Jump to content
Electronics-Lab.com Community

0-30 Vdc Stabilized Power Supply


Sallala
 Share

Recommended Posts

well ... i changed the orientation of the Q2( so that the emitor is pointing towards 8th pin( the base of Q4 ) ) and now the situation is following ... output voltage is 0.5V, u2 is still heating like mad ;), the voltage on Q2 collector is ~34V ...

Q4 base volt.  = 0,5V
      collector volt. = 34V

Q3 emmiter volt. = 34V

U2 output = ~200mV


any directions?

Link to comment
Share on other sites

Its output voltage (emitter of Q4) is 0.5V and also is ithe base of Q4. Therefore Q4 is shorted base to emitter.
The output voltage of U2 is less than the output voltage when it should be about 1.4V higher, and it gets very hot. Therefore U2 is destroyed.
Q2 might be damaged from being connected wrong, it probably damaged U2. 

Link to comment
Share on other sites

Hello,

it's me again.
I've replaced burned output transistor with 2 new ones.
I still belive that current regulation is not working, because load isn't reacting on pot 2 changing no mather how, just, like I said before, only drops the voltage down and turns the LED on at max setting.
Can you please give me some advices or measuring points to check this out?
Or, could it be something else wrong?

Please, help me!

Thanks, Triple.

Link to comment
Share on other sites

Hi Triple,
I think your project works fine.
The current regulator causes the voltage to be reduced until the set amount of current flows in the load. Try a 10 ohms load resistor:
1) 30V/10 ohms= 3.0A.
2) If you turn down the current setting then only 20V/10 ohms= 2.0A.
3) Turn it down more then only 10V/10 ohms= 1.0A.
4) Turn the current setting all the way down and the output voltage should normally be nearly zero.

Link to comment
Share on other sites

Hi!


I think your project works fine.

Yes, you're absolutely right! My power supply is working!  :D


I think the pot should have its wiring reversed so that it gives max current when it is turned to max.

Yes, you were right about that too. My apologies that I claimed contrary.

What else to say... Everything is working fine, just one thing I don't understand: I have exactly 30,8 VDC and 3,08 ADC at max setting over 10 Ohm resistor, but I have only 100VA transformer.  ???

And, of course, big big thanks to designer of this circuit, everyone who helped with it's modification, and especialy Mr. Audioguru for doing such a great job. Thank you. 
Link to comment
Share on other sites

Hi!

When I buy proper heatsink I will test PS for a longer period and then report. It's robust, it will keep it up.  ::)

And a question about measuring output current as voltage over R7: wont R7s tolerance and temperature change ruin correct reading?

Thanks, Triple

Link to comment
Share on other sites

Hi audioguru and others,

Very much looking forward to building this but want to be 100% sure of the approach from the start. I'm fairly new to electronics though so please excuse the stupid questions... in replacing D1-D4 diodes with a single bridge and Q1/Q2 with TIP31A,

Link to comment
Share on other sites


replacing D1-D4 diodes with a single bridge

Yes. Then it can be bolted to the chassis as its heatsink.

don't we need an updated PCB diagram? Not sure Q2's TO220 style case and heatsink would fit well on the original board design. Unfortunately I already have the etched board!

The same pcb is fine. A few larger resistors will need to be spaced above the board. Q1 will need its pins bent around. Q2 can be bolted to the heatsink and won't need an insulatopr and the Q4's are also without insulators. Then insulate the heatsink from the chassis.
Link to comment
Share on other sites

A 24V transformer has a peak voltage of 34V. The rectifier bridge will drop about 2V at full output. The two darlington-connected output transistors, R15, R7 and U2 will drop about 8V at full output leaving about 24VDC for max output.

The transformer's peak voltage of 34V will have an average current of 4A which is 136VA. Your transformer is way overloaded when the project has a 4A output.

100VA/34V= 2.94ADC. When the load draws slightly less than 3A then the transformer will be operating at an average of 100VA.
Add them up:
24V/3A for the load. P= 72W.
8V/3A for the parts. P= 24W.
2V/3A for the rectifiers. P= 6W.
Total power= 102W.

Link to comment
Share on other sites

At first I'm quite newbie in electronics and  me and my firend decided to make our course project about this power supply. And there is needed some analysis and calculations. So in schematic is C4, R8 and R9 ... why are these needed? I understand why C5 is nessesery, but isn't better to use a ceramic cap. (we use mod list - [email protected])? And there is that ripple... How I can make somekind of calculations about ripple in U2 output (pin 6.) and how about calculations about ripple smoothing by Q2 and Q4 (we going to use two 2N3055). And how about C9? (I undestand that, it puts U2 working also like integrator). C6 is understandable in high frequencieis...

can anyone help us about these problems ? :-)

Link to comment
Share on other sites


At first I'm quite newbie in electronics and  me and my firend decided to make our course project about this power supply.

Hi Muffits,
Welcome to our forum. ;D

So in schematic is C4, R8 and R9 ... why are these needed?

R8 and C4 are a filter to remove high frequency noise from the D8 reference voltage circuit. R9 isolates C4 so that D9 can reduce the voltage of the project quickly when current regulation needs it to be reduced.

I understand why C5 is nessesery, but isn't better to use a ceramic cap?

It doesn't matter because the opamps don't operate at radio frequencies when a ceramic cap would be necessary.

And there is that ripple... How I can make somekind of calculations about ripple in U2 output (pin 6.) and how about calculations about ripple smoothing by Q2 and Q4 (we going to use two 2N3055).

U2, Q2 and Q4's amplify the reference voltage output of U1. U1 has hardly any ripple because it is the current source for D8 and buffers D8 from the unregulated supply.

And how about C9? (I undestand that, it puts U2 working also like integrator). C6 is understandable in high frequencieis...

C9 reduces the high frequency gain of U2 so that it doesn't oscillate when Q2 and Q4 are added in its negative feedback loop.
Link to comment
Share on other sites


A silly question: in wich frequency is low pass filter (R8, R9 and C4) calculated on? I mean does it block from 100Hz to ...?

The resistor R8 has part of P1 in series with it so the total resistance is about 32k ohms. Therefore the cutoff frequency of 32k ohms with C4 is about 50Hz.

post-1706-14279142876613_thumb.png

Link to comment
Share on other sites

A small thing that I've noticed... I simulated C4, R8 and R9 and only R8 and C4... and thing is that with R9 isn't it 3 order LPF (as without it it's 2 order LPF)? The thing is with with R9 it cut's harder frequencies, then without R9..... it's just a remark...btw at 100Hz it drops ripple voltage something like 30 times...c :-)

And a question.... I understand why C6 and C9 are neended, but can anyone explain they workin principals.... Sry about all my silly questions, but I can't get rest, unless I know enough :-D

PS. If anyone intrested, then this is how should U3 output voltage look like, then it is working. (the red one is u3 output)

post-19683-1427914287905_thumb.gif

Link to comment
Share on other sites

About C4, R8 and R9 forget it.... my bad, R9 only takes down a bit amplitude of ripple voltage at low frequencies (something like to 10Hz). So I still don't understand R9 fuction.... I remeber that audioguru said that it isolates D9... does it mean that for D9  R9 seems to be source?

Link to comment
Share on other sites

The input of the opamp U2 is megohms so R9 doesn't affect how it works, until the current regulator U3 needs to quickly reduce the voltage at the input of U2. Then the resistance of R9 isolates C4 from the input of U2 so that D9 can quickly reduce the voltage at the input of U2 without waiting for the filter C4 to discharge.

This project is regulated and shouldn't have ripple. If R4 is the original too high value of 4.7k instead of 1k, and D8 is a high current zener diode instead of a 5mA one, or if C1 is the original too small 3300uF instead of 12,000uF then the output will have ripple. 

Link to comment
Share on other sites


i replace the Q2 and the u2(new one is still heating like mad) and now the output voltage is ~0.4V .. the voltage in q4 E,C and B is 32V+ ... so what is wrong now?

Q4 is shorted and might have destroyed the new U2 and Q2.

Remove R15 and get Q2 and Q4 working properly first. As a test, connect two 10k resistors to the base of Q2. Connect one resistor to +34V and the other resistor to the output's 0V. Without a load, the output voltage should be about +15.7V. Without R15, U2 should be cool.
Link to comment
Share on other sites

Hi MonSSter,
In your reply #931 all 3 terminals of Q4 were at +32V and the output of U2 was at 0.4V. So I thought that Q4 was shorted.

Now with R15 disconnected the base and emitter of Q4 dropped to +0.5V which is normal.
The output voltage of U2 is high which is normal without R15.
Is U2 still hot?
Which part number is U2?

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

×
  • Create New...