 ### Network # Opamp circuit clarifications

#### electronicsLearner77

Jul 2, 2015
306
I am trying to solve the opamp circuit as below (V2 - Vopamp)/1.5e3 + (3.3 - Vopamp)/22e3 = Vopamp / 2.2e3 + C1 * dVopamp/dt ->1
V2 - pulsed input
C1 - 14e-9 F

0.66*(V2 - Vopamp) + 0.045*(3.3 - Vopamp) = 0.45*Vopamp + 14e-6*dVopamp/dt ->2
0.66V2 - 1.155Vopamp + 0.1485 = (14e-6)dVopamp/dt
(14e-6)dVopamp/dt + 1.15 Vopamp - (0.66V2 + 0.1485) = 0 -> 3

dVopamp/dt + 82e3 Vopamp - e3 *(47V2 + 10) = 0 -> 4
dVopamp/dt + 82e3Vopamp - c=0
c = 47000V2 + 10000
The solution of the equation from wolfram is Vopamp(t) = c/82000 + k1 * e^{-82000t}

For a pulse input of V2 how does my output look like? It is basically a current shunt sensing circuit. I have doubts on the capacitor to be placed or not? Hence i am analyzing. How do i conclude?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,456
For a pulse input of V2 ...
Let's assume your pulse input is defined as
V2 = 0 V for t <= 0
V2 = X V for t > 0 where X V ist the amplitude of the pulsed voltage.

Start with the steady state for t <= 0. In the steady state you can ignore the capacitor (it will be charged to the steady state voltage). Therefore Vopamp will be given by the voltage sources (V1 = 3.3 V, V2 = 0 V) and the resistors R1, R2 and R3. You can easily calculate this voltage.
To find the factor k1 in your equation, set t = 0 and solve for k1 using Vopamp as calculated before for the steady state t <= 0.
Vopamp(t) = c/82000 + k1 * e^{-82000t}
You then have c, k1 and x (x = t, I presume) and can calculate the exponential rise of Vopamp for t > 0.
Compare your calculations with the simulation (as you already have the circuit in LTSPCE, obviously). They should match closely. Otherwise there is a fault in youit calculations (which, frankly, I have not checked in detail).

Feb 19, 2021
672
There is a problem attaching a large cap directly to an OpAmp input, depending
on OpAmp architecture. That is discharge of the capacitor when supplies are
removed, can cause input stage damage, such as leakage current in jfet input
opamps, and or damage in ESD protection diodes internal to OpAmp.

Regards, Dana.

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