Series compensating stabilizer with specific outcome

Thread Starter

Ester315

Joined Mar 22, 2021
3
Hello everyone, I try to do my university project, which I have to create a simulation of series compensating stabilizer in MULTISIM NI with following parameters: output voltage should be 6.3V and max output current 1.8A. I've already created a circut, calculated the dependency between diode voltage, resistances and output voltage, but it still has wrong values. Could anyone give me some hints? Why it's wrong?

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crutschow

Joined Mar 14, 2008
34,450
That's commonly called a "series linear voltage regulator".

As AH noted, if you want DC out, then you need to have a DC source, not AC, as well as DC power and ground to the op amp.

And a Zener voltage of 2.4V is a little marginal for the input voltage range of the 741 (worst-case of 3V above the negative supply rail, which here is ground).
Better to use a 4.7V Zener, which has better voltage regulation and a lower temperature coefficient.
 

Thread Starter

Ester315

Joined Mar 22, 2021
3
Hi, thank you for your reply. I've added some adjustments, as you mentioned and I got quite accurate results, but I'm wondering if there is any way to make output current a little bit more like 1.8A? Or is it physically impossible?
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AlbertHall

Joined Jun 4, 2014
12,346
With a 6.3V output, ohms law will tell you what value the load resistor should be to get 1.8A output current.
Also note that the maximum collector current for the 2N5657 is just 0.5A.
 

crutschow

Joined Mar 14, 2008
34,450
I got quite accurate results
Don't see how that's possible. (?)
You have the supply voltage on both terminals of the op amp so it is not being powered.
One terminal needs to be positive (goes to V+) and the other terminal is negative (goes to ground).
Look at the 741 data sheet.

Why are R1 and R2 should low value resistors. They waste a lot of current.
Values in the 1k to 10k range are more typical for that.

To get 1.8A you will need a Darlington or Sziklai pair capable of carrying at least 2A.
A single NPN won't work since that op amp can't provide enough current needed by the transistor base at that current.

You do realize that, from Ohm's law, you can't get 1.8A through a 3.9Ω load resistor with a 6.3V supply voltage?
 

AlbertHall

Joined Jun 4, 2014
12,346
Don't see how that's possible. (?)
You have the supply voltage on both terminals of the op amp so it is not being powered.
One terminal needs to be positive (goes to V+) and the other terminal is negative (goes to ground).
In the simulator it may work without power connections.
 

Thread Starter

Ester315

Joined Mar 22, 2021
3
Yeah, now I can see it, U = R * I -> R = U / I = 3.5. I've adjusted the input voltage to 9.92V and it's ok. Thank you. :)

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