# Op-amp circuit analysis without input signal

In the following circuit i need to find the OUTPUT and the current that flows in the zener diode D1, which works at the "zener" breakdown region with $$Vzener=5,6V$$ The opamp is considered to be ideal.

I cannot understand how this circuit works since there is not an input signal to be applied to the inverting or to the non-inverting terminal of the op-amp and we only have the power supply. I have never seen such a case before and i am a bit confused. Which is the right way to approach it? Any help is appreciated!Thanks in advance! You can assume that at -Vin (pin 3) you have 5.6 volts. If this wasn't true then, because of the massive open-loop op-amp gain, the output would be against one of the power rails.

If you have 5.6 volts across the 220 kohm then it takes 25.45 uA and that also flows through the 100 k so, the output voltage MUST be 8.145 volts.

This means that (8.145 - 5.6)V/2200 amps flows through the 2k2 = 1.157 mA.

• Nice and easy explanation!Upvoted!Just 2 more questions. I didn't understand exactly what happens if our assumption is not true(If D1 doesn't conduct).Which is the input of the op-amp then that leads to a saturation? And the second one is about the use of this circuit. What's the application of this circuit? Where can be used?
– MJ13
May 1 '18 at 10:40
• If D1 isn't conducting then the op-amp output voltage would equal +Vin and -Vin would be lower (because of the potential divider) hence the op-amp output would grow and grow rapidly until it hit the rail or D1 finally started conducting. May 1 '18 at 12:21
• And which operation does this circuit serve?I mean where can we use this certain circuit?
– MJ13
May 1 '18 at 12:28
• Dunno, you picked it. What did the site say about it. I can turn requirements into a circuit and I can explain a few circuits but I can’t reverse engineer a requirement from such a simple circuit. May 1 '18 at 12:44
• @MJ13 I suppose if the zener diode was used to provide a 5.6 volt power rail to a circuit that also needed another but proportionally higher power rail, then the op-amp output could be used in this respect - the 220k and 100k set the higher rail. In your circuit it would be 8.145 volts but there's no reason (within power supply limits for the op-amp) why this couldn't be 10 volts or 11.2 volts (double). Any diode drift would also drift the higher supply and this might be useful in some circuits. May 1 '18 at 16:36

Think about the steady state effects this circuit has, what will go on in the long term.

There is only one voltage you can guarantee in this circuit: the Zener diode.

V+ = Vzenar

The next thing that we know is that the OpAmp will keep increasing or decreasing its output voltage until it hits its rail or V- = V+

so the next thought process is How do you get V- = V+ (if it is at all possible)

Because we know that V+ is Vzenar

We need the voltage across the 220k resistor on the negative size to be equal to Vzener

We know that:

• Voutput = VR100k + VR220k
• VR220k = 5.6v
• Current for both Resistors are the same
• Voutput = I(R100 + R220k)

So Current for both resistors is:

5.6/220k = 25.45uA

Voutput = 25.45uA* (100k + 220k) = 8.14volts

• Any idea about where can we use the certain circuit? I mean in which applications?
– MJ13
May 1 '18 at 16:32

If there is a solution within the linear range of the opamp the following must apply:

Vn=Vp (ideal opamp assumed).

Vn=Vout*220/(100+220) and Vp=5.6 volts

Hence: 5.6=Vout*0.6875 and

Vout=5.6/0.6875=8.14545 volts.

• Do you know in which applications can this certain circuit be used?
– MJ13
May 1 '18 at 16:31
• likely when you want to maintain a specific voltage but don't have a voltage regulator, want and OK efficiency and you're stuck with an ill-fitting zener diode. Other than that it seems like a pretty theoretical question. May 2 '18 at 21:31
• yes - using this circuit you can create each desired DC voltage source (at a low-resistive output) by selecting suitable resistor values.
– LvW
May 3 '18 at 9:46