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?