Electronics-Lab.com Community

# 0-30V 3A - I need some answers

## Recommended Posts

The U1 is clear - a reference voltage source 11,2V. U2 also - opamp reinforcement 3, so that the maximum voltage at the input of the output equal to 11,2V receive up 33,6V (using an appropriate transformer, but we assume that it is good).
One of the features is not very clear U3 - I know it is responsible for the current protection and opamp works here as a comparator (I think). However, from my calculations in section 6 of the potentiometer P2 maximum voltage of about 1,7V (P2 and R18 are a voltage divider, R17 can be omitted because it is of little value, we put it becouse on the lowest setting to P2 amplifier input can be shorted to ground). So potentiometer P2 can set the voltage from 0 to 1,7V. However, at maximum current consumption measurement resistor R7 hangs at most 1,41V (0,47R * 3A). Which implies the difference? Comparator can detects a voltage difference already tiny.
But another question - if U3 inverting input (-) have less potential than the non-inverting input (+) the output voltage of U3 is potential of leg 7 of U3, and according to that diode D9 does not conduct. However, if power consumption exceeds the voltage at the output of U3 have potential of leg 4 of U3 (so it is about -5V) and diode D9 becomes conductive. But what happens next? Because that's the part I do not understand.

Another problem is the presence of R8 and R9. I think in terms of the setting of gain opamp U2, right? What the presence of C4, C6 and C9?

And the last question - why current limiter isn't linear? I mean when we change output voltage with the same load, the current limiter change value. Why? Maybe you can give me a example power supply we can set output current which doesn't depend of output voltage?

I know that on this site is to explain the action, but does not respond to my questions. I hope someone enlighten me, because a long time I break their heads.

##### Share on other sites

One of the features is not very clear U3 - I know it is responsible for the current protection and opamp works here as a comparator (I think). However, from my calculations in section 6 of the potentiometer P2 maximum voltage of about 1,7V (P2 and R18 are a voltage divider)

The original circuit has errors. So we fixed it in the forum.
A maximum voltage from P2 is 1.7V when R18 is WRONG at 56k. When we fixed the original circuit we changed R18 to 33k and added a 100k calibration trimpot in series so when the trimpot is about 46.7k then the total with 33k is 79.7k then the maximum voltage from P2 with a 3A load is exactly 1.41V. 1.41V/0.47 ohms= 3.0A.

But another question - if U3 inverting input (-) have less potential than the non-inverting input (+) the output voltage of U3 is potential of leg 7 of U3, and according to that diode D9 does not conduct. However, if power consumption exceeds the voltage at the output of U3 have potential of leg 4 of U3 (so it is about -5V) and diode D9 becomes conductive. But what happens next? Because that's the part I do not understand.

The voltage across R7 caused by the load current in it goes to the inverting input pin 2 of U3, and the pot P2 voltage goes to the non-inverting input pin 3. Usually pin 3 voltage is higher than pin 2 voltage so the output of the opamp pin 6 is high and D9 is reverse biased and does not conduct.
If the current in R7 exceeds the setting of pot P2 then the opamp output drops low enough for it to reduce the output voltage through D9 that reduces the output current until the current stays at the setting of pot P2. If the output of U3 went low like a comparator then there will be NO OUTPUT CURRENT! Instead the output voltage of U3 drops only low enough so that the load current is the same as the setting of p2.

Another problem is the presence of R8 and R9. I think in terms of the setting of gain opamp U2, right? What the presence of C4, C6 and C9?

No. The voltage gain of the voltage amplifier is 1+ (R12/R11) where R11 has a calibration trimpot to allow you to set the maximum output voltage to exactly 30.0V.
R8, R9 and C4 simply filter the 11.2V voltage reference. C6 and C9 compensate for phase shifts in the amplifier so it does not ring nor oscillate.

And the last question - why current limiter isn't linear? I mean when we change output voltage with the same load, the current limiter change value. Why? Maybe you can give me a example power supply we can set output current which doesn't depend of output voltage?

No. The voltage setting does not affect the current setting and the current setting does not affect the voltage setting EXCEPT when the load current exceeds the setting of the current pot P2 then the output voltage drops low enough that the load current is the same as the pot P2 setting. Then the LED lights to warn that the output voltage is being reduced by the current regulator.
Here is the fixed voltage amplifier schematic:

##### Share on other sites

• 3 years later...

Hello dear inventor, I need a voltage adjustment of 0-30V and 0-15 ampere, I have 2 transformer AC output 25.5v to 8 am now, when I assembled the circuit on IC IC34071 chip I did not get the circuit, please help how to solve the problem ?? ? !!!

##### Share on other sites

15A output current is much too high for this circuit designed for a maximum of 3A using two output transistors to share the heat. Your schematic shows only one 2N3055 output transistor so its maximum output current will be only 1.5A.

There are three MC34071 ICs. Which one do you have a problem?

You said, "I did not get the circuit". It is not English, what do you mean to say??

## Join the conversation

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

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

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