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# Making a [email protected] Power Supply using linear regulators (LM350)

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Got a question.

I am working on a 0-28V, 0-3A power supply using several LM350 linear regulators.  I'm including necessary capacitors to minimize ripple and ringing and implementing short circuit protection via some standard rectifier diodes.  I'm wondering if the following schematic makes sense.  D2 and D3 drop the reference 1.25V out of the regulator to allow for a minimum Vout of 0V.  R3 allows a variable Vout up to ~28V with a 32VDC source (which is nothing more than a transformer and rectifier bridge).  D1 and D4 offer short circuit protection.

My questions:

1. Should D1 and D4 be of the 1N5401 type as well (3A maximum as opposed to 1A maximum 1N4002s)?
2. Do I need any kind of load resistor after D3 (or maybe it has to be east of D4, or even east of C3)?
3. Do I need any kind of minimal resistor before R1?
4. What will happen if, for instance, I set R1 to near 0 resistance?  There should be a maximum of 3A coming out of the regulator no matter what R1's resistance is, right?  This is in relation to question 3.  The problem with placing a minimal resistor in series with R1 is that, even at 1 Ohm, it would cap the maximum at 1.25A.  I'd need something rated at ~0.4 Ohms to get a maximum of 3A.
5. Should D1 really be 2 diodes (i.e. sould D1 end to the left of the LM350 on the right, and then should I place another diode just east of R1)?  I don't think so because I can't imagine any current flowing from the LM350 on the right's input to the LM350 on the left's output (i.e. a short circuit).
6. D2 and D3 on average drop about 1.4V together.  The reference voltage is 1.25V, and since the diodes are in place, they effectively drop the reference voltage to 0V (no negative voltage is possible due to the diodes, right?)
7. In terms of a transformer, does anyone know what kind I should get?  I want about 32VDC as input to the regulator.  I read that 32VAC will be rectified to something greater than 32VDC, but I'm not sure.

Thanks a bunch!!  I'm just starting out and learning.

Oh, and I know that both LM350s will have to be properly heatsinked.  Anyone know how hot they can get with this circuit?

--Niksun

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Hi Niksun,
Look carefully at the datasheet for the LM350:
1) With more than 10V across it, it reduces its max current to protect itself. With 30V across it then its output current minimum is only 250mA, 1A is typical. Look at the Current Limit curve. The LM317 and LM338 operate similarly.
2) It needs a minimum of 3V to 5V across it to meet its specs. Its dropout voltage (no longer regulating) is typically 2.3V across it. So the total voltage needed for both is 6V then an additional 1.25V for the current regulator and up to 2V for the two diodes, so the max output voltage of your circuit is about 22.75V at 3A with 32V input.
3) The TO-220 package can dissipate only 25W with a perfect heatsink then its junction is at its max temperature and it will shut down. Then if your project had a low output voltage the current would be only 1A max but the current limit would reduce it anyway. You won't have a perfect heatsink so it might have a max output current of only 800mA.
4) The diodes in series with the output ruin the voltage regulation. They will cause the output voltage to change up to 0.8V, typically 0.6V with changes of load current. 0V will change a lot with load current changes. Usually a low current negative 1.25V supply is made as "the ground" for the voltage setting pot for good voltage regulation.

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hello Niksun

You know th LM338 can handle 5amps, and is verable from 1.2 to 37 volts.
heres the data sheet

gogo

LM338K.pdf

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You know the LM338 can handle 5amps

Only sometimes, read the details in the datasheet.
5A output current if it has between 3V and 10V across it. With more than 10V across it, it also reduces its max current to protect itself. It doesn't have a guaranteed minimum output current at higher voltages like the LM317 and LM350, but is typically a little more than 1A with 30V across it and typically 3A with 20V across it.
It also can't dissipate more than only 25W for the TO-220 package.

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Hello audioguru

I have a LM338 in a T3 package. I read the data sheet and it really dosn't show any difference between the 220 and T3. I salvaged one out of a printer and it had a real nice heat with it. What I had in mind to do was use the Lm338 in a power supply project with 3amp output at 24 volts. Guess I have more reading to do.
thanks for the info
gogo

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I have a LM338 in a T3 package. I read the data sheet and it really dosn't show any difference between the 220 and T3.

The datasheet specs the TO-220 case can dissipate a max of 25W and the TO-3 case can dissipate a max of 50W with a perfect heatsink.

What I had in mind to do was use the Lm338 in a power supply project with 3amp output at 24 volts.

Then it will need a 27V minimum input, and if the voltage is set low like 2V to 5V, then its max current will be reduced by its protective current limit to typically 2A, but it could be lower. It will also be at its max temperature so it might shut-down.
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OK, so if I understand the LM317 (I decided to switch to the LM317) data sheet correctly, so long as the Vin/Vout difference is within 15V, then I should 1) be able to get 1.5A out of the LM317 (so long as I have a Vin/Vout difference of at least 3V or so), and 2) have manageable heat so long as I use a decent heatsink.

I'm thinking of supplying the current limiting LM317 with [email protected], which should supply the voltage regulating LM317 with a minimum of 15V (up to 3V is used by the current limiting LM317).  Since the minimum Vout is 1.25V, then my Vin/Vout difference here is 13.75V which is fine.  I figure I can get a maximum of about 12V (forget the diodes I mentioned previously to allow for an output of 0V).  I figure I'll replace R3 with a 5k potentiometer.

For higher voltages, I figure I'll just supply my variable PSU with [email protected]  Again, that will probably supply the voltage regulating LM317 with 27V which will give me an output maximum of about 24V.  In order to ensure that I do not go below 12V (which would generate a Vin/Vout difference of 15V), I can replace R3 with a 1.8k resistor in series with the 2.5k potentiometer.  If the pot's resistance is 0 ohms, then the 1.8k resistor ensures that my minimum Vout is about 12V.

Does this make sense?

Another issue is that I wish the 1.8k resistor to only be used if I am using the 30V supply.  I think I'll put a switch to select which AC adapter I use, but without using another switch to select the resistor, how can I make sure that the 1.8k is included in the circuit if the 30V adapter is selected?

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Hi Niksun,
Your LM317 power supply design is much better.
For up to 1.5A output, R1 will be set to 1.25V/1.5A= 0.833 ohms. A tiny portion of its resistance track will dissipate 1.9W, get extremely hot and maybe burn. Therefore R1 must be a wire-wound rheostat rated to carry 1.5A at low resistance setting. I hope you can find one.

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A tiny portion of its resistance track will dissipate 1.9W, get extremely hot and maybe burn. Therefore R1 must be a wire-wound rheostat rated to carry 1.5A at low resistance setting. I hope you can find one.

I think I found something that may work.  I couldn't find a data sheet on it though.  It's a 20 ohm wire-wound linear taper potentiometer rated at 2-4W.  The part description:

"Nice quality screwdriver adjust full size 20 ohm 2-4Watt potentiometer by Clarostat features 3/8" panel mount bushing. Enclosed case wirewound potentiometer is marked A43-20ohm. Comes in factory box with panel mounting nut. Linear taper."

I know it's only 20 ohms, but at this point I don't think I mind having a 62.5mA minimum current limiter.

I also found a 50 ohm (or even 1kOhm or 2kOhm) wire-wound 5W potentiometer.
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Hi Niksun,
Turn the 50 ohm pot down to 1 ohm for 1.25A output. It will dissipate 1.56W in only 2% of its physical resistive track. 100% of its track is rated for 5W so you will need to investigate how well it can dissipate heat from only 2% of its track.
A 20 ohm/4W pot will dissipate 1.56W from 5% of its track so might not burn as much as the other.

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OK, so here's what I have so far.  I have added an analog voltmeter and ammeter.  I've also updated the notes (primarily for my newbie self) :P

I figure I should have the heat managed with properly heatsinked LM317s.  Also, I should have the full 1.5A available so long as I stay within the 15V Vin/Vout difference.

I have to find some power rheostat which can handle 1.57W at its 1ohm setting.  I was able to find such an item, but it was very expensive (\$30 or so).  I don't really want to spend that much, so I'll try to find a better way to limit the current... maybe.

If anyone has any more suggestions, please post them.  Thanks!

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I keep getting stuck on the current limiting rheostat.  No matter how I do it, it's going to potentially dissipate a lot of power.  I tried placing a 0.833 ohm resistor in series, but that still could mean 0.85W between both if the rheostat is set at 1 ohm.  Sure, a 100W 100 ohm rheostat would then work, but again, those are fairly expensive.  Plus, this may also limit my upper current limit to less than the 1.5A the LM317 can provide.

So, after some quick research, I have another idea.  Since I want to include both a voltmeter and an ammeter, perhaps a slightly different design can be beneficial.  In my last schematic, I included analog meters.  This new design incorporates digital meters and limits the current without using an LM317.  Essentially, I remove the current limiter (which includes the LM317 and the 500 ohm rheostat).  My digital meters are powered via a 9V battery and do not share the PSU ground.

What do you folks think?  Will it work?

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Product Description
2K Ohm 2 Watt Pot
2K ohm by PEC/Canada. Perfect for LM317, LM350 circuits. 3/4" long shaft. With hardware. RES1352....

Price: \$2.49
Part No.: RES1352

Put resistor in parallel to get value down to 500R

500 Ohm 10 Turn Pot Price: \$9.95
Limited availability. 2 watts
Part No.: RES1355
In Stock: Yes

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Hi Niksun,
You forgot to show where the load connects on your meters schematic. I think it is the LO terminal of the A meter.
I don't know why R3 is in series with the voltmeter.

Now it current-limits only when the LM317 has its max of about 2.2A, or when it gets too hot and shuts down. The resistors won't limit the current.

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Ack!  Yes, you are correct.  If you draw a horizontal line which starts at the right end of R3 and goes right, that will be the regulated Vout.  PSU Ground can also be extended to the right for the regulated ground.

In terms of R3: can you not simply set your voltage, short the output of the PSU, then adjust the current limit with R3?  Perhaps another (safer) way is to add a normally open 3A push-button switch in series with a 30 ohm, 10W resistor across the regulated Vout and ground and use that to short the output of the PSU when the switch is closed, then adjust the current limit.  Perhaps I can refer you to a web site, but I'm not sure the rules on this forum (some do not like to have people posting URLs).

Thanks for all your help, by the way.

Oh, and possibly I can do this better with just a FET and a potentiometer to limit the current, but I'm not exactly sure how to do this yet.

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If R3 is in series with the load then the voltage drop across it will change when the load current changes resulting in no voltage regulation.
Even the 0.1 ohm shunt resistor for the current meter degrades the voltage regulation to a 150mV drop with a 1.5A load. The LM317 without the 0.1 ohm resistor is much better because its output voltage drops typically only 1.5mV with a 1.5A load.

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Audioguru,

OK, so if it were up to you, would you go with my double LM317 design with the high watt rheostat to limit the current?  I'm not sure the best way to go about placing a potentiometer there to vary the current limit (due to the potentially large amount of power dissipated by the pot), but I do know that I want to have a ~15mA to 1.5A variable current limiter.

Any suggestions?

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Look at our 30V variable power supply project. It uses a fixed power resistor to sense the current outside the voltage regulator so it doesn't cause a reduction of voltage regulation.
Then it uses an ordinary inexpensive low power pot to set the reference voltage for an opamp that regulates the current.

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