Jump to content
Electronics-Lab.com Community

0-30V Stabilized Power Supply


redwire

Recommended Posts


Ok, then it`s working rigth but I have a doubt, I have using PSUs and all of them have the possibility to limit the output current shorting the PSU and then selecting the max current, is it possible to do something like that here?

Link to comment
Share on other sites

Hi, reading your message again i realized that i misunderstood you, P2 don't limit the current but is used to adjust this limit, i have checked that when i shorted the output the led turn on but the current isn't limited by anything, if i measured the current sense in R7 it reach 8A i think that U3 is not working right.

Link to comment
Share on other sites

Guber,
The (+) input of U3 has a voltage divider including P2 from the reference voltage of 11.2V.
The (-) input of U3 has the voltage across R7 created by the output current.

When the voltage on the (-) input exceeds the voltage on the (+) input then the output of U3 goes low enough to reduce the output voltage through D9 then the output current is also reduced.
When the output of U3 goes low then it turns on Q3 which lights the LED. Your LED turns on so U3 is working correctly.
Maybe your D9 is burnt open or is connected backwards.

Link to comment
Share on other sites

Hi, I see U3 is working as comparator between the voltage set by P2 and R7, I have checked D9 and seems to be broken I have replaced it but I still have the same problem, I set P2 to a very low voltage then I connect some LEDs to the output and the indicator of current limit turn on but the current is not limited....

Link to comment
Share on other sites

hi, i have been studying the circuit and i see that the voltage at the input of U2 must be the 11,2 established by the zener diode and when U3 is turned on because of the Voltage set by P2 is greater than the one in R7 D9 must be open, but this happen is the Voltage in cathode is greater than the Voltage in anode less 1V, I mean, Voltage in D9 have be minor than 1V, then the Voltage in the output of U3 must be greater than 10,2V when is turned on, isn't it?

Edit: of course when P1 is at his Maximum Value.

Link to comment
Share on other sites


hi, i have been studying the circuit and i see that the voltage at the input of U2 must be the 11,2 established by the zener diode and when U3 is turned on because of the Voltage set by P2 is greater than the one in R7 D9 must be open, but this happen is the Voltage in cathode is greater than the Voltage in anode less 1V, I mean, Voltage in D9 have be minor than 1V, then the Voltage in the output of U3 must be greater than 10,2V when is turned on, isn't it?

Edit: of course when P1 is at his Maximum Value.

When the voltage setting pot, P1 is set to maximum then the input to U2 is 11.2V and the output of the project is 30.0V if it is not regulating the current. Then the (+) input of U3 is higher than its (-) input causing the output of U3 to be high at about +29V. Then Q3 is not turned on, the LED is not lighted and D9 is reverse-biased so it does nothing.

If the output is shorted then there will be a fairly high current in R7 causing the (-) input of U3 to be higher than its (+) input so its output goes low enough that D9 reduces the input voltage to U2 to almost zero. Of course then the output of U3 must go one diode voltage drop below zero volts which is fine because it has a negative supply in addition to its positive supply.

Maybe your -1.3V negative supply is not working.
Link to comment
Share on other sites

hi, finally i understand the functionality of U3 is the opposite of i was saying, when voltage in R7 is greater than the one set by P2 the Output of U3 goes to -1,3 and this active D9, after that i realized that the route to D9 was broken i fixed it and now finally is working.

Thanks a lot.

Link to comment
Share on other sites

Hi,

It seems that every week i have a new failure..

Now i have very little current at the output, few mA, the output voltage seems to be ok then i suppose that BD139 is ok, i have test the 2n3055 out of the circuit and also seems to be ok but i suspect from one of this components, i have a doubt about where the circuit gets the ouput current, i think that current is the one which  flows through emitter to collector of 2n3055 but i'm not sure.

Link to comment
Share on other sites


Now i have very little current at the output, few mA, the output voltage seems to be ok then i suppose that BD139 is ok, i have test the 2n3055 out of the circuit and also seems to be ok but i suspect from one of this components, i have a doubt about where the circuit gets the ouput current, i think that current is the one which  flows through emitter to collector of 2n3055 but i'm not sure.

The fixed and improved version of this project is supposed to have TWO 2N3055 output transistors to share the heat. The BD139 and the output transistors are emitter-followers. The small base current into the BD139 from U2 causes it to pass up to 86mA through its collector-emitter to the bases of the 2N3055 transistors that pass the positive unregulated supply up to 3A at their collectors to the output though their collectors-emitters.

Ohm's law says that if the output current is reduced by something then the output voltage will also be reduced.
Then find what is reducing the output voltage when there is a load.
Link to comment
Share on other sites




Kevin, the improved and fixed project uses opamps with PNP bipolar transistor inputs, not Jfet inputs so that they work with an input voltage as low as the negative supply voltage.
The input voltage of opamp U2 goes from 0V (its negative supply voltage) to +11.2V and never goes as high as the positive supply where it will have its output saturated as high as it can go.

Recently I used a JFET in a small circuit. The gm specification recommended is around 0VGS at 15V VDS. I used a very low value source resistor and drain resisitor for gain. Should they be good for low resistance loads?
Link to comment
Share on other sites

  • 3 weeks later...

Hello everyone, I'm new here, I'm from Peru (south america). I've been reading the entire topic all this week, and want to thank Audioguru for his patience, very nice explaining about this circuit. I'm gonna start building this power supply (REV 3 I think) from PicMaster's PCB, I've got the TLE2141 opams.

At the moment I have one question: How can I control (set) de voltage and current from a microcontroller (PIC), I know I have to replace the pots, but how does it affect the entire circuit?.

I have been studying another circuit wich is controlled by a PIC (PWM, D/A converter, etc), but at the end it has a lot of problems because the regulator circuit. So I'm all for this one :).

Thanks again to all in this forum. I hope for some advice about my question  ;D.

Link to comment
Share on other sites


How can I control (set) the voltage and current from a microcontroller (PIC), I know I have to replace the pots, but how does it affect the entire circuit?

The existing voltage-setting and current-setting pots are fed from 11.2V. But most digital pot ICs have a maximum of 5V. Then the project must be re-designed for them to be used.
Link to comment
Share on other sites

Thanks for your reply audioguru. And with another opam (with its own supply) with a gain of aproximately 2.24, connected at the output of the PIC, will that work in theory right?.

thanks again, i'm already building the 3A circuit  ;D

Link to comment
Share on other sites


Thanks for your reply audioguru. And with another opam (with its own supply) with a gain of aproximately 2.24, connected at the output of the PIC, will that work in theory right?.

Yes, if the DAC in the PIC has an output from 0V to 5.0V then to get 11.2V the gain of another opamp must be 2.24.
The current-setting pot has an output of from about 5.6mA for an output max of a few mA to 1.41V for an output max of 3A, then an extra opamp is not needed. 
Link to comment
Share on other sites



Yes, if the DAC in the PIC has an output from 0V to 5.0V then to get 11.2V the gain of another opamp must be 2.24.
The current-setting pot has an output of from about 5.6mA for an output max of a few mA to 1.41V for an output max of 3A, then an extra opamp is not needed. 


Ok, I'll keep that in mind  8). I'm already making my PCB, I will test it and post my results and pics as soon as I can. If anyone already has a modified circuit for digital voltage control, I'll really apreciate it, I read that Picmaster did something like that before  ???.

Anyway, thank you again :)
Link to comment
Share on other sites


The current-setting pot has an output of from about 5.6mA for an output max of a few mA to 1.41V for an output max of 3A, then an extra opamp is not needed. 


What did you mean by a range from 5.6mA to 1.41V? 5.6mV to 1.41V?
Link to comment
Share on other sites

  • 2 weeks later...

Hello again!!!! Well i have bad news i built the circuit and i don't get any output....The only thing i can see is the led to light up when i turn the p2.I measuered 40 volts in and -1.1 ....Also i used a 2n3722 instead of a 2n3055.Any ideas what to look??

Regards Panagiotis

Link to comment
Share on other sites


Hello again!!!! Well i have bad news i built the circuit and i don't get any output....The only thing i can see is the led to light up when i turn the p2.I measuered 40 volts in and -1.1 ....Also i used a 2n3722 instead of a 2n3055.Any ideas what to look??

Regards Panagiotis

Since you used only one output transistor then it sounds like you made the original circuit that has many errors instead of making the improved and fixed circuit that has two 2N3055 high current output transistors to share the heat. The new driver transistor is much better than the old one.

I could not find a detailed datasheet for the 2N3722 transistor but its spec's say its max output current is too low for this project so maybe yours is blown up. Then maybe the driver transistor is also blown up.
Opamp U2 , the driver transistor and output trtansistors are a simple almplifier with a voltage gain of 30V/11.2V. It amplifies the 0V to 11.2V from the voltage-setting pot. This amplifier has a max output current of 3.0A.
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...
×
  • Create New...