 ### Network # Opamp Filter circuit

Feb 1, 2023
161

#### crutschow

May 7, 2021
789
why there is no gain?
It does have gain (of 1).
Because at that simulated frequency, C1 has a very low impedance which basically shorts out R1, so the op amp is acting like a non-inverting follower amp with a gain of 1.
As the frequency is lowered to the corner transition frequency of C1 and R1, the gain will increase to a maximum value of 5.

What frequency are you trying to filter?
Is that just a random value for C1?

• bertus

#### bertus

Moderator
Nov 8, 2019
3,103
Hello,

Looking at the V(vopamp) the signal seems to rise a bit.
When you would have a longer timescale, it will show more if the rising.
The rising will likely be faster when you lower the value of C1.

Bertus

#### electronicsLearner77

Jul 2, 2015
306
It does have gain (of 1).
Because at that simulated frequency, C1 has a very low impedance which basically shorts out R1, so the op amp is acting like a non-inverting follower amp with a gain of 1.
As the frequency is lowered to the corner transition frequency of C1 and R1, the gain will increase to a maximum value of 5.

What frequency are you trying to filter?
Is that just a random value for C1?
I just used a random value for C1.

#### crutschow

May 7, 2021
789
I just used a random value for C1.
Then random results should be no surprise.

• davenn and AnalogKid

#### electronicsLearner77

Jul 2, 2015
306
It does have gain (of 1).
Because at that simulated frequency, C1 has a very low impedance which basically shorts out R1, so the op amp is acting like a non-inverting follower amp with a gain of 1.
I am really worried of my understanding with the explanation, since i considered the pulse input can impedance come into analysis for the capacitor, it is done only for AC signals?

#### crutschow

May 7, 2021
789
since i considered the pulse input can impedance come into analysis for the capacitor, it is done only for AC signals?
The pulse is an AC signal consisting of many Fourier frequencies.
What exactly do you expect to see (or want to see) with the pulse input, as random component selection obviously didn't work?

Last edited:

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,461
The pulse frequency is 1 kHz.
The 10 µF capacitor has an impedance of ~ 16 Ohm. Let's call this Z1 and let's ignore the phase shift incurred by the capacitor.
Therefore R1 is negligible and the gain of the opamp is $gain = \frac{R2 + Z1}{R2}= \frac{10 kOhm + 16 Ohm}{10 kOhm}= 1.0016$ which is as close to 1 as you can get I just used a random value for C1.
A simulator is not a tool to design a circuit. It is a tool to verify that your design is correct.
Component selection is part of the design, so it is your task, not to be left to the simulator.

• electronicsLearner77

#### electronicsLearner77

Jul 2, 2015
306
The 10 µF capacitor has an impedance of ~ 16 Ohm. Let's call this Z1 and let's ignore the phase shift incurred by the capacitor.
But for step or pulse inputs the analysis will be different compared to the sinusoidal signals, the impedance is never used in my understanding, and a pulse input has several frequencies how do we consider that?

Feb 19, 2021
673
The easiest way is to use LaPlace transforms, compute T(s) and stimulate it
with a step function, and if needed reciprocal transform back to time domain
to look at step response. Impedances of elements, like L and C, are used in
the transforms.

Just google "op amp step response laplace", lots of video and solutions done.

Also google this "op amp step response laplace analog devices"

Regards, Dana.

Last edited:
• electronicsLearner77

#### crutschow

May 7, 2021
789
pulse input has several frequencies how do we consider that?
Besides using Fourier analysis, which can be a little complicated, you can use piece-wise linear transient analysis.
If the pulse rise-time is fast, then the capacitor will have close to zero impedance and you can replace it with a short for the pulse rise (giving a gain of 1 in your circuit).
Then for the flat top part of the pulse, the capacitor will charge to a new voltage equal to the output without the capacitor when the capacitor is fully charged (gain of 5 in your circuit), with an RC time-constant as determined by the output capacitor and resistor values.
So the output will look like a pulse with an initial value equal to the input pulse charging towards a final value equal to 5 times the input pulse.

You can see this in your simulation, if you use a single voltage step instead of pulses.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,461
But for step or pulse inputs the analysis will be different compared to the sinusoidal signals,
Which makes it even worse.
Of course you can do a Laplace analysis or analyze the circuit using differential equations. This way you will get the correct frequency response. For every frequency.
The simplified consideration in my post #9 does use the impedance for the capacitor for a sinusoidal signal, right. Your pulse can be considered as the addition of many different sinusoidal signals, also right, which consist of a fundamental frequency of 1 kHz and other frequencies much higher. I used the 1 kHz fundamental to calculate the impedance of the capacitor. For the other components of the pulse, their frequency being even higher, the impedance is consequently even lower and gain is even more near 1.

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