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0-30V Stabilized Power Supply


redwire
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The opamp having a very high voltage gain should regulate it's output at the voltage set by the offset RV1. The 0.85 gain of the negative feedback loop transistors combined with R12 is where the loss in load regulation is.

No. The load regulation is excellent. The line regulation is also excellent.

The very high open-loop voltage gain of the opamp driving the driver and output transistors allow the negative feedback from R11 and R12 to keep the output voltage at almost the same voltage as at the voltage-setting pot.
If the output voltage tries to drop 1V then the opamp output rises almost 1V to cancel the output voltage drop.
In my previous post I calculated that a 1V attempted drop actually causes only an 8.3uV drop.

RV1 adjusts the output voltage only a tiny amount, about plus and minus 0.05V to adjust the output to exactly 0.0V when the voltage setting pot is at minimum. The maximum input offset voltage of the TLE2141 opamp is only 0.9 milli-volts (0.0009V).

R12 from the output of the project to the negative input pin of the error opamp plus R11 to ground set the voltage gain of the entire amplifier to 1+(56k/33.4k)= 2.68 times so that when the voltage setting pot is at maximum then the reference of 11.2V is amplified to 30.0V.
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The voltage regulation looks very good at around a gain of 3.

The voltage regulation is good at any amount of gain. A gain of 2.7 is used because the reference zener diode D8 is 5.6V which has the best regulation and opamp U1 is a current source for it that doubles the reference voltage to 11.2V. Then for a maximum output of 30V a gain of 30V/11.2V= 2.6786 is needed.

RV1 doesn't need any offset adjustment unless the regulated output voltage is more than 15 volts.

RV1 has nothing to do with output regulated voltages. Opamp U2 has a small input offset voltage when its input is zero then its output has a voltage that is called its offset voltage. With an input voltage of zero from the voltage setting pot then RV1 is adjusted until the output voltage is zero.
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audioguru,   Why is U1 and the associated negative and positive feedback necessary to output 11.2 V, used instead of a simple voltage regulator in the 11-12 volt range?

A single 11.2V zener diode powered from a resistor has much worse voltage regulation than a 5.6V one and the 5.6V one is powered by a constant current opamp circuit that makes it even better.
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It will be fine if it has a load of at least 5mA and an input from 19V to 35V.


Then why not eliminate D8, U1, R5, R6, and R4  and replace with one T0-92 that can handle a 40V input voltage that would get power from the 10V zener?  These voltage regulators are pretty indestructible with thermal, voltage and current cutoff features.

L7812AC.pdf

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hello and  HAPPY NEW YEAR to all !!!!!!! Well i finished my psu and now i am facing a new problem.I have output of 36v no matter what. The pontesiometer doesn't change the output voltage.Can you imagine what it might be wrong?

Regards Panagiotis

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hello and  HAPPY NEW YEAR to all !!!!!!! Well i finished my psu and now i am facing a new problem.I have output of 36v no matter what. The pontesiometer doesn't change the output voltage.Can you imagine what it might be wrong?

You might be using old TL081 opamps that have failed because the unregulated supply voltage is too high for them.
You might have the pins on the driver or output transistors connected backwards.

When the problem is fixed then you might find opamp U2 destroyed from excessive heat.
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i use the mcp chips.The bd transistor is placed correctly.Can you tell me what i can measure with my multimeter (Voltages) To spot the problem? Thoughts of what might be how can i test the op amps?on ic1  i get +5,5v on pin 2 and 11,2v on pin6

Regards Panagiotis

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i use the mcp chips.

What is "mcp"? It should be an MC34071 or a TLE2141.

on ic1  i get +5,5v on pin 2 and 11,2v on pin6

Perfect. The +11.2V feeds the voltage setting pot whose slider moves from 0V to +11.2V. The slider feeds opamp U2 and it has negative feedback from the output of the project setting its voltage gain at 2.68 times.

The driver and output transistors are simple emitter-followers with an output that is about 1.5V less than their input and a voltage gain of about 0.95 times.

Then if the slider of the voltage-setting pot is +5V then the output of the project is 5V x 2.68= +13.4V and the output of U2 is about +14.9V.

Does the output of U2 go from about +1.3V to about +31.5V when the voltage setting pot is turned from zero to maximum?
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Guest DrinkH2SO4

Hello people, i want to ask you something.  First i started building the original project but since reading this nice thread, i started building the revised version. I now got the rev 5 parts list, but i already had made the pcb from Picmaster. Now there are some layout inconsistency, i cant find R17 anywhere, or it says R16 instead R17 and in the place of Q1 which is not used there is a 3 pin gap. Is there a revised version of this pcb, or i need to make the another one, that is in the zip with rev 5? Any help would be appreciated. (still learning)

Thanks

post-105887-14279144597201_thumb.jpg

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Hi DrinkH2O4,    I don't know if anyone is going to try to match up parts on the board layout provided, to see what Picmaster changed from the design being discussed on this thread.  It would certainly be easier with the sketch layout but you may need to pick a design and either ask Picmaster or this board for help.  I thought Picmaster used a PIC chip to do some of the control work.

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i get constantly 28,8v on pin 6 of u2 although i turn the pontesiometer.

But the output of your project is at 36V which is 7.2V higher.

If the base or emitter of one or both of the output transistors is shorted to the positive unregulated supply then the emitter-base diode of the BD139 is reverse-biased and acts like a 7.2V zener diode.
After fixing it then maybe the U2 opamp is destroyed and needs replacement. 
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Guest DrinkH2SO4

Hi again, and thx for the replies. I think the pcb in the latest zip file (the brd file) is double layered pcb, for which i dont have both sides copper plated board at the moment. Also i wouldn't want the pcb i already made (from the layout picture i posted before) to go to waste.  :)

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Hello well great news...finally got it work.I had a 2n3055 in my drawer that was "off" and caused all the problem.(Can't figure out if i had tested before solder it...)
It blown up (i think not sure ,but i changed it)the bd139 and i think that burn one of the inputs of the mc34071 and thats because when i changed the transistors, i noticed that when i was turning the pontesiometer from the half and clockwise the voltage was increasing from 15-30 from the half and anticlockwise it was happening the same...on the center position of the pontesiometer was the lowest voltage value which was 15 and turning it clockwise or anticlockwise the voltage was increasing from 15-30v.I am writing this down because somone would might face a similar problem.

Regards and thank you all for your precious help and ESPECIALLY audioguru which i propose you to change it To --->electronics_guru..  :)

P.S I haven't  test it but what will happen if i short circuit the output of the power supply?  Will C.C take over or it will burn anything?

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Hello well great news...finally got it work.I had a 2n3055 in my drawer that was "off" and caused all the problem.

Good, you fixed it.

I haven't  test it but what will happen if i short circuit the output of the power supply?  Will C.C take over or it will burn anything?

If the current regulation circuit works properly and the BD139 and 2N3055 transistors have enough heatsinking then the output current will be what the current-setting pot makes it from a few mA to 3.0A and the LED will light.
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audiogury sorry for asking that but i need to solve it to my head (I am newbie so show understanding    ;D)  Why when the current limiter is enabled the voltage drops???.I know that is has to do with the ohms law but i need more spesific details... I mean why the voltage drops instead of staying steady ...
For example we have a device that needs 12v 3,2A to operate if we set the voltage pot to 12 and the current pot to 2,5 then we will notice that when the current will reach the 2,5A the voltage will start falling (don't know how much) why is that happening instead of preserving the voltage staedy to 12V and the current to 2,5Amps ????

Regards.

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Why when the current limiter is enabled the voltage drops???.I know that is has to do with the ohms law but i need more spesific details... I mean why the voltage drops instead of staying steady ...
For example we have a device that needs 12v 3,2A to operate if we set the voltage pot to 12 and the current pot to 2,5 then we will notice that when the current will reach the 2,5A the voltage will start falling (don't know how much) why is that happening instead of preserving the voltage staedy to 12V and the current to 2,5Amps ????

You want 12V at 2.5A. Then Ohm's Law says the load is 12V/2.5A= 4.8 ohms.
You set the voltage to 12V and you set the current regulator to 2.5A and load it with 2 ohms. Then Ohm's Law says the voltage MUST be 2.5A x 2 ohms= 5V as shown with Ohm's Law again.

If the voltage stayed at 12V then the current in 2 ohms would try to be 12V/2 Ohms= 6A which is impossible in this supply with a maximum current of 3A.

Set the voltage to anything and set the current to 2A. Then short the output. The voltage will be zero and the current in the short will be 2A. 
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Thank you very much!!!!It seems that you won't get away easy from me..... ;D  I am joking.....
Something i had forgot...Why we use 3 2n3055 for this power supply where 1 2n3055 can provide this current?

regards!

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The 3A version of this improved project uses two 2N3055 output transistors. The 5A version uses three of them.
They share the heat.

When the output is shorted or set to a low voltage and the current is 5A then the output transistors must dissipate a total of about (40V - 1.35V - 0.55V) x 5A= 191W. Each transistor must dissipate 191W/3= 64W.

One 2N3055 transistor can dissipate a maximum of 115W if its case is held to only 25 degrees C with some kind of freezer. But then its chip will be extremely hot so maybe 90- degrees C would be more reliable.
Where will you find three suitable freezers or one big one?

We use heatsinks to transfer heat from a power transistor to the ambient air. A heatsink is aluminium (conducts heat well) and it has many fins for a large surface area in the ambient air. Thermal grease is used between the power transistor and the heatsink to fill the microscopic dents with a heat conducting material.
A heatsink is not perfect since it and the power transistors will still get hotter than the ambient air but they work pretty well if each transistor dissipates only 64W.
Then a pretty big heatsink can cool the three power transistors. A smaller heatsink can be used with a fan. 

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(40V - 1.35V - 0.55V) x 5A= 191W.

can you explain me the 2 values??? 1.35V and 0.55V...


And i can understand the part....
" But then its chip will be extremely hot so maybe 90- degrees C would be more reliable."

Do you want to say the it would be safer for the transistor to work in a temperature of 90 degrees C? What do you mean when you say "its chip" ?

Regards 

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