf and If.

[TheElectricChicken]

http://rvroadtrip.us/graphics/12v_4batteries.jpg
http://ww3.wholesalesolar.com/battery-folder/parallel-batts.gif

here are two diagrams for wiring batteries in series parallel combination in such a way as the losses in the cables are evenly balanced same way that the diodes should be wired.

Sorry, Vf depends on the load current. Most diode datasheets provide a nice diagram showing the relation between Vf and If.

"Most diode datasheets" ...you can't provide one, or don't have one.

[madires]

Sorry, Vf depends on the load current. Most diode datasheets provide a nice diagram showing the relation between Vf and If.

"Most diode datasheets" ...you can't provide one, or don't have one.

Ok, let's take a classic diode: https://www.fairchildsemi.com/datasheets/1N/1N4001.pdf Figure 2. Based on your lack of basic knowledge I have to assume you're a beginner.

[SharpEars]

Please see http://toshiba.semicon-storage.com/ap-en/design-support/faq/diode.html for "Is it OK to connect multiple diodes with the same part number in parallel?" for example.

So not caring. this is like saying you require matched pairs of transistors or CPU's for a backyard job. You do not need oxygen free cable for this power supply either. Do you need oxygen free cable ? what do you think ? ridiculous. It's for people with lots of dollars and NO SENSE at all. same with quoting toshiba as if to suggest one diode will carry 4 percent more current than the next one. The amount of difference is proportional to the IQ of people who demand such precision for a DIY power supply. Hey, what do you think of my treating the stereo that way ? MwahHAhaHAhaHAHahAHAHhahaha

If you want to make ridiculous claims that  the diodes will have significant differences then quote the specification. It will state the tolerances of the diodes. A bag of diodes does not have enough variation to be measured at home.

Here is the circuit for above power supply.

Wait, how many amps do you expect this thing to put out continuously, before you smell some diode smoke? Those diodes are rated for 1 A of forward current.

I guess if your intent was to build a 1 A power supply, then yes, possibly mission accomplished...

[PSR B1257]

batch has nothing to do with it.

Well, it has. Did you think, semiconductor manufacturer producing A, B, C-grade transistors intentionally?
They don't. They producing just one type, but due to imperfection during manufacturing the acutal parameters of each exemplar spread more or less.
The same is true for diodes of course.

hot leads have higher resistance, in the real world it works fine.

They have. But who wires dozens of diodes in parallel using separate wires? Long and thin wires providing a voltage drop which compensates for the decreasing forward voltage - true.

BETTER is to give newbies good advice they can always use.

25 diodes in parallel rather than one diode bridge capable of carrying the whole current - best advice i've ever read...

It also very popular, to put LEDs in parallel on a circuit board. There you have almost no trace resistance - hence the LEDs die quite often (self-experienced due to lack of knowledge).

[SharpEars]

It also very popular, to put LEDs in parallel on a circuit board. There you have almost no trace resistance - hence the LEDs die quite often (self-experienced due to lack of knowledge).

But wait, if you put them in parallel at the current entry point of their 220-330 ohm current limiting resistor, things will be OK, because the current limiting resistor will also act as a ballast resistor I think.

Maybe I'm wrong.

What I am saying is if two or more LEDs are connected in the following parallel configuration, things should be OK:

       /-\/\/\/--|<--\
___/                       \____
      \                       /
       \-/\/\/\--|<--/

..., things are much worse if instead you use the following configuration with a single current limiting resistor:

                  /--|<--\
__/\/\/\__/            \____
                 \            /
                  \--|<--/

[TheElectricChicken]

Wait, how many amps do you expect this thing to put out continuously, before you smell some diode smoke? Those diodes are rated for 1 A of forward current.

I guess if your intent was to build a 1 A power supply, then yes, possibly mission accomplished...

They don't put out smoke !, they work perfectly because I know what I'm doing. Yes, it's just a few amps because it goes from zero to well over 100Volts DC. If you have two 300 Watt supplies and one is 10 volts and the other is 100 volts then one will have 30 amps and the other just three.

I believe they were 3A diodes but I should have to fine the top of the pack wherever it went.  You just double them up until you have enough to handle the total current. Even if one diode is out by 2 % tolerance it's completely irrelevant considering you will get a few milliamps difference between diodes at 1amp each for example. If it worries you, just add one more diode for paranoia's safety's sake.

somone posed a datasheet which proves my points and that there won't be any smoke.
https://www.fairchildsemi.com/datasheets/1N/1N4001.pdf

For the 4001 you can overload it without the smoke, up to 30a instead of 1amp, for short periods. the smaller the overload, the longer it can take it. It also shows that the current will distribute evenly by itself, there is a graph in there showing that the voltage drop increses with increasing current, comparable to a resistor. So as one gets 'overloaded' it's voltage drop increases and the current shunts to other diodes.

Take into account that if all the elaborate studies and calculations and advice is all wrong, then one of the string of diodes is going to blow up, oh the humanity, and like a fuse, protect the rest, so you lose one 5c diode which you can salvage from any circuit board. Eventually you should say 'this is my fourth year of theorizing what these diodes could do, I've given up family, social life and my health, I know some people say I should just go ahead and risk a diode, but they are all crazy ~!!!!!'

I really think you should think four times before you try any of this insanity because what would a person who has built one know anyhow.

[TheElectricChicken]

25 diodes in parallel rather than one diode bridge capable of carrying the whole current - best advice i've ever read...

I'm sorry you must be looking for the thread titled " Using bridges to reduce voltage by 1-2 V under a very high load (e.g., 20A)

http://imgur.com/gallery/PZRWvjd

[TheElectricChicken]

But wait, if you put them in parallel at the current entry point of their 220-330 ohm current limiting resistor, things will be OK, because the current limiting resistor will also act as a ballast resistor I think.

Maybe I'm wrong.

What I am saying is if two or more LEDs are connected in the following parallel configuration, things should be OK:

       /-\/\/\/--|<--\
___/                       \____
      \                       /
       \-/\/\/\--|<--/

..., things are much worse if instead you use the following configuration with a single current limiting resistor:

                  /--|<--\
__/\/\/\__/            \____
                 \            /
                  \--|<--/

you are correct. The thing to remember with LEDs is, if you want them to last for ages, then run them at less than their rated maximum. If it works well at half or a quarter it's rated current, then give it just that.

Leds have great differences in voltage drop according to colour and type, much much more than say power diodes.

[SharpEars]

25 diodes in parallel rather than one diode bridge capable of carrying the whole current - best advice i've ever read...

I'm sorry you must be looking for the thread titled " Using bridges to reduce voltage by 1-2 V under a very high load (e.g., 20A)

http://imgur.com/gallery/PZRWvjd

Since I created this thread, I will chime in here.

Bridge rectifiers are made from diodes, so bridge rectifiers are an acceptable response as was the suggestion of using power resistors with the caveats given.

I am open to suggestions outside of the strict boundaries implied by the thread title.

[SharpEars]

For the 4001 you can overload it without the smoke, up to 30a instead of 1amp, for short periods. the smaller the overload, the longer it can take it. It also shows that the current will distribute evenly by itself, there is a graph in there showing that the voltage drop increses with increasing current, comparable to a resistor. So as one gets 'overloaded' it's voltage drop increases and the current shunts to other diodes.

So, if the voltage drop increases with the current flowing through the diode, why is it a bad idea to connect them in parallel?

It seems like the rising forward voltage drop (with current) will balance them all out (i.e., no thermal runaway scenario).

[TheElectricChicken]

Bridge rectifiers are made from diodes, so bridge rectifiers are an acceptable response as was the suggestion of using power resistors with the caveats given.

I am open to suggestions outside of the strict boundaries implied by the thread title.

Yes it would work, I am confident that the respondent may have possibly been having a bit of a troll with some of their responses. There are times when you just have to pick up the soldering iron, make it, and then watch the entire world NOT come to an end as the device simply works as advertised.,

[AF6LJ]

There isn't really enough information to make intelligent suggestions...

What is the nature of the load?
Why are you stuck with a 9V supply?
What enviroment is this gear used in?
What is it for that matter?
How many 5V loads are there and why not use distributed regulators?
What about air flow, or is this water cooled?
EMI considerations?

[TheElectricChicken]

So, if the voltage drop increases with the current flowing through the diode, why is it a bad idea to connect them in parallel?

It seems like the rising forward voltage drop (with current) will balance them all out (i.e., no thermal runaway scenario).

Exactly correct. But I liked tricking them in to getting a datasheet which they thought would prove their point but actually proves mine.

For power diodes soldered together in parallel, it would be enormously difficult to measure any discrepancies in the differing currents. Now different colour LEDS, yes, simple and easy and some won't light. But power diodes ? forget it.

[PSR B1257]

It seems like the rising forward voltage drop (with current) will balance them all out (i.e., no thermal runaway scenario).

The thermal runaway is exactly the problem which TheElectroicChicken supposedly simply ignores.

The bridge rectifier gets rid of this issue, since all diodes are on the same substrate and therefore at the same temperature. And they are manufactured literally at the same time, so they is basically to parameter drift.

That's all I say.

[TheElectricChicken]

It seems like the rising forward voltage drop (with current) will balance them all out (i.e., no thermal runaway scenario).

The thermal runaway is exactly the problem which TheElectroicChicken supposedly simply ignores.

The bridge rectifier gets rid of this issue, since all diodes are on the same substrate and therefore at the same temperature. And they are manufactured literally at the same time, so they is basically to parameter drift.

That's all I say.

I'm not ignoring it, I'm making fun of it. I drew a picture didn't you see ?

Locked to the same temperature makes it worse because it doesn't allow the temperature coefficient of resistance even anything out.

[madires]

So, if the voltage drop increases with the current flowing through the diode, why is it a bad idea to connect them in parallel?

It seems like the rising forward voltage drop (with current) will balance them all out (i.e., no thermal runaway scenario).

Exactly correct. But I liked tricking them in to getting a datasheet which they thought would prove their point but actually proves mine.

As I said before, Vf decreases when the junction temperature increases. And of course you're smarter than a company producing semiconductors   

[TheElectricChicken]

As I said before, Vf decreases when the junction temperature increases. And of course you're smarter than a company producing semiconductors   

Oh, I'm smarter than several for sure.

[c4757p]

I see we're having fun with the "screw learning and understanding, let's all just derp our way to a working circuit" brigade.

[TheElectricChicken]

I see we're having fun with the "screw learning and understanding, let's all just derp our way to a working circuit" brigade.

All I'm saying is you won't find videos of diodes having runaway nuclear meltdowns and you won't have to dynamically load share between them with coils because in the real world you are going to have a hard time finding this ever happen. ever. connect a Laboratory power supply to a half dozen identical diodes from a packet, twist and solder the leads together and send the total rated current through. See what (doesn't) happen. You can make caps, leds, batteries, and lots else explode, but diodes running in parallel at or below rating ? Nope. not easy. Just build it. I'd bet all my money on it, so long as you don't go over the total for half a dozen.

Anyhow, it's whether you can use diodes, not diodes in parallel that's the topic, and the answer either way is yes, and I have uploaded the photos to prove diodes are fine to drop voltage.

[c4757p]

Let's do this like engineers, not derps. Engineers use numbers. I took a 6A-rated diode and attached it to a power supply set to 1.5A (overhead for a bit of thermal mass, mostly to give myself time to write down numbers as my GPIB setup's a bit fucked right now). I included the diode's leads in the voltage measurement, since you think that's important.

Voltage drop immediately at power-on: 0.855 V
Voltage drop after the diode heats up and stabilizes: 0.819 V.


Okay, I'll let it cool and try again at 3A:

Voltage drop immediately at power-on: 0.877 V
Voltage drop after the diode heats up and stabilizes: 0.800 V.

Bullshit busted. Diodes have negative temperature coefficients, even accounting for lead resistance.

By the way, note that the voltage drop increased by 22mV when I doubled the current, which is quite close to the 18mV I calculated earlier.

Turns out diodes are actually fairly predictable.

[TheElectricChicken]

Let's do this like engineers, not derps. Engineers use numbers. I took a 6A-rated diode and attached it to a power supply set to 1.5A (overhead for a bit of thermal mass, mostly to give myself time to write down numbers as my GPIB setup's a bit fucked right now). I included the diode's leads in the voltage measurement, since you think that's important.

Voltage drop immediately at power-on: 0.855 V
Voltage drop after the diode heats up and stabilizes: 0.819 V.


Okay, I'll let it cool and try again at 3A:

Voltage drop immediately at power-on: 0.877 V
Voltage drop after the diode heats up and stabilizes: 0.800 V.

Bullshit busted. Diodes have negative temperature coefficients, even accounting for lead resistance.

By the way, note that the voltage drop increased by 22mV when I doubled the current, which is quite close to the 18mV I calculated earlier.

Turns out diodes are actually fairly predictable.

Yes, yes real world testing and real world results !!! finally. Great work. I don't have a decent lab supply at the moment to demo it myself. The original question is about using pre-regulator diodes to unload the main regulator, which is good logical thinking and works well. The variations in drop are cleaned up by the main regulator, so it ends up just doin' it.

Edit: oh P.S. forgot to ask, are you still ok, Hole in the desk, fire department called, or nothing happened ? Would you say diodes are a great pre-regulator to dump some of the load or not?

[TheElectricChicken]

I took a 6A-rated diode and attached it to a power supply set to 1.5A (overhead for a bit of thermal mass, mostly to give myself time to write down numbers as my GPIB setup's a bit fucked right now).
...
Okay, I'll let it cool and try again at 3A:
...

Bullshit busted. Diodes have negative temperature coefficients, even accounting for lead resistance.

are you willing to lose half a dozen smaller ones to test the parallel theory ?

[c4757p]

are you willing to lose half a dozen smaller ones to test the parallel theory ?

Working on that now.

[TheElectricChicken]

are you willing to lose half a dozen smaller ones to test the parallel theory ?

Working on that now.

  Excellent !!!!! just three words, Video, Video and Video ? at what point on the meter do they go up ? can you crank it slowly and see ? This is so cool.

[TheElectricChicken]

I would do this myself but I can't find my packet of 100 diodes I had a while ago, I can't guess where they all are 

[madires]

Bullshit busted. Diodes have negative temperature coefficients, even accounting for lead resistance.

By the way, note that the voltage drop increased by 22mV when I doubled the current, which is quite close to the 18mV I calculated earlier.

Turns out diodes are actually fairly predictable.

Therefore a diode can be used as a temperature sensor or to compensate the positive temperature coefficient of a Zener diode.

[AF6LJ]

Let's do this like engineers, not derps. Engineers use numbers. I took a 6A-rated diode and attached it to a power supply set to 1.5A (overhead for a bit of thermal mass, mostly to give myself time to write down numbers as my GPIB setup's a bit fucked right now). I included the diode's leads in the voltage measurement, since you think that's important.

Voltage drop immediately at power-on: 0.855 V
Voltage drop after the diode heats up and stabilizes: 0.819 V.


Okay, I'll let it cool and try again at 3A:

Voltage drop immediately at power-on: 0.877 V
Voltage drop after the diode heats up and stabilizes: 0.800 V.

Bullshit busted. Diodes have negative temperature coefficients, even accounting for lead resistance.

By the way, note that the voltage drop increased by 22mV when I doubled the current, which is quite close to the 18mV I calculated earlier.

Turns out diodes are actually fairly predictable.

I also don't think with any self respect wouldn't use a diode that way to begin with.
My opinion....

[CatalinaWOW]

Electric Chickens approach will work most of the time.  It is probably fine for careful use in a hobby application.  But tell me if you want to buy gear that works "most of the time".  Fails or even goes up in smoke one time in 100 or one time in 1000 or whatever your personal standard for reliability is. 

That is why the engineers here are telling you that it isn't a good design approach.  Thermal runaway is real.  Many of us have observed it. 

Paralleling diodes is much like connecting capacitors in series to get a higher voltage rating.  The engineers will tell you that it isn't great practice, and that if you do choose to do it you should have a resistive voltage divider with appropriate values to assure that each capacitor stays within its assured ratings.  Does that mean that every time you hook up a capacitor chain that the magic smoke will come out.  Of course not.  But remember that Murphy saves his mischief for the time when it hurts the most.

[SharpEars]

are you willing to lose half a dozen smaller ones to test the parallel theory ?

Working on that now.

I am very curious to see the results, one way or the other.

[c4757p]

And, to return: you win this one. By grabbing a handful of diodes, directly paralleling them, and then pumping a bunch of current through them, I was not able to blow them up.

I will make these observations, however:

• V_F was chaotic, implying the action of positive feedback. There was positive feedback, just not enough.
• The diodes were very closely coupled, with their somewhat heavy leads twisted together. This close thermal coupling would have helped smooth temperature differences. So if you insist on doing this, make sure you repeat that: keep the diodes as close to touching as possible.
• Thermal runaway still isn't a myth just because I couldn't make it happen now. I've seen it happen. It's a well known problem in particular in poorly designed power amplifiers. The mechanism of it is known, and I already proved that mechanism to be present here. So while you might be able to get away with it nine times out of ten....just don't. Sure, for a quick throw-together it'll work, but don't actually make something with it.
• As I said, you have to keep them thermally coupled. This will cause power dissipation issues! A 1W resistor can't take 1W if it's hugging twenty other 1W resistors and holding in its heat. The same applies to diodes.
• Single diodes that can take 20A are hardly prohibitively expensive.
  

Conclusion? It'll work, sometimes, if you're careful. Do it if you like living on the edge.

[SharpEars]

And, to return: you win this one. By grabbing a handful of diodes, directly paralleling them, and then pumping a bunch of current through them, I was not able to blow them up.

I will make these observations, however:

• V_F was chaotic, implying the action of positive feedback. There was positive feedback, just not enough.
• The diodes were very closely coupled, with their somewhat heavy leads twisted together. This close thermal coupling would have helped smooth temperature differences. So if you insist on doing this, make sure you repeat that: keep the diodes as close to touching as possible.
• Thermal runaway still isn't a myth just because I couldn't make it happen now. I've seen it happen. It's a well known problem in particular in poorly designed power amplifiers. The mechanism of it is known, and I already proved that mechanism to be present here. So while you might be able to get away with it nine times out of ten....just don't. Sure, for a quick throw-together it'll work, but don't actually make something with it.
• As I said, you have to keep them thermally coupled. This will cause power dissipation issues! A 1W resistor can't take 1W if it's hugging twenty other 1W resistors and holding in its heat. The same applies to diodes.
• Single diodes that can take 20A are hardly prohibitively expensive.
  

Conclusion? It'll work, sometimes, if you're careful. Do it if you like living on the edge.

Thanks for doing this test and the advice on diode spacing for anyone crazy enough to go TheElectricChicken route. As for me I'll stick with a bridge rectifier or other alternatives that can be properly heat sinked.

Having said that, I do appreciate some practice being thrown into the fray on top of all of the theory that this forum generates  .

I think most of the theoretical advice is great and we should all learn and know it, but we're not always building pacemakers and in a pinch have to settle for "second best." In my particular case, given that I will be dealing with high currents, I'll take a pass on the "let's just chance it" route and do this the right way.

[edavid]

It seems like the rising forward voltage drop (with current) will balance them all out (i.e., no thermal runaway scenario).

The thermal runaway is exactly the problem which TheElectroicChicken supposedly simply ignores.

The bridge rectifier gets rid of this issue, since all diodes are on the same substrate and therefore at the same temperature. And they are manufactured literally at the same time, so they is basically to parameter drift.

That is not correct, power bridges use separate die or pellets in the same package.  The only monolithic bridges are parts like the old CA3019 and CA3039.

[TheElectricChicken]

And, to return: you win this one. By grabbing a handful of diodes, directly paralleling them, and then pumping a bunch of current through them, I was not able to blow them up.

I will make these observations, however:

• V_F was chaotic, implying the action of positive feedback. There was positive feedback, just not enough.
• The diodes were very closely coupled, with their somewhat heavy leads twisted together. This close thermal coupling would have helped smooth temperature differences. So if you insist on doing this, make sure you repeat that: keep the diodes as close to touching as possible.
• Thermal runaway still isn't a myth just because I couldn't make it happen now. I've seen it happen. It's a well known problem in particular in poorly designed power amplifiers. The mechanism of it is known, and I already proved that mechanism to be present here. So while you might be able to get away with it nine times out of ten....just don't. Sure, for a quick throw-together it'll work, but don't actually make something with it.
• As I said, you have to keep them thermally coupled. This will cause power dissipation issues! A 1W resistor can't take 1W if it's hugging twenty other 1W resistors and holding in its heat. The same applies to diodes.
• Single diodes that can take 20A are hardly prohibitively expensive.
  

Firstly and foremost, thank you for doing real world experimentation for all of us. Secondly thank you for sharing your findings with us even though they didn't support your own view. I'll admit I didn't know about the runaway effect too. I'll own that, however I'll say I have been soldering since I was a child and I've seen and blown many a LED, see many, many parts blow like caps, resistors, diodes (but not for runaway), transistors, just about everything. The runaway effect is far too rare to worry about, and there are not many ways you can solder diodes in parallel without one hot one heating up the rest and sharing the load. They do is what I know and what the real world shows too.

I think it'll be impossible to get the prophets of doom to admit that your real world findings apply to this planet and they'll still say that doom awaits everyone and risking a hideously expensive diode is just not worth the risk. We should comfort these people by explaining that an inductor should be inserted into each series string of diodes so that the parallel legs will dynamically share the load. I've never seen this in real life, but it does exist if you google it, and it surely exists on the planet where these people live. So lets help them move forward to the bright sunny day when they'd actually consider building something.

AU $1.19 free postage
http://www.ebay.com.au/itm/Brand-New-8-Value-40pcs-Rectifier-Diode-kit-1N4001-1N4007-1N5408-Hot/321768544480
Brand New 8 Value 40pcs Rectifier Diode kit 1N4001 ~1N4007 ~ 1N5408
Type    Quantity
IN4001    5PCS
IN4002    5PCS
IN4003    5PCS
IN4004    5PCS
IN4005    5PCS
IN4006    5PCS
IN4007    5PCS
IN5408    5PCS

AU $1.20 free postage
http://www.ebay.com.au/itm/KBPC5010-1000-Volt-Bridge-Rectifier-50-Amp-50A-Metal-Case-1000V-Diode-Bridge-/181803791770
KBPC5010 1000 Volt Bridge Rectifier 50 Amp 50A Metal Case 1000V Diode Bridge

You can live dangerously like I do, and repurpose equipment, risk burning your fingers occasionally desoldering, like, if there is not a full power supply like I posted pics of earlier.

Conclusion? It'll work, sometimes, if you're careful. Do it if you like living on the edge.

Yes, let's live on the bleeding edge.

[c4757p]

I think it'll be impossible to get the prophets of doom to admit that your real world findings apply to this planet and they'll still say that doom awaits everyone and risking a hideously expensive diode is just not worth the risk.

I did actual tests. You just blurted out your gut feeling, which is the exact same thing the "prophets of doom" did, except that the "prophets of doom" based their gut feelings on real physics rather than wishful thinking. You'll note that the prophets are still correct! There is still a possibility of the diodes crapping themselves, and I advise you try some more diverse tests than this.

On the other hand, you insisted that the very well established negative temperature coefficient (even in practice very close to -2mV/K) of silicon PN diodes didn't exist.

It'd do you well to realize that the prophets are cautious because they have some experience and perhaps some pride in their work.

We should comfort these people by explaining that an inductor should be inserted into each series string of diodes so that the parallel legs will dynamically share the load.

An... inductor? You surely have no clue what you're talking about. It'll work... due to its resistance. You could just use a resistor.

Conclusion? It'll work, sometimes, if you're careful. Do it if you like living on the edge.

Yes, let's live on the bleeding edge.

Do so intelligently, please. If you're going to grab a fistful of diodes and twist them together like I did, it'll probably work due to the thermal coupling. If you try paralleling them on a PCB, your chances will be lower. And try it with 20A - I strongly suspect there will simply be too much heat in one place for those poor diodes to get rid of it.

[TheElectricChicken]

I did actual tests. You just blurted out your gut feeling

It's not called gut feeling sonny, it's called a hell of a lot of experience. A hell of a lot of actual tests in the real world included.

On the other hand, you insisted that the very well established negative temperature coefficient (even in practice very close to -2mV/K) of silicon PN diodes didn't exist.

where did I say that ?

We should comfort these people by explaining that an inductor should be inserted into each series string of diodes so that the parallel legs will dynamically share the load.

An... inductor? You surely have no clue what you're talking about. It'll work... due to its resistance. You could just use a resistor.

the following is from
http://engineering.electrical-equipment.org/electrical-distribution/series-and-parallel-connected-diodes.html

"Consider two diodes connected in parallel configuration. Current would be shared among the two diodes. To make this sharing equal, inductors (with same inductances) are connected. When current at D1 increases, the voltage drop across L1 increases, generating an opposite polarity value at L2. "

[retrolefty]

Would it be possible to use diodes to reduce voltage by 1-2 V for a high load circuit. Let's say I have a supply that puts out 9 V 20 A (max) which feeds several 5 V linear voltage regulators. I want to reduce the input voltage to the regulators as much as possible, because that excess voltage above the minimum needed for a 5 V output will only result in heat within the regulator.

 I would question the viability of this 'solution' in a real world situation. What is the normal AC line voltage variation your power supply will see over a 24 hour period?

 When trying to maintain a constant (low) voltage drop across the linear voltage regulator's series pass transistors when the source voltage may have variation more then just a fixed voltage drop across a series diode. That just moves the operation point, not compensate for variation of the AC voltage source driving the power supply?

[madires]

Some example for the load sharing issue: https://library.e.abb.com/public/34663f6e6f61d4acc12573400023ae32/High%20power%20rectifier%20diodes_5SYA%202029.pdf section 3.2 "Current sharing issues at paralleling of devices". Don't worry about the 6kA, the same rules apply to 1A diodes too.

You might be lucky with the next best bag of 1A diodes and an 1A diode will not explode at 1.001A suddenly. And other things, like the load profile for example, need to be taken into account. But it's bad design. There is a huge difference between "it's working, done that a hundred times" and good design. Even cheap Chinese lab PSUs got series resistors for each paralleled pass transistor (similar thermal runaway issue).

Also slighting and trolling EEs isn't a good idea. The next time you need some help and would love to get an answer from a grey beard, he'll just think "sorry son".

[wraper]

"Consider two diodes connected in parallel configuration. Current would be shared among the two diodes. To make this sharing equal, inductors (with same inductances) are connected. When current at D1 increases, the voltage drop across L1 increases, generating an opposite polarity value at L2. "

But how are you going to dynamically share current at what is almost DC (particularly this case or rectifying at low frequency)?
As of thermal runaway: here are setup with 6x 1N4007 from the same tape (still on the tape actually), 2.5A total current.
You can see 10^oC difference even between the diodes from the same batch. And IMO it is almost the best case you can get without selecting the diodes.

[TheElectricChicken]

As of thermal runaway: here are setup with 6x 1N4007 from the same tape (still on the tape actually), 2.5A total current.
You can see 10^oC difference even between the diodes from the same batch. And IMO it is almost the best case you can get without selecting the diodes.

That's the most impressive thing I've seen yet. It reminds me of Dr Karl and how excited he gets when people get out there and actually DO science experiments in real life. Outstanding work.

But how are you going to dynamically share current at what is almost DC (particularly this case or rectifying at low frequency)?

I have no idea, I just posted up that inductors can be used, though I haven't seen it. I expect that the currents move initially with noise and react with each other to resonate, I haven't come across it myself. My intuition would suggest it may be more to do with industrial mercury rectifiers or large currents because you really would never need to balance at the currents that we are using, you'd never come across the need. A few extra diodes is easy at .001c each, but worth an inductor if they cost $6,000 each.

I am on the DO IT side of things, and as far as low currents go, I'm saying I've DONE it and it works. Even posted the pics of the last time I did it (used diodes as a voltage drop, but in series only). It works.

I do wonder if putting a big twist in the leads on each side of the diodes, so the leads are all twisted together, would both even up the temperatures and better balance the voltage drop across the solder. But I don't think that the results could be seen so very well.

I would say over all, you can parallel diodes for this purpose, but maybe use 10 ? or 20 % extra diodes to further prevent the possibility of what we haven't yet seen, that is, a diode blowing up. fair conclusion ?

Edit: actually you're running it at roughly 40% so that would be twice the diodes needed. But 10Degrees is not a lot, and it can't be taking all of the current as the others are at 60 Degrees Celsius. So it would be a negligible effect because all diodes are operating and the temperature difference is not significant (explosion wise or anything else)

[wraper]

What happens if pushing all 6A through them (as of 6x of 1A rated). Temperature measured is the max temp in the area inside the box. As you can see, the difference is 28^oC and was still growing.

[TheElectricChicken]

What happens if pushing all 6A through them (as of 6x of 1A rated). Temperature measured is the max temp in the area inside the box. As you can see, the difference is 28^oC and was still growing.

Oooh, this is so good.  I get too excited mythbusters-style 8-) I am rapidly wiring up a power supply that I posted a picture of before (salvaged from a printer) so that I can do my own experiment maybe until destroy the diodes. I expect they should go off like fuses of course, that is, when one blows, they all will in the parallel portion (if series and parallel were used, not all would blow). I want to see.

I theorise the temperatures would get quite hot, and after some time, so long as there is no fan and no breeze to cool them, they should eventually go bust. But I expect that cooled by air, they won't go off at their rating, which you are showing.

[TheElectricChicken]

Oh this is terrible. I have hit a big snag. I collect from my messy desk the last few diodes that I didn't use in the power supply I uploaded pictures of in the first few pages of this thread. I find 5 !! I think this is awesome because if only one or two, they cannot react with each other.

But the type is 1N5404 they are 3 amp and I have only 4.2 or maybe 10 amp supply if I find a big load for it. But then I find on http://www.mouser.com this

Forward Voltage Drop:    1.2 V    
Forward Continuous Current:    3 A    
Max Surge Current:    200 A    
Maximum Operating Temperature:    + 150 C    

not figures that look good I think. Like Tim the toolman Taylor, I need more power, argh argh arrrrr. And then I find this even worse !!!

https://www.fairchildsemi.com/products/discretes/diodes-rectifiers/rectifiers/1N5404.html
3.0 ampere operation at TA = 75°C with no thermal runaway.

argh.!!! I need smaller diodes to torture !!! oh my, do I say that ? what about their families !!!!



I think I shouldn't kill the small diode, it has many 3 AMP cousins backing it up.

[TheElectricChicken]

Do a test to destruction.

YES. NO !!!! YES !!! NO!!!!  YES !!! NO!!!!  YES !!! NO!!!!  YES !!! NO!!!!  YES !!! NO!!!!  argh !!!!!!!!!



Or rather should I calm down and say MMmmm ok !!!! Meh !!!! MMmmm ok !!!! Meh !!!! MMmmm ok !!!! Meh !!!! MMmmm ok !!!! Meh !!!!

[CatalinaWOW]

Hey Chicken.  Real world trumps theory any day, in my book and many others.  But theory is a real benefit in understanding what you saw in the real world.  Don't dismiss it.  If you had the theory background you would understand what circumstances make inductors useful in load sharing and understand that it is not a general solution.  You and others would understand when paralleling diodes is risky, and when it is not.  You would also realize that many times it is easier to avoid a configuration than it is to either figure out if it really is ok, or to do a test comprehensive enough be convincing in the absence of theory.

Also remember that you are using theory in your approach to electronics.  If you weren't you would just grab parts randomly off the bench and see if something good happens.  You are just using much simpler theories that may be harder to generalize to new situations.

[wraper]

I would say over all, you can parallel diodes for this purpose, but maybe use 10 ? or 20 % extra diodes to further prevent the possibility of what we haven't yet seen, that is, a diode blowing up. fair conclusion ?

Even with 50 diodes in parallel, you only need one diode with significantly lower forward voltage drop to cause the failure as it will take too much current on itself and eventually short out. I'd say that the more diodes, the worse, you'll more likely will get one that is significantly different from the crowd.

[CatalinaWOW]

I would say over all, you can parallel diodes for this purpose, but maybe use 10 ? or 20 % extra diodes to further prevent the possibility of what we haven't yet seen, that is, a diode blowing up. fair conclusion ?

Even with 50 diodes in parallel, you only need one diode with significantly lower forward voltage drop to cause the failure as it will take too much current on itself and eventually short out. I'd say that the more diodes, the worse, you'll more likely will get one that is significantly different from the crowd.

Remember, it could also fail open.  Maybe after initially shorting out.  And in this application a short may or may not be detected as a fault, and an open almost assuredly will not be detected.  So it "works".  Might even be entertaining.  If all the circumstances were "right" (or wrong depending on your worldview), you could see a diode fail every few minutes, with each failure dumping two transients into the system, one when it shorted, the other when it opened.  The interval between failures would tend to get shorter as each diode picks up more of the load.

This entertaining outcome would be rare and hard to duplicate.  As Chicken says, in a great many cases nothing exciting will happen at all.

[TheElectricChicken]

Even with 50 diodes in parallel, you only need one diode with significantly lower forward voltage drop to cause the failure as it will take too much current on itself and eventually short out. I'd say that the more diodes, the worse, you'll more likely will get one that is significantly different from the crowd.

If it was so easy to have such a fault why is it NOT showing up, ever? you tried to reproduce it, but it WORKED instead. Now you are saying that in order to have a problem, you FIRST need to have a faulty part, THEN that faulty part will cause further problems. I'm sure if you FIRST smashed hell out of the diodes with a hammer THEN they'd have a hard time working properly too.  What's the difference ? If you don't go in search of a unicorn diode or use a hammer, then diodes work fine.

[Chris C]

Now you are saying that in order to have a problem, you FIRST need to have a faulty part, THEN that faulty part will cause further problems.

I wouldn't consider a diode with an unusually low Vf a faulty part.  Under normal circumstances, low Vf would be a highly desirable characteristic.

I recall someone who needed a 1% tolerance resistor.  He didn't have any of those on hand, but he had lots of 5% resistors.  And figured that since tolerance is a range, at least some of them would be 1% or less.  All he had to do was test until he found one.

He was a bit surprised to find them all near -5% or +5%, with NOTHING in the middle.  Apparently the manufacturer was using the same idea.  They were all tested, and those with better tolerance were labeled and sold as such, at a higher price.  The automated sorting was presumably cheaper than running a separate manufacturing process to specifically produce precision resistors.

What if diodes undergo a similar sorting process?  With all the "unicorn" (low Vf) diodes removed and sold as a different part, then that would be awesome for running them in parallel.  And you might do so successfully, and repeatedly.  But suppose the manufacturer ends up with more low Vf diodes they can sell, and decides to temporarily skip the now unnecessary sorting; selling all the diodes regardless of spec as plain 400X series.  You try paralleling these, and what worked many times suddenly doesn't.

And it would be no one's fault but your own.  You broke the rules, using a part in a way that is technically questionable.  If it's for a hobby project, chances are it's no big deal; I've done my share of rule bending there.  But if it's for a product, and many products start blowing out in the field, you're in big trouble.

Then there's recommending something that's technically questionable to others.  As a general rule I try to avoid this.  On occasion I have not.  There was one case where I suggested something, for which there was some risk, but on average the reward outweighed it.  I explained all that based on my own personal experience, expecting it to scare most away, but surprisingly people tried it en masse.  Turns out that while the reward still outweighed the risk, the risk was much higher than I'd represented.  Fortunately most people didn't blame me for their losses, but I still feel responsible.

[SeanB]

That is done with diodes to select reverse voltage. You might have say 4 different diodes, all identical in all respects aside from reverse voltage. Tested during production ( after the die test to weed out those which were either short circuit or otherwise not worth packaging) and then binned according to reverse leakage at selected test voltages. Thus you will get the cheapest part at say 10V, a slightly more expensive one at 20V, a more expensive 30V and finally the premium 40V parts. During manufacture they might improve yield such that all devices pass 20v test after a while so the 10V and 20V parts now are equal, and still later on they might creep up so there are only 30V and 40V devices, which means the 3 lower parts are supplied from the basic lot with only a label change.

Now, if you are using the 10V part, relying on it breaking down at some point between 10 and 20V as cheap overvoltage protection, with a load that will survive 20V for a while, but now you are getting devices that will withstand at least 30V, and up to 45V ( your 40V part will be capable of more than 40V, probably over 50V on most parts before the leakage gets to the limit values), so now you find the equipment it is in is now failing from this. Vf also varies from unit to unit, even on parts next to each other on the reel. Not much, but you get a variation around a nominal value.

[wraper]

Even with 50 diodes in parallel, you only need one diode with significantly lower forward voltage drop to cause the failure as it will take too much current on itself and eventually short out. I'd say that the more diodes, the worse, you'll more likely will get one that is significantly different from the crowd.

If it was so easy to have such a fault why is it NOT showing up, ever? you tried to reproduce it, but it WORKED instead. Now you are saying that in order to have a problem, you FIRST need to have a faulty part, THEN that faulty part will cause further problems. I'm sure if you FIRST smashed hell out of the diodes with a hammer THEN they'd have a hard time working properly too.  What's the difference ? If you don't go in search of a unicorn diode or use a hammer, then diodes work fine.

WHAT THE HELL WORKED??? I had shown that the current through the diodes is very different and one of them heats up significantly more than others  . What the fail should I reproduce the hell??? Should it fail after a few minutes I was running the test? Have you made any real electronics ever? Overloaded parts don't fail immediately but wait to fail at customers site at the worst moment they could  . When pushing 6A through them, that diode which were the hottest, operated with current significantly over spec while others were underloaded. I could also figure out what current was flowing through each of them by running the diode individually and measuring the current needed to rich particular temperature, but I'm to lazy to do this now. Don't forget that those are just 6 diodes from the same tape. In real life, if you assemble like 100 devices with 6 diodes in parallel in each device, I'm sure that in some if them imbalance would be much worse than in this case.

[Chris C]

[SeanB],  I'd heard normal variations in reverse voltage preclude using a rectifier as a zener, but I had no idea they could be this large.  Very informative, thanks!

[TheElectricChicken]

WHAT THE HELL WORKED??? I had shown that the current through the diodes is very different and one of them heats up significantly more than others  . What the fail should I reproduce the hell??? Should it fail after a few minutes I was running the test?

I don't see what there is to be upset about. Sure, one gets hotter, use resistors and one has to be hottest, I doubt perfect even heating is an easy thing to get. The runaway effect was the goal, to show it, but I think it is rare, so rare that it is difficult to come across. Would resistors have one hotter than the rest without proving resistors have a runaway effect ?

End of the day, what I am saying is that yes, diodes can be used for voltage drop, to unload the heat from a regulator even at 20A. Nothing says they have to be parallel, but I'll say that if you use enough it will work, and using 20 % more than you need to in parallel is going to work and I would say that it will not fail in reality any more than other devices in the home which fail often even with 'perfect' engineering.

Have you made any real electronics ever?

what kind of question is this ? I posted pics in this thread of a recent effort of mine, which works fine, using diodes to drop voltage.

Overloaded parts don't fail immediately but wait to fail at customers site at the worst moment they could  .

even with standard engineering.

When pushing 6A through them, that diode which were the hottest, operated with current significantly over spec while others were underloaded. I could also figure out what current was flowing through each of them by running the diode individually and measuring the current needed to rich particular temperature, but I'm to lazy to do this now. Don't forget that those are just 6 diodes from the same tape. In real life, if you assemble like 100 devices with 6 diodes in parallel in each device, I'm sure that in some if them imbalance would be much worse than in this case.

Agreed on all points. I don't see that this proves that failure is imminent, more than any other design, I do see that the hotter a diode is, the shorter the life, same as LEDs, but if the power supply has 20% extra or doesn't run maxed out all day every day for 40 years, then it just works. Is it meant to go up into orbit and last forever, then sure, use the very best. Is it meant to work at home and get the job done today ? then use diodes that you have on hand and unload the regulator and save the regulator instead of the diodes..

I can't see how endlessly arguing to have me agree with your points when I agree with you ALREADY is going to help. Some people just won't take yes for an answer I have found out in life.

[c4757p]

It's not just running hotter. Do you even understand what positive feedback is?

[TheElectricChicken]

It's not just running hotter. Do you even understand what positive feedback is?

Yes, the hotter it gets the lower the voltage drop and the larger the current flowing through it and therefor it gets hotter still. A function that in the case of the diodes that we've tried, and the diodes on my desk are having to a degree that doesn't blow them up or make them go open.

The greenhouse effect and methane hydrate under the oceans is the one to worry about.

[c4757p]

Just warn me if you ever design a product I might end up buying...

[TheElectricChicken]

Just warn me if you ever design a product I might end up buying...

Why would it be different to any other product you buy. Are you the only person in the universe for whom an item has never broken down ?

[c4757p]

Certainly not! I've repaired tons of test equipment, in particular. Huge numbers of failures are due to manufacturers trying to get away with things like this to save a bit of money.

[bills]

Did Mr. Chicken ever indicate he would use diodes In  parallel (or series)  in a consumer product?
I think the point here is there are many ways to do things as a hobbyist, Right or wrong finding out is a learning experience .
I do agree that if you design a product you better be sure that it is safe,and semi idiot proof. 

[TheElectricChicken]

I do agree that if you design a product you better be sure that it is safe,and semi idiot proof.

That's like feeding antibiotics to chickens, it just makes stronger idiots., more resistant to the steps you take to defeat their foolishness. Same as herbicide resistant weeds. They just get stronger and take over the government.

[bills]

Here is one of my favorite quotes.
"I'am not saying go kill all the stupid people. I'am just saying let's remove all the warning labels and let the problem work it self out."  

[oldway]

Paralleling diodes is often used in bridge rectifiers for welding machines and battery chargers.
This let use low cost 35A press-fit diodes for high current output (100A and more).
But the diodes are mounted on the same dissipator, what reduces temperature differences and risks of thermal runaway.

[TheElectricChicken]

Paralleling diodes is often used in bridge rectifiers for welding machines and battery chargers.

how could they, that's terrible 

This let use low cost 35A press-fit diodes for high current output (100A and more).

  This makes me think old welders are an excellent source of high power diodes.   

But the diodes are mounted on the same dissipator, what reduces temperature differences and risks of thermal runaway.

Hmm, a bit like twisting leads together.

[madires]

Paralleling diodes is often used in bridge rectifiers for welding machines and battery chargers.
This let use low cost 35A press-fit diodes for high current output (100A and more).
But the diodes are mounted on the same dissipator, what reduces temperature differences and risks of thermal runaway.

Not by itself. The diodes need to be matched too to get a low unbalance in the load sharing. A few posts ago I've posted a link to a document with a sample calculation of the unbalance. Paralleling diodes for a low cost solution might be more expensive than most think. If a diode needs to be replaced one would have to find a matching one, i.e. buying a large bag and/or replacing all diodes with a new set of matching diodes.

[oldway]

The diodes need to be matched too to get a low unbalance in the load sharing.

I don't agree...These press-fit diodes are not matched, they are only current derated.
Sometime, the lead is crushed and drilled to reduce the section of the conductor and create a low-value resistor.
This reduced section of lead also serves as a fuse.

[TheElectricChicken]

These press-fit diodes are not matched, they are only current derated.
Sometime, the lead is crushed and drilled to reduce the section of the conductor and create a low-value resistor.
This reduced section of lead also serves as a fuse.

That's what my intuition tells me too. If you popped 30 1A diodes in parallel and ran 20A nothing would happen. If there was a fault with one of them, I wonder if it wouldn't just blow and then leave the rest just fine. It is often the case that faults can be burnt out. It's a practice I have heard of many times and I wouldn't be surprised if it did it itself in the scenario of parallel diodes. While one is shorted out, the rest are protected because of the low voltage across them and not carrying any current.

[madires]

The diodes need to be matched too to get a low unbalance in the load sharing.

I don't agree...These press-fit diodes are not matched, they are only current derated.

Actually it's the same Either you match the diodes or calculate the unbalance and derate the maximum current accordingly.

Sometime, the lead is crushed and drilled to reduce the section of the conductor and create a low-value resistor.
This reduced section of lead also serves as a fuse.

Classic series resistor for compensation

[wraper]

These press-fit diodes are not matched, they are only current derated.
Sometime, the lead is crushed and drilled to reduce the section of the conductor and create a low-value resistor.
This reduced section of lead also serves as a fuse.

That's what my intuition tells me too. If you popped 30 1A diodes in parallel and ran 20A nothing would happen. If there was a fault with one of them, I wonder if it wouldn't just blow and then leave the rest just fine. It is often the case that faults can be burnt out. It's a practice I have heard of many times and I wouldn't be surprised if it did it itself in the scenario of parallel diodes. While one is shorted out, the rest are protected because of the low voltage across them and not carrying any current.

Power semiconductors usually become short, not open. And all your circuit becomes busted.

[TheElectricChicken]

Power semiconductors usually become short, not open. And all your circuit becomes busted.

according to the encyclopedia of you ? no seriously, where do you get your guesses from ?

[retrolefty]

Power semiconductors usually become short, not open. And all your circuit becomes busted.

according to the encyclopedia of you ? no seriously, where do you get your guesses from ?

 Well it was true in the case of a large 7 megawatt variable speed drive at my refinery. The main power SCRs were assembled in series strings of 6 for a total of 144 SCRs for the 3 phase variable frequency output. They were designed to allow continuing running even with a shorted SCR in the string. The OEM stated that the SCRs if failed, it would be shorted and log the need for replacement at the next scheduled maintenace turnaround.

 Most failed open condition results from first shorting out and drawing enough current to 'burn open'. If one can maintain adequate heatsinking the short condition can be maintained.

[TheElectricChicken]

Well it was true in the case of a large 7 megawatt variable speed drive at my refinery.

Wait, I need to go outside and check the 7 megawatt variable speed drive on my cordless drill. I'll be back in 5 minutes.

[wraper]

Power semiconductors usually become short, not open. And all your circuit becomes busted.

according to the encyclopedia of you ? no seriously, where do you get your guesses from ?

LOL, I have my guesses from expedience. Your guesses seem to be out of thin air. BTW this is not guess but how the things do happen in the real world.

[madires]

Power semiconductors usually become short, not open. And all your circuit becomes busted.

according to the encyclopedia of you ? no seriously, where do you get your guesses from ?

That's common EE knowledge! You will see that quite often when you repair broken stuff. For example, I've fixed a broken remote controlled mains adapter recently. The fault was caused by a shorted Zener in the power supply circuitry.

[TheElectricChicken]

Ok, my 7 megawatt variable speed drive on my drill is fine. Actually maybe I should have taken my glasses because I don't think that a 7 megawatt variable speed drive is the sort of equipment that beginners will come across. Certainly has nothing to do will "ALL SEMICONDUCTORS FAIL SHORT" . Maybe I'm missing the connection without my glasses because it looks like you're just making up guesses with statements like that.

I can make up better crap, like "open circuit" gets 5 million hits on google whereas "short circuit" gets 20 million hits therefor 75% of faults are short circuit. See? how will you ever start a company that sells battery life extenders if you can't sound more convincing with your wild guesses.

[TheElectricChicken]

That's common EE knowledge! You will see that quite often when you repair broken stuff. For example, I've fixed a broken remote controlled mains adapter recently. The fault was caused by a shorted Zener in the power supply circuitry.

Well there WAS that 150 Megawatt drive on that old ladies wheelchair that shorted out, boy was she pissed by the time the batteries ran out. Luckily she still hasn't made it back to my workshop to claim on the warranty I gave her, I hear she's in Uzbekistan last time NASA picked her up on radar.

Sorry, but that is STILL "encyclopedia of YOU" with nothing at all to back it up.

[wraper]

Sorry, but that is STILL "encyclopedia of YOU" with nothing at all to back it up.

Sorry, I have no intention to back up anything for people not seeing further than their arse.

[TheElectricChicken]

http://images.sodahead.com/polls/004375463/2955483819_not20surprised20_3e3394ec866826de6153b04b99d9a297_xlarge.jpeg

[wraper]

http://images.sodahead.com/polls/004375463/2955483819_not20surprised20_3e3394ec866826de6153b04b99d9a297_xlarge.jpeg

Congratulations! You won in this contest. Continue to live happy in your imaginary world.

[TheElectricChicken]

there was no contest, you were simply making up the idea that most semiconductors fail shorted, and presenting it as though it was fact. It is not fact. This is the beginners section so presenting bull&** as though it was fact and being serious about it, is misleading the beginners. I just don't find it cool.

And no, people were never able to send me to the store to get a box of amps or a left handed spanner. I just think it's inappropriate to give poor information dressed up as something it is not. I always say "I would guess" or "my intuition tells me..." or something like that, or better, give a link to fact.

[c4757p]

Chrissake, dude, do you know anything? Yes, semiconductors almost always fail short. Why? Well, why wouldn't they? Think about the way they are made. Almost the whole thing is conductive, except for that extremely thin depletion region in the middle...

[SharpEars]

Well it was true in the case of a large 7 megawatt variable speed drive at my refinery.

Wait, I need to go outside and check the 7 megawatt variable speed drive on my cordless drill. I'll be back in 5 minutes.

Chicken, I may not agree with your arguments, but I do love your humor.

[TheElectricChicken]

Chicken, I may not agree with your arguments, but I do love your humor.

Thank you SharpEars. I do very well when I have good quality questions being fed into my brain, however, when poor quality is fed in... well the output is ... DODGEY!!

I always like to be the sort of person who, if I had to argue, you'd shout food and drink just so that you could have a good argument with me all afternoon.

[TheElectricChicken]

Chrissake, dude, do you know anything? Yes, semiconductors almost always fail short. Why? Well, why wouldn't they? Think about the way they are made. Almost the whole thing is conductive, except for that extremely thin depletion region in the middle...

Oh I'm sure you know more than anyone, myself included. I'm a chicken who learned to type after all.

There are only two choices when you send 20 AMPS continuous through a 1A diode. It will either melt, blow off the plastic coating and splutter all over the place and go OPEN CIRCUIT,

OR

You've invented a new way to distribute power across the country in a way that engineers have been perusing for decades, they can dump their ridiculous and expensive pursuit of room temperature super-conductors and go with a long string of diodes instead. Because if they fail, they fail as a short circuit and therefore, you can continue sending the fault current through them forever. 20x 50x 100x doesn't matter because they always fail shorted, according to the encyclopedia of you. which I don't believe btw. I use it to prop up a lop sided table.

[wraper]

You were simply making up the idea that most semiconductors fail shorted, and presenting it as though it was fact. It is not fact. This is the beginners section so presenting bull&** as though it was fact and being serious about it.

This is the fact, if you were a tiny bit smart, you'd asked where you could find a more information about it. Yet you disagree with every good practices an widely known information people suggesting to you, claiming they got it in their own "encyclopedia". You even managed to completely upend the idea in those thermal images  .

[wraper]

...according to the encyclopedia of you. which I don't believe btw. I use it to prop up a lop sided table.

Believing if for religious nuts. Smart people prefer knowledge.

[c4757p]

There are only two choices when you send 20 AMPS continuous through a 1A diode. It will either melt, blow off the plastic coating and splutter all over the place and go OPEN CIRCUIT,

OR

No, typically it will fail short. Yes, if you're pumping an insane 20x overload through it, it will then continue to fail in new and interesting ways, usually ending in an open circuit, but most semiconductor failures are not due to 20x overload! Try a 1.5x overload for an extended period of time.

[madires]

Yesterday a friend brought a broken TV, obviously bad power supply (standby LED not lit, doesn't turn on). A voltage check confirmed that the the main voltage is missing. The secondary of the SMPSU has three diodes (SB5B0: 5A, 100V, Schottky) in parallel for the main voltage. Actually there are four on the PCB, one unpolulated. One failed short and the fault also released the magic smoke of the SMPSU controller (integrated MOSFET) and two resistors (current path to ground for the SMPSU controller's MOSFET). A quick search showed that Vestel, the manufactuer of the SMPSU, is known for that problem. Of course there's also a video: