1

I am currently working to make a 30V 5A linear power supply and came up with this circuit so far. The circuit is supposed to give twice the reference voltage Vref.

enter image description here

This is the output voltage that I am getting which is the expected output.

enter image description here

But to be able to achieve 0V(or very close to 0V) at the output, I decided to take the output voltage as reference for the opamp supply voltage and the reference voltage(Vref). This is the circuit that I came up with to achieve this.

enter image description here

This is the output voltage that I am getting.

enter image description here

Changing the reference voltage(Vref) has a negligible effect on the output in the second case. In first circuit everything works as expected.

What mistake am I making here?

tinkerer
  • 170
  • 11

3 Answers3

4

How can you drive the ballast transistor with a voltage that can not be higher than 12 V (output of op-amp -> base of Q2) (first picture)?

EDIT : As @VictorTito pointed out, it is (arguably) better if the op amp power supply is "fixed" and not floating ... Ok, I haven't done a full stability study in this case. This may have "advantages"?

enter image description here

Seems that the emitter of Q2 would return to the ground, with some resistor between the Q3 base and Q2 collector. More effective. example of behavior, but not very linear.

enter image description here

Another example... more interesting and "linear".

enter image description here

Antonio51
  • 11,004
  • 1
  • 7
  • 20
  • 2
    My poor eyes thank you for using a white theme for your images. – Elliot Alderson Dec 05 '21 at 20:35
  • What "color" would you like? – Antonio51 Dec 06 '21 at 12:35
  • 1
    Antonio, circuits and plots with white background are much easier to visualize than with dark gray or black. I'm pretty sure there was no irony in @ElliotAlderson 's comment, but a sincere *"thank you"*. In fact, +1. ;) – devnull Dec 06 '21 at 13:19
  • 1
    No problem. I really understand. This remark was actually really very interesting. Maybe another color would be perhaps more appropriate and also maybe more user-friendly. :-) – Antonio51 Dec 06 '21 at 13:41
  • 1
    Yes, I was seriously grateful. The usual thin dark green lines on a black background make my head hurt. – Elliot Alderson Dec 06 '21 at 16:07
  • Sometimes also, some "pictures" are really unreadable ... It is perhaps a "recall" to our "old" times ... when we used a "blackboard" ... which is now "whiteboard" ... :-) – Antonio51 Dec 06 '21 at 16:15
2

I’m confused by your second circuit, and here is why:

  1. The reference voltage of one volt has its negative connected to the output Vout. This will make your reference Vout + 1 Volt. The reference voltage will increase for each interaction of Vout + 1 volt until it reaches the maximum input voltage of 32 V minus any drop in the transistors Vbe and diode Vf. That is probably why you are getting Vout = 30 V.

  2. The op-amp power supply V+ is 12 V plus Vout and V- is 12 V minus Vout. This will change the op-amp power supply for each interaction of Vout until the op-amp will have no effect in controlling Vout, wich will reach its maximum of 32 V minus any drop in the transistor Vbe and diode Vf.

Move the op-amp power supply back to 12 V and ground, not 12 V and Vout.

To reduce the 1 Volt reference voltage use a voltage divider.

Null
  • 7,448
  • 17
  • 36
  • 48
VictorTito
  • 400
  • 1
  • 4
1

Using KVL for the Virtual Null, OA input, it appears as your Vref needs to be reversed.

Let Vo/2 = Vo-Vref not Vo+Vref

Tony Stewart EE75
  • 1
  • 3
  • 54
  • 182