Author Topic: 4-output (but very basic) Power Supply  (Read 1047 times)

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Offline SkrillBill

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4-output (but very basic) Power Supply
« on: November 04, 2017, 07:26:13 AM »
Hey everyone,
   I'm working on small project; a simple bench power supply with 4 outputs (3 fixed and one variable) using LM-series regulators. it's not finished yet, but I wanted to share what i had in case there are any obvious mistakes. I noticed im using a lot of capacitors for line filtering, around the LMs, nto sure i need all of that.

Anyways, here is my schematic so far, drawn on digi key:
https://www.digikey.com/schemeit/project/powersupply-v1-1-GDI647G303I0/

Looking forward to feedback, both good in bad. This is my first attempt at making a schematic of anykind, so im sure i messed it up.
 

Offline Kleinstein

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Re: 4-output (but very basic) Power Supply
« Reply #1 on: November 04, 2017, 07:46:34 AM »
There is no need for the diode D1. There should a large electrolytic cap for filtering the raw voltage - this one seems to be missing.
Just a single cap at the output of the regulators is OK on principle. The second (electrolytic one) might improve things if not to large.

The regulator might want a diode from the output to the input side : the 78xx don't like a higher voltage at the output than input very much. That is one way to break them.

The LM317 part does not look right, especially the pot. A diode across the outputs might be a good idea for the other regulators too - not an absolute must, but helpful if another power source is used in combination.

A 7824 suggests that the input voltage can be rather high. Remember that there is an upper limit. For the lower voltage outputs the power dissipation and maybe SOA limit for the regulators can become a problem.  So they might like to start from a lower voltage.

The regualtors need a minimum load to work well.

I have some doubt the 7805 would like a short towards the 7824 - this might be though to get around.
 

Online trys

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Re: 4-output (but very basic) Power Supply
« Reply #2 on: November 04, 2017, 08:23:57 AM »
Hi SkrillBrill,

I'm new here too - so welcome.

Funnily enough, I finished building a PSU just today using a 7805 (a "L78S05CV") to feed a DDS arbitrary function generator.

You need to fit a fuse on the live side of the AC input to transformer T1. If you're using a metal case or if there are exposed metal bits then you need to use mains earth to the chassis (but check with the regs in your country).

I agree with Kleinstein's comment about needing a large electrolytic cap after the rectifier. The one I chose was 4700uF, at a voltage a fair bit bigger to allow for a higher voltage when the transformer isn't under load.

The other things that almost caught me out was to remember that the rectified DC voltage of an AC voltage is 1.41 times higher. So a "9V transformer" produces about 12.7V.

Going on from Kleinstein's comment about the 7824. What that essentially means is that if you want to get a 24V regulated supply from that, you need a minimum input voltage of 33V. But for the smaller regulators on the circuit they have to "burn off" as heat (as it were) all that excess voltage. So, considering the 7805 in that circuit if it's faced with 33V then compared with the 5V that you want from it, it has to dissipate about 28W of heat if you are pulling 1 Watt off it. (Voltage drop multiplied by amps).

This brings me onto what I learnt today, was the importance of a heat sink. If you bear in mind these little beasties warm up by about 50 or 70'C per watt of power and their temperature limit is about 150'C then you can see that you need to keep them happy with an adequate heat sink. The good thing is though (although not sure about the LM317), they all automatically and gracefully drop volts if they get too hot, so they do look after us.

I spent about an hour today sawing away and bending a big strip of aluminium to cool the 5V 7805 power supply, then trying to get it to fit into a case that was a bit too small really. It all worked out well in the end.

The other thing I learnt many years ago is to be ridiculously careful near live mains electricity, and plan ahead in your head any measurements or movements you are about to make when putting your hand in. I've had two nasty jolts in my life, both my own stupid fault. You're not as stupid as me I'm sure!

All the best,
Trys
« Last Edit: November 04, 2017, 08:26:50 AM by trys »
 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #3 on: November 04, 2017, 08:28:29 AM »
There is no need for the diode D1. There should a large electrolytic cap for filtering the raw voltage - this one seems to be missing.
Just a single cap at the output of the regulators is OK on principle. The second (electrolytic one) might improve things if not to large.

The regulator might want a diode from the output to the input side : the 78xx don't like a higher voltage at the output than input very much. That is one way to break them.

The LM317 part does not look right, especially the pot. A diode across the outputs might be a good idea for the other regulators too - not an absolute must, but helpful if another power source is used in combination.

A 7824 suggests that the input voltage can be rather high. Remember that there is an upper limit. For the lower voltage outputs the power dissipation and maybe SOA limit for the regulators can become a problem.  So they might like to start from a lower voltage.

The regualtors need a minimum load to work well.

I have some doubt the 7805 would like a short towards the 7824 - this might be though to get around.

I completely forgot the single large filter cap on the input, and just straight up mis-wired the 317. I fixed both of those and added a diode from output to input. The schematic looks weird because it doesn't have 'cross overs' for traces.  Same link will get you there.

I'm still undecided on the LM317. The input will have to be 30v to account for drop, but honestly i could use the 317 for anything higher than 12v.
 

Offline stmdude

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Re: 4-output (but very basic) Power Supply
« Reply #4 on: November 04, 2017, 08:30:31 AM »
A diode across the outputs might be a good idea for the other regulators too - not an absolute must, but helpful if another power source is used in combination.

I'd very much recommend a diode across the LM317s. They'll work for a while without it, but the bulk-decoupling caps on your device-under-test is enough to fry them after a while.
It's in the datasheet for the LM317, and I'm guessing it's here for the LM78xx ones as well.
 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #5 on: November 04, 2017, 08:37:10 AM »
Hi SkrillBrill,

I'm new here too - so welcome.

Funnily enough, I finished building a PSU just today using a 7805 (a "L78S05CV") to feed a DDS arbitrary function generator.

You need to fit a fuse on the live side of the AC input to transformer T1. If you're using a metal case or if there are exposed metal bits then you need to use mains earth to the chassis (but check with the regs in your country).

I agree with Kleinstein's comment about needing a large electrolytic cap after the rectifier. The one I chose was 4700uF, at a voltage a fair bit bigger to allow for a higher voltage when the transformer isn't under load.

The other things that almost caught me out was to remember that the rectified DC voltage of an AC voltage is 1.41 times higher. So a "9V transformer" produces about 12.7V.

Going on from Kleinstein's comment about the 7824. What that essentially means is that if you want to get a 24V regulated supply from that, you need a minimum input voltage of 33V. But for the smaller regulators on the circuit they have to "burn off" as heat (as it were) all that excess voltage. So, considering the 7805 in that circuit if it's faced with 33V then compared with the 5V that you want from it, it has to dissipate about 28W of heat if you are pulling 1 Watt off it. (Voltage drop multiplied by amps).

This brings me onto what I learnt today, was the importance of a heat sink. If you bear in mind these little beasties warm up by about 50 or 70'C per watt of power and their temperature limit is about 150'C then you can see that you need to keep them happy with an adequate heat sink. The good thing is though (although not sure about the LM317), they all automatically and gracefully drop volts if they get too hot, so they do look after us.

I spent about an hour today sawing away and bending a big strip of aluminium to cool the 5V 7805 power supply, then trying to get it to fit into a case that was a bit too small really. It all worked out well in the end.

The other thing I learnt many years ago is to be ridiculously careful near live mains electricity, and plan ahead in your head any measurements or movements you are about to make when putting your hand in. I've had two nasty jolts in my life, both my own stupid fault. You're not as stupid as me I'm sure!

All the best,
Trys

I'm still unsure of the 24v regulator. The presence of the 317 almost makes it obsolete. However that doesn't solve the problem of the transformer & input voltages. I will definitely have heat sinks on all of these; i have a few for LM's that i pulled out of scrap PCBs.

Never considered a fuse on the live side but now that you mention it it seems pretty obvious. For now its just going to go inside of a project box, tho future versions may eventually have a metal case. Would it be worth it to have the earth ground exposed with a stud or jack of kind on the outside of the plastic case?
 

Online trys

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Re: 4-output (but very basic) Power Supply
« Reply #6 on: November 04, 2017, 08:44:59 AM »
Bill,

You're on the right lines with that cap, but you just need to reposition it (see attached).

You could have the higher regulators feeding the smaller ones if you are only using one at the time, and share a common large heatsink (check datasheet for what the heatsink connector is connected to - I *think* it's ground).

As for the earth, if it's a metal chassis, then over here in the UK the practice is to also connect the earth to the chassis. I'm not familiar with how the earth is wired for US mains supplies, so hopefully somebody else can help here.

Trys
 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #7 on: November 04, 2017, 08:48:06 AM »
Bill,

You're on the right lines with that cap, but you just need to reposition it (see attached).

You could have the higher regulators feeding the smaller ones if you are only using one at the time, and share a common large heatsink (check datasheet for what the heatsink connector is connected to - I *think* it's ground).

As for the earth, if it's a metal chassis, then over here in the UK the practice is to also connect the earth to the chassis. I'm not familiar with how the earth is wired for US mains supplies, so hopefully somebody else can help here.

Trys

Ahhh, yeah. Moved it to the proper location now.
 

Online trys

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Re: 4-output (but very basic) Power Supply
« Reply #8 on: November 04, 2017, 08:51:46 AM »
Great - that's better.  :)

Just a quick thought - you could use a transformer with dual secondary outputs.

Let's say you had a 0-15 0-15 secondary transformer you could run the two smaller regulators off the first 0-15v tap then the other two off the two taps in series (so they run off 0-30v).

Edited to add: You'd need separate rectifiers and caps for these two sets.

Trys
« Last Edit: November 04, 2017, 08:53:42 AM by trys »
 

Offline Kleinstein

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Re: 4-output (but very basic) Power Supply
« Reply #9 on: November 04, 2017, 09:04:56 AM »
One can still get 2 x 15 V DC from a split winding transformer with a single rectifier - but it need 2 fitler caps. Just use the same circuit commonly used for +-15 V, with GND to the center of the transformer winding.

A split transformer is a good idea to keep the power consumption low.  Another way is having a separate filter cap for the 5 volt regulator and have a resistor or inductor (though likely impractical large in size) between the filter cap and rectifier. This shifts some loss to the resistor and also improves the power factor and reduces overall loss a little.

One might be able to protect the 7805 output with a transsorb or similar again too much voltage at the output. Not sure this is enough. Just in case someone thinks he gets 19 V between the 24 V and 5 V outputs.
 

Online mariush

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Re: 4-output (but very basic) Power Supply
« Reply #10 on: November 04, 2017, 09:13:07 AM »
C2 is not correct there.  It shouldn't be in series, it should be between the + and - outputs of the bridge rectifier.

Depending on the maximum current you want the power supply to output, 2200uF may be too small... you can approximate the capacitance required with the formula

C = Current / [ 2 x Mains Frequency x (V dc peak - Vminimum) ]

so for example, if you have a 24v AC transformer, that means the bridge rectifier will convert that to DC voltage with a peak voltage of Vpeak = Vac x 1.414 = ~ 34v and if you want to have at least 30v DC after the capacitor at 1A of current, then you'll want :

C = 1A / [2 x 60 Hz (if you're in US) x (34v - 30v) ] = 1 / 120x4 =  0.002174 Farads or 2174 uF

In order to output the right voltage, a linear regulator in the 78xx series needs around 2v above the output voltage on the input pin. So, you must size that capacitor after the bridge rectifier so that the 7824 linear regulator will always receive at least 26v.

However, that's a problem for the other linear regulators.

The way you made the circuit, all linear regulators are connected directly to that high voltage (26v or whatever you'll have) after the C2 capacitor.

But, linear regulators produce the desired output voltage by eliminating the difference between the input voltage and output voltage as HEAT.  So, for example, with your 5v linear regulator 7805, if you connect an Arduino or something that uses 100mA (0.1A) , then this linear regulator will produce this amount of heat:

Power dissipated = (26v - 5v )  x 0.1 A = 2.1 watts

Remember that these linear regulators are capable of up to 1A of current, but you can see that with just 0.1A of current, the regulator would dissipate over 2 watts of energy as heat. For reference, for anything above around 1 watt, you'll need a heatsink, and you really shouldn't use a linear regulator if the dissipated power will be over 10 watts, because the heatsink would be too large, or you'll actually need a fan blowing air over the heatsink to help move away the heat.

So basically, it's in your best interest to keep the voltage difference between input voltage and output voltage as small as possible.

If you're not going to use both 24v output and 12v or 5v output, you could buy a transformer with two secondary windings, and depending on which output voltages you plan to use, connect just one winding or both windings to the bridge rectifier.

If you want to learn more, reply and i'll explain in more detail.


Also, these linear regulators don't really need, but it would help to have a small capacitor on the input of each regulator, very close to the input pin, for example a 10uF to 100uF electrolytic capacitor with voltage rating higher than the maximum input voltage you'd ever have (i'd suggest 50v or 63v or even higher, capacitors with such low capacitance value should be cheap either way)
There's no need for 2 capacitors on the output of each regulator, they only need around 1uF of capacitance and again, you could use the same 10uF to 100uF capacitor you choose for the input (some stores will give you discounts if you buy packs of 10 or bigger, so take advantage of that)


If you already have a transformer with a single winding, have a look at regulators like LM1085 ... if you can make sure the input voltage will always be below 29v , you can use one to output 24v at maximum 3A, then connect a second LM1085 to output 12v from 24v input , then connect a third LM1085 or 7805 to output 5v from 12v input.

This way, you have (29v in , 24v out ) x current amount ,  (24v in , 12v out ) x current , (12v in , 5v out) x current

So if you want 5v at 0.5A, you'll have this :

lm1085  (30v - 24v) x 0.5A = 3W
lm1085 (24v - 12v) x 0.5A = 6A
7805 / lm1085 (12v - 5v) x 0.5A = 3.5w 
 = 12.5w spread across three linear regulators that can be placed on a wide heatsink.

« Last Edit: November 04, 2017, 09:20:17 AM by mariush »
 

Offline rdl

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Re: 4-output (but very basic) Power Supply
« Reply #11 on: November 04, 2017, 10:25:23 AM »
The wiper of the LM317 adjustment pot needs to be connected or it won't adjust.
 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #12 on: November 04, 2017, 10:30:12 AM »
The wiper of the LM317 adjustment pot needs to be connected or it won't adjust.

Thats just how the schematic icon displays it, not sure why.


C2 is not correct there.  It shouldn't be in series, it should be between the + and - outputs of the bridge rectifier.

Depending on the maximum current you want the power supply to output, 2200uF may be too small... you can approximate the capacitance required with the formula

C = Current / [ 2 x Mains Frequency x (V dc peak - Vminimum) ]

so for example, if you have a 24v AC transformer, that means the bridge rectifier will convert that to DC voltage with a peak voltage of Vpeak = Vac x 1.414 = ~ 34v and if you want to have at least 30v DC after the capacitor at 1A of current, then you'll want :

C = 1A / [2 x 60 Hz (if you're in US) x (34v - 30v) ] = 1 / 120x4 =  0.002174 Farads or 2174 uF

In order to output the right voltage, a linear regulator in the 78xx series needs around 2v above the output voltage on the input pin. So, you must size that capacitor after the bridge rectifier so that the 7824 linear regulator will always receive at least 26v.

However, that's a problem for the other linear regulators.

The way you made the circuit, all linear regulators are connected directly to that high voltage (26v or whatever you'll have) after the C2 capacitor.

But, linear regulators produce the desired output voltage by eliminating the difference between the input voltage and output voltage as HEAT.  So, for example, with your 5v linear regulator 7805, if you connect an Arduino or something that uses 100mA (0.1A) , then this linear regulator will produce this amount of heat:

Power dissipated = (26v - 5v )  x 0.1 A = 2.1 watts

Remember that these linear regulators are capable of up to 1A of current, but you can see that with just 0.1A of current, the regulator would dissipate over 2 watts of energy as heat. For reference, for anything above around 1 watt, you'll need a heatsink, and you really shouldn't use a linear regulator if the dissipated power will be over 10 watts, because the heatsink would be too large, or you'll actually need a fan blowing air over the heatsink to help move away the heat.

So basically, it's in your best interest to keep the voltage difference between input voltage and output voltage as small as possible.

If you're not going to use both 24v output and 12v or 5v output, you could buy a transformer with two secondary windings, and depending on which output voltages you plan to use, connect just one winding or both windings to the bridge rectifier.

If you want to learn more, reply and i'll explain in more detail.


Also, these linear regulators don't really need, but it would help to have a small capacitor on the input of each regulator, very close to the input pin, for example a 10uF to 100uF electrolytic capacitor with voltage rating higher than the maximum input voltage you'd ever have (i'd suggest 50v or 63v or even higher, capacitors with such low capacitance value should be cheap either way)
There's no need for 2 capacitors on the output of each regulator, they only need around 1uF of capacitance and again, you could use the same 10uF to 100uF capacitor you choose for the input (some stores will give you discounts if you buy packs of 10 or bigger, so take advantage of that)


If you already have a transformer with a single winding, have a look at regulators like LM1085 ... if you can make sure the input voltage will always be below 29v , you can use one to output 24v at maximum 3A, then connect a second LM1085 to output 12v from 24v input , then connect a third LM1085 or 7805 to output 5v from 12v input.

This way, you have (29v in , 24v out ) x current amount ,  (24v in , 12v out ) x current , (12v in , 5v out) x current

So if you want 5v at 0.5A, you'll have this :

lm1085  (30v - 24v) x 0.5A = 3W
lm1085 (24v - 12v) x 0.5A = 6A
7805 / lm1085 (12v - 5v) x 0.5A = 3.5w 
 = 12.5w spread across three linear regulators that can be placed on a wide heatsink.




I definitely want to learn more! Do tell.
 

Offline rdl

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Re: 4-output (but very basic) Power Supply
« Reply #13 on: November 04, 2017, 10:43:26 AM »
The wiper of the LM317 adjustment pot needs to be connected or it won't adjust.

Thats just how the schematic icon displays it, not sure why.


It displays it that way because it isn't connected to anything.
 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #14 on: November 06, 2017, 03:32:22 AM »
The wiper of the LM317 adjustment pot needs to be connected or it won't adjust.

Thats just how the schematic icon displays it, not sure why.


It displays it that way because it isn't connected to anything.

Well in this case what should it be connected to in the schematic?
 

Online trys

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Re: 4-output (but very basic) Power Supply
« Reply #15 on: November 06, 2017, 03:37:26 AM »
Without checking, the variable bit should be connected to its upper leg.
 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #16 on: November 07, 2017, 02:46:35 AM »
quoting for tag

1A is currently the maximum I planned to draw. I'm not really adding in any additional circuitry to handle or control anything higher in this version of the power supply, and your formula is more or less exactly what i would be using; so a 2200uF cap should be a sufficient.

Speaking of, i did have couple questions about the formula.
C = Current / [ 2 x Mains Frequency x (V dc peak - Vminimum) ]

In this example, is VDC Peak the max output of the transformer, and vMin is the minimum requirement needed to leave the filter?
 

Online mariush

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Re: 4-output (but very basic) Power Supply
« Reply #17 on: November 07, 2017, 02:33:21 PM »
edit: it's actually 230v + 10% and -6% in EU, not +/- 10% ... so the voltage range is 216.2v ... 253v AC but everything else remains valid.
I initially wrote the post with 210v AC minimum voltage, and redid the numbers now to consider 215v AC as minimum voltage. If i missed some edit somewhere, hope this makes it clear.

edit2 : regarding too high voltage at idle ... some will use 35v rated capacitors in such designs and a lot of capacitors will actually tolerate some voltage above their rating at such low currents, so they won't damage easily or you won't notice the capacitors swelling or going bad (right away at least) , but that doesn't change the fact that you should pick the proper voltage rating for capacitors. or at least have some mechanism to block those high voltages at idle (like maybe a zener diode that can be removed from circuit when it's not required, or a bunch of diodes in series which can be shorted out) .. but generally it's more convenient to just use the proper voltage rating.
--

That formula only approximates the capacitance required, but it's a good enough approximation.
The Vdc peak is supposed to be peak DC voltage the capacitor sees on its leads, but you should read it more like "the peak dc voltage the capacitor will see under any conditions, the safe minimum peak dc voltage"

Let's say you're in Europe where you have this :

 230v 50 Hz  -> transformer  -> 24v AC - > bridge rectifier - > capacitor  - > some DC voltage.

First of all, your 230v will not always be 230v, it's allowed to have +10% -6% . By making the standard 230v +10% -6% they were able to keep everyone happy, some countries have 230v, some 240v and some French colonies and African places have 220v and everyone can use 230v designed products in theory.
 
At 2-4 am in the mornings, I sometimes have 245v AC on the mains socket and that's still perfectly acceptable from the power company's point of view because it's within those 10% ... 230v +10% -6% is 216v .. 253v AC.

Your transformer will have a fixed ratio more or less, so for example in the case of 230v in <-> 24v out, that ratio is 9.58 : 1 , so with 216v.. 253v in, you're going to have  22.5v ... 26.35v AC on the output of the transformer.

So right from the start, you have to decide on the input voltages your product is more likely to meet during real world use.

For this example, let's say you're confident that your product will never see less than 215v AC and will never see more than 250v AC (because they're nice round numbers) , in which case you will know the output of your transformer will be 22.44v (let's round it up to 22.5v to keep things simple) and 26.09v  (let's round this down to 26v to keep things simple)

So right from the start, our transformer is a bit more complicated, it's suddenly 215v..250v IN , 22.5v AC ... 26v AC

Now, it gets a bit messier. You see, transformers due to how they're designed, don't really output that exact AC voltage no matter what's connected to them.  The smaller the transformer is, there will a higher percentage of ... the term doesn't come to mind right now ... basically at low power consumption, the transformer's output will be higher.

For example, let's say you have a 24v AC transformer rated for 10VA ... if you connect a 24v 10w incandescent light bulb to it and measure the voltage coming out of the transformer, it will probably be very close to 24v AC because the incandescent bulb consumes the rated power the transformer is rated for.   However, if you disconnect the bulb or use let's say a 1w incandescent bulb, the transformer's output voltage may go up by even as much as 10-15%.

This percentage of 10-15% is common for low VA transformers but it typically gets smaller for higher VA transformers, for example a 150VA or higher transformer may only have a 5% variation.

You need to be aware of this because think of the case where your power supply idles on the desk with nothing connected on the outputs of your linear regulators. The only power consumption would be the quiescent current of the regulators (a few mA), maybe a power on led (a couple mA) and the losses in the bridge rectifier... so with just a few mA of load on the transformer, the output voltage could be quite a bit higher than the advertised voltage (what says on the label).

Here's some random examples I picked from Digikey

12VA transformer with 2 x 14v secondary windings  ( Voltage Regulation: 25% TYP @ full load to no load )  : http://catalog.triadmagnetics.com/Asset/FS28-420.pdf
50VA transformer toroidal with  2 x 15v secondary windings ( Voltage Regulation: 12% TYP from  full load to no load ) : http://catalog.triadmagnetics.com/Asset/VPT30-1670.pdf
100VA transformer toroidal with 2 x 24v secondary windings ( Voltage Regulation: 9% TYP from  full load to no load ) : http://catalog.triadmagnetics.com/Asset/VPT48-2080.pdf

So let's go with 15% as a safe assumption for the transformer used.
In this case, at idle or low power consumption, our transformer could suddenly change from 22.5v AC .. 26v AC to  25.9v AC ... 29.9v AC   but at high power consumption, they'll sag down to the 22.5v .. 26v AC range, depending on mains input.  So really, our transformer's output range has to be considered 22.5v .. 30v AC if we must take in account all the possibilities.


Now we have the bridge rectifier, which converts the AC voltage to DC voltage, and it does this by moving the electricity through diodes. At any point in the process, there's two diodes that are active so as the AC waveform is converted to DC waveform you get a DC voltage with the peak value of :

Vdc  =  sqrt(2) x Vac  - 2 x Vdiode  where the Vdiode is the voltage drop on one of those diodes inside the bridge rectifier..

This is another small gotcha ... this voltage drop on individual diodes varies with the current going through the diodes and the temperature of the diodes.
As the temperature of the diodes goes up, the voltage drop on the diodes falls down just a bit.

Let's take some cheap and generic bridge rectifier you may harvest out of some ATX power supply or buy from the store, a 400v 25A bridge rectifier  GBJ2504 from Diodes Inc (but you'll find something similar from other manufacturers) : https://www.diodes.com/assets/Datasheets/ds21221.pdf

Now if you go in datasheet and scroll down to page 2 you will see Figure 2, Instantaneous Forward Voltage (v) , typical forward characteristics per element.

So what happens when the current is a few mA, when your circuit is idle? At let's say less than 0.1A the forward voltage per element will be less than 0.5v voltage drop per diode.
But at high currents, let's say at 1A the voltage drop goes up to 0.8v per diode. At more than 1A, the voltage drop only goes up by a bit, graph says around 1.1v at 10A... but let's say 1v for 2-3A of current or more.

So now we can calculate the peak dc voltage with the above formula at idle and at load, keeping in mind that our transformer may output anything between 22v and 30v as we discovered above :

At idle,  we're using the 25.3v and 30v AC in order to account for that estimation of up to 15% increase in output:

idle :  Vdc  minimum =  1.4142 x 25.3v AC - 2 x 0.5v = 34.78v DC    Vdc maximum = 1.4142 x 29.9 - 2 x 0.5v = 41.28v

At load, the transformer's output should be closer to the rated value, so that 15% increase can be more or less ignored but we still have to account for variations in mains voltage :

load : Vdc minimum = 1.4142 x 22.5v AC - 2x1v = 29.8v  , Vdc maximum  = 1.4142 x 26v AC - 2 x 1v = 34.77v

So what this tells you is that the peak DC voltage will be between 29.8v DC and 41.28v so you should use a capacitor rated for 50v or higher (35v rated capacitor shouldn't be used based on our estimations)
However, during regular operation, you can only rely on the peak DC voltage to be within 29.8v and 34.77v  and if you want to be absolutely safe no matter the mains voltage, you should really use 29.8v as peak DC voltage in that formula. Just for super extra safety and all that, I'd probably round it down to 29.5v

Therefore, going back to the formula :

C = Current / [ 2 x Mains Frequency x (Vdc peak - Vdc minimum) ]

since we did everything above assuming 230v +/- 10% and 50Hz (Europe) and we have the safe Vdc peak of 29.5v (again, not the absolute peak, but what's going to be achievable at any time during the operation of the product) and let's say we want a minimum of 26v DC after the capacitor, now we can calculate a minimum capacitor :
I'll go with 1A of current since it makes things easy.

C = 1A / [ 2 x 50Hz x (29.5v - 26v)] = 1 / 100 x 3.5 = 1 / 350  = 0.002857 Farads  or 2857uF

Now 2857 uF is not a common value, so you could go with maybe 2700uF and say "meh, close enough" but 3300uF is what I would use) .You can go higher to 3900uF or 4700uF ... more capacitance wouldn't hurt too much (too high capacitance can cause higher current spikes from the transformer when you first plug the product in the mains socket, and too high current spikes could cause your fuse to blow - that's why you usually use time delay fuses with transformers) but at 50v rating (or better) the capacitor would be bulky and expensive.


In US, with 60Hz mains frequency, you may need slightly less capacitance to account for everything.

And remember, this just sort of guarantees that your DC voltage will always be at least 26v, but based on the math we did above, at idle or low power consumption with the data above, our DC voltage could be as high as 41.2v DC .. at just a few mA of current, such a big capacitor will easily charge and maintain the voltage much closer to the peak possible.

That's important if you're going to use voltage regulators which have maximum voltages like let's say LM1085 with a maximum of 29v input voltage. You would have to be careful about using one of these because your voltage will be anything from 26v to 35v at various loads and up to 40v at idle.

Hope it helps and it's easy to understand what I explained above.


« Last Edit: November 07, 2017, 02:59:21 PM by mariush »
 
The following users thanked this post: SkrillBill, trys

Offline xani

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Re: 4-output (but very basic) Power Supply
« Reply #18 on: November 07, 2017, 06:19:53 PM »
I'd advise to do put a little bit more effort and have at least one output with current limit (preferably having range starting from 10mA)

It is good way to prevent magic smoke from escaping in case of a mistake ;D

 

Online trys

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Re: 4-output (but very basic) Power Supply
« Reply #19 on: November 08, 2017, 08:06:41 AM »
Mariush,

I've just got to thank you for such a detailed and well written response to SkrillBrill's question. I've learnt quite a few things from your post that I hadn't realised. I knew about the voltage of a transformer off load compared to on-load, but I did not have an idea of how much, nor the variance of mains voltage and how to allow for it.

It's amazing how much time you have taken in that post (and many of your other ones too), and you explain things very well. Brilliant.

Trys

 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #20 on: November 08, 2017, 08:52:49 AM »
Thank you @mariush for the information. That was a very informative and educational post.  :-+

Now onto current limiting as mentioned by @xani; i do plan add that into the design. Maybe not the first run of the board but it will be in there. The next step after that is current contro; being able to adjust the current.

Is there a certain way to adjust current in which you don't drop voltage? Say, i have my 317 set to 20v @2a. I set the 317 to 20v output then set the current adjust to 2a. Would that affect the voltage going out of the circuit at all?

 

Online trys

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Re: 4-output (but very basic) Power Supply
« Reply #21 on: November 08, 2017, 10:43:08 AM »
Come on SrkillBill, you can give him more than fourteen words of feedback on that epic post (1,841 words) of his surely? He's spent ages on that.

Edited: added actual word count of 1,841 words
« Last Edit: November 08, 2017, 10:46:56 AM by trys »
 
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Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #22 on: November 09, 2017, 04:07:59 AM »
*ahem*

Mariush, i greatly appreciate the time you took to put that post together. That was more or less a course in electronics engineering. You could have easily just said "this formula means this and this value is this and that value is some volts, based on your description here is what you want", but instead i know *understand* what i'm working with. I've got a long road ahead on the path of understanding EE, but every post like that will help.
====


Should have been my original reply, but i was at work and kept minimizing the window and coming back to it between projects.
 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #23 on: November 23, 2017, 08:41:04 AM »
Kinda reviving this.

I updated the schematic with a dual secondary transformer(dual 15v or 16v, depending what i find) and a second rectifier, so i can reduce the voltage being pumped into the regulators. I'm not sure the proper way to tie the rectifiers together for the 24v and LM317. Right now i have the neutral separated and the positives tied together with a diode for the LM7824 and LM317. Not sure if the neutrals should also be tied together or if that would cause electrical problems.

Edit: Added GNDs to the negative side of the rectifiers.
Edit 2: Realized i would not tie the rectifiers together. Noobie mistake!

« Last Edit: December 08, 2017, 07:51:12 AM by SkrillBill »
 

Offline SkrillBill

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Re: 4-output (but very basic) Power Supply
« Reply #24 on: December 08, 2017, 08:06:56 AM »
Bringing this back up, rather than making a new thread.
Assuming i have a 36v transformer (dual secondary, 18v each), if i want to hook it up to a rectifier so that all 36v is going to that, how would i do that? Would i wire one winding into the other then wire to the rectifier, or would I run both windings directly to the rectifier?

I know this is probably a basic question, but i'm kinda struggling with this for some reason..  |O
 


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