For must of the circuits, I know how to analyze them but I am having difficulty with this one, and with the understanding of its purpose.


The switch changes state every 0.5 msec. VZ is 5.6V

The problem I am having with this circuit, which is where I think I am making my mistake, is that I can't decide if this is a negative feedback or positive feedback.

Because of my fail to analyze the circuit correctly, I can't understand what is it for. I think it supposed to be astable multivibrator but I can't confirm it.


The output of the circuit is the output of the opamp.


Assume a general ideal op-amp. I used the TL because it was the first op-amp I saw.

Edit 3: I am trying to draw the timing diagram and this is what I get and I can't understand why I am wrong here.

Vs = Vswitch. Vx= Voltage point between R1 and R2.

enter image description here

  • \$\begingroup\$ What is the original source of the circuit? Please provide a link. Where is the output? It looks like the op-amp and diode contribute nothing. \$\endgroup\$
    – Andy aka
    Apr 8, 2017 at 10:16
  • \$\begingroup\$ Image link is broken \$\endgroup\$
    – Macit
    Apr 8, 2017 at 10:23
  • \$\begingroup\$ added a link to the image in case you can't see the image . the output of the circuit is the same is the output of the opamp \$\endgroup\$
    – Maor
    Apr 8, 2017 at 10:46
  • \$\begingroup\$ Cap current is spurious Relaxation Oscillator due to +ve FB depending on Input waveform with slow + ramp out. What is input duty cycle? \$\endgroup\$ Apr 8, 2017 at 12:09
  • \$\begingroup\$ According to the datasheet This is a 5V1 zener not 5V6. Which is correct voltage or part number? \$\endgroup\$ Apr 8, 2017 at 12:58

4 Answers 4


This circuit produces a linear ramp output. The zener diode and R1 produce a constant voltage across R2 (since the op-amp output tracks the capacitor voltage as it increases). The output slope is \$+\frac{V_Z}{R_2 C_2}\$ (once the switch opens, before that it sits near 0V, depending on the op-amp and supplies).

The output ramp will deviate from linearity when either the op-amp output saturates or the zener is no longer able to maintain regulation. The op-amp should have bipolar supplies for this to work well, or use a 'single supply' op-amp that has an output that can get close to the negative rail.

For better results you can replace the zener with a shunt reference such as the LM4040.

Edit: Here, in the below simulation, you can see the action- the output voltage vs. time is very linear for the first millisecond or so. The switch opens at t=0+, allowing the current to flow into the capacitor.


simulate this circuit – Schematic created using CircuitLab

enter image description here

  • \$\begingroup\$ 1) shouldn't the diode work for both ways ( the 0.7 and the 5.6?) 2) for the calculation of the output you don't need to use this img2 formula? \$\endgroup\$
    – Maor
    Apr 8, 2017 at 11:02
  • \$\begingroup\$ The zener is connected through R1 to +15V so we know it is reverse biased. If you connected R1 to -15V the -0.7V would come into play- the equation above (ideally, anyway) would apply with Vz = -0.7V. That means the ramp would be negative and probably much slower. \$\endgroup\$ Apr 8, 2017 at 11:05
  • \$\begingroup\$ thanks for 1. 2) for the calculation of the output you don't need to use this img2 formula? \$\endgroup\$
    – Maor
    Apr 8, 2017 at 11:07
  • \$\begingroup\$ The exponential ramp formula would apply if the zener was missing (or not regulating). In this case the voltage across R2 (ideally) is constant so the current into the capacitor (or switch when closed) is just Vz/R2. \$\endgroup\$ Apr 8, 2017 at 11:11
  • \$\begingroup\$ sorry I am still missing something . as Vc increase isn't the current on R2 changes according to IR2 = (Vz-Vc)/R2? \$\endgroup\$
    – Maor
    Apr 8, 2017 at 11:31

I think this is a ramp generator. The op-amp is used as a voltage-follower.

When the switch is pushed, C2 is “instantly” discharged to 0V. The output of the op-amp is also 0V, and D2 limits the voltage of the midpoint between R1 and R2 to 5.6V.

When the switch is released, there’s 5.6V on R2 and C2 is charged with a 1mA current. Then C2’s voltage raises, the output of the op-amp follows it and D2 keeps the voltage on R2 at 5.6V.

All of this assumes that the op-amp’s output can be as low as the low power rail (I don’t think this is true for a TL081 op-amp) and that C2’s voltage will not reach (or exceed) 7.4V.

  • \$\begingroup\$ shouldn't the voltage on R2 increase as the voltage of the op-amp output increases to keep the 5.6v difference? \$\endgroup\$
    – Maor
    Apr 8, 2017 at 10:59

Here is the behavior of this circuit.

S1 is a switch 1 Ohm / 1 Meg, for this purpose, closed at start (Vc=0).

It has to be analyzed in two steps.

1- Until Zener is well reversed polarized (current in Dz), output is linear -> constant current source for capacitor.

2- When Zener "switch" off, regular charge to 15 V.

v(Vo) is behind v(Vc) (~ same). The formulas are yours ...

enter image description here


What you failed to do is define the source impedance of the generator.

It is not a 0 Ohm switch.

If it were a 0 Ohm switch , the zener does nothing

Now fix the question with a 2k source impedance you see the voltage follower only clamps down to the R ratio but with the Zener feedback it clamps down to Vout=0 obvious from positive feedback in the unity gain negative feedback voltage follower.

So the purpose is to make a slow attack, fast decay pulse from a source impedance transistor switch with a series impedance of 1k<= 2.5k

Thanks for the time wasted in your missing input. ROTFL

I proved this is true here

I added 2V zener on input to simulate CM input range invalid within 2V of gnd. I modified duty cycle to show a triangular pulse. If you pull Zener Vol rises. But if you reduce source impedance of switch , it wont rise.

enter image description here

I included 1K Zout due to open loop internal impedance of Op Amp. Which could be as low as 300 Ohms which affects fall time here from current limit.

  • 1
    \$\begingroup\$ BTW I learned how to do this in my R&D days in the 70's and now I have 40 yrs experience and can do it in my head, but I wish I had the Falstad simulator then I could have been home for dinner more often rather than breadboarding. \$\endgroup\$ Apr 8, 2017 at 13:54

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