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I am new to electronics and have been exploring easy to make DIY things for a while now.

I am using the following circuit to make a voltage regulated high current DC power supply.

The problem is, I can't adjust the potentiometers to get an output voltage beyond 11-12 volt. The DC output is always between 11 - 12 volts, neither less, nor more. By the way, the transformer is 12-0-12 5 amp, and I am measuring no-load voltage with a digital voltmeter.

My questions are:

  1. Is the circuit correct, regulated voltage at the base of 2N3055 will regulate the voltage at its emitter - is this principle true?
  2. Is keeping one of the terminals of the potentiometers open (not connected) a good idea? Should they serve the purpose in this circuit?
  3. If I want to troubleshoot, what are the test points and what are the expected voltage or other attributes I should look for?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Nice how the diode and Vbe drops compensate for the 1.25 Vref. You should be able to regulate from 0 to 32V with those pots and a LM317 but limited to about 5.6V below your supply as this is the dropout voltage of the reg-diode-transistor chain. \$\endgroup\$ – KalleMP May 9 '17 at 11:13
  • \$\begingroup\$ Could one move the sense point to the output if one were happy to have the minimum as 1.25V? It would eliminate any drift due to transistors heating and such. \$\endgroup\$ – KalleMP May 9 '17 at 11:15
  • \$\begingroup\$ And what is your transformer voltage? And the very first thing you should do is follow Andy_aka's advice about putting a load on the output. \$\endgroup\$ – WhatRoughBeast May 9 '17 at 12:50
  • \$\begingroup\$ The trafo is 12-0-12 Center tapped. I am getting almost 34V DC after the filter since I am using the 2 outside terminals of the output (not the center one). \$\endgroup\$ – sribasu May 9 '17 at 19:25
  • \$\begingroup\$ I saw another version of the circuit, where the transistor base is connected to a different position. See the last diagram on this page. Now I am in doubt. Is the diagram in this link, correct? Or the diagram I have have drawn is correct? \$\endgroup\$ – sribasu May 10 '17 at 13:15
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The use of PNP pass transistor(s) will improve the circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

Paralleling multiple transistors is perilous because the warmest of them passes more current (and heats, and goes into thermal runaway). Emitter resistance here is good, promotes equal current sharing. The base of those power transistors SHOULD have a base-emitter resistor to turn them OFF reliably if they get hot.

The series diode and NPN base-emitter junction decouple the output voltage from the feedback sensing (which lowers the regulation accuracy), but the PNP driven off the input pin does not interfere with the normal output sensing.

For low currents, the PNP transistors just turn OFF; at high current, they pass at most 4x the regulator's current limit (LM317, about 1.5A).

D2, D3, D4 are optional, to prevent reverse voltages when power is turned off.

Leaving one end of the potentiometer open is not recommended, because adjustment during operation can open-circuit the wiper for a short time, and limiting that glitch (or adding a capacitor to limit its slew rate) is good practice.

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Simply "hanging" a diode and emitter-follower on the output of a voltage regulator ruins the excellent voltage regulation of the regulator IC. The circuit with the PNP transistor should be used since the PNP transistor is inside the negative feedback loop which cancels its voltage variations.

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The biggest problem here is that you have 2N3055 output transistors (that are NPN) and you are assuming they are PNP. I would go for the emitter connected to the output so that they act as current amplifiers. Hence: -

I am correcting the diagram

Test the LM317 disconnected from the 2N3055s and verify what supply range you can achieve. With 2N3055s connected as parallel emitter followers it should work just fine but please test with a 1 kohm load instead of an open circuit.

Regarding the pots, one end left open is fine but do calculate the maximum current that can flow through the wipers and verify that the pot's data sheet confirms it can handle it.

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  • \$\begingroup\$ Really sorry, the diagram is incorrect. Please assume, that its' NPN. I am correcting the diagram \$\endgroup\$ – sribasu May 9 '17 at 11:00
  • \$\begingroup\$ How to calculate the current flow through the POTs? I am assuming voltage drop across R1+R4 is 1.25volts, is that correct? And what is the drop for R2 and R3? \$\endgroup\$ – sribasu May 9 '17 at 11:05
  • \$\begingroup\$ Isn't the 1.25 volts across R1+R2? I haven't checked - just a memory thing. \$\endgroup\$ – Andy aka May 9 '17 at 11:33
  • \$\begingroup\$ Vref is between Vout and Adj, so across R2 and R3 for the current schematic. \$\endgroup\$ – Peter Smith May 9 '17 at 14:49
  • \$\begingroup\$ @PeterSmith well spotted. \$\endgroup\$ – Andy aka May 9 '17 at 14:55
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Is the circuit correct, regulated voltage at the base of 2N3055 will regulate the voltage at its emitter - is this principle true?

the answer depends on your definition of "regulate": the output voltage will vary slightly with the load current. Something inherent to this design.

Is keeping one of the terminals of the POTs open (Not Connected) a good idea? Should they service the purpose in this circuit?

No. an open circuit should be avoided at all cost here.

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  • \$\begingroup\$ What is 'slightly' in your opinion? Not between a range of say 2v to 30v as I am expecting? \$\endgroup\$ – sribasu May 10 '17 at 11:48
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Circuit is good. I simulated it on Proteus and was able to change the output voltage by changing the variable resistors. I used a direct 24V 50Hz alternator as source. I did not however, perform any current/power calculations on the circuit. I did notice the voltage dropping once I connected the output to a resistive load. May be that is your problem; you might be trying to source too much current from the circuit or you might have a short somewhere on the output. As far as testing goes, I would suggest taking the diode off and measuring there and go on forward from there.

As for your other questions,

  1. Yes, this is a correct circuit
  2. There is no problem in keeping one end of the pot open. In this configuration it works as a variable resistor, the resistance between the two points can be changed by sliding/rotating the knob. The other configuration will be to use it as a voltage divider, you can connect one end to higher voltage and the other to ground. Measuring voltage between the viper and ground will give you the divided output.

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    \$\begingroup\$ This answer is totally wrong. 1) It's really bad to connect in parallel two power BJTs in this way, without current sharing resistors 2) the good way to increase output current of an LM317 (or also 78xx) is shown in the datasheet (see texas instruments datasheet, page 16 and 17). 3) it's a bad practice of leaving one terminal of the pot open. It's better to short it to the tap. In this way, if dust accumulates (or if the pot wears out) you don't end with a open, but with a fixed maximum value resistor. \$\endgroup\$ – next-hack Sep 21 '17 at 19:36

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