Circuit Analysis of Opamp Ckt

[electronicsLearner77]

i am trying to analyze the below circuit

It is part of the actual and bigger schematic. I tried to simplify the circuit as below, i am not sure of the effect of capacitor C1 hence as first step i removed it and analyzed


Assuming ideal op amp configuration V+=V-, I+ = I-=0; And let V0 be the voltage at the output
V1 - I1*R2 - I1R1+ V0 = 0 --> eq1
V2 - I2*R3 - I2R7 - Vref = 0 --> eq2
V1 - I1*R2 = V2 - I2*R3 --> eq3
Simplifying eq3
V1 -V2 -I2 * R3 = I1*R2 => I1 = (V1 - V2 - I2*R3)/R2 --> eq4
Substitute eq4 in eq1
I1(R1+R2) = V1+V0
(V1-V2-I2*R3)(R1+R2)= R2*(V0+V1)
V1 - V2 - I2R3 = R2(V0+V1)/(R1+R2)
{V1-V2 - R2(V0+V1)/(R1+R2)}/R3 = I2 -->eq5
substitute eq5 in eq2
I2(R3+R7) = V2-Vref
{V1-V2-R2(V0+V1)/(R1+R2)}(R3+R7)/R3 = V2 - Vref
The final output comes as
V0 = V1(R1/R2) - {V2(R1+R2)/R2(2R3+R7)}/(R3+R7) + {Vref*R3*(R1+R2)}/{(R3+R7)R2} -->eq6
I am not sure it is a very big equation, i cannot directly understand the dependence of V0 on the input signals? And what is the effect of capacitance?

 

[Harald Kapp]

V1 - I1*R2 - I1R1+ V0 = 0 --> eq1

Why "+Vo"?
When I follow the loop I have "-Vo" in this position.

This is a classic difference amplifier as analyzed e.g. here. Set V_ref = 0 V to make your circut equal to the one analyzed on that website. Adding in V_ref <> 0 V is the your next exercise.

A few more notes:

• Simply removing R4 and R5 changes the gain of the circuit. If you want to remove C1 from the analysis to simplify matters, then R2 and R3 in the new equivalent circuit need to be changed to 2 kΩ.
• R4, R5 and C1 form a low pass filter to remove noise from the input signals. If you want to include C1 in your analysis, you need to go complex and take the frequency dependece of C1 into account.

 

[electronicsLearner77]

Why "+Vo"?

Yes, sorry my mistake

Adding in V_ref <> 0 V is the your next exercise.

I have done it, the equations are
V1 - I1(R1+R2) = Vo ->eq1
V2 - I2R1 - I2R2 = Vref ->eq2
V1 - I1R1 = V2 - I2R1 ->eq3
Solving for V0 the answer is
V0 = (V2-V1)(R2/R1) + Vref; ->eq4

A few more notes:

• Simply removing R4 and R5 changes the gain of the circuit. If you want to remove C1 from the analysis to simplify matters, then R2 and R3 in the new equivalent circuit need to be changed to 2 kΩ.

• I will put some effort to understand why the resistance increased when i remove C1

  [*]R4, R5 and C1 form a low pass filter to remove noise from the input signals. If you want to include C1 in your analysis, you need to go complex and take the frequency dependence of C1 into account.

I try to solve this and see if i can do it.

 

[AnalogKid]

R1-C1-R5 form a 1.5 MHz lowpass filter. Why? The only thing I can think of is to filter out interfering AM broadcast signals or some other medium-frequency interference in the area.

ak

 

[WHONOES]

R1-C1-R5 form a 1.5 MHz lowpass filter. Why? The only thing I can think of is to filter out interfering AM broadcast signals or some other medium-frequency interference in the area.

ak

It's probably there to maintain common mode performance with rising frequency.

 

[electronicsLearner77]

R4, R5 and C1 form a low pass filter to remove noise from the input signals. If you want to include C1 in your analysis, you need to go complex and take the frequency dependece of C1 into account.

It does seem to be a complex expression, is the approach in the right direction?

 

[Harald Kapp]

Your equations work for DC only as you omit C2.
But for DC I1=I3 (no DC current through C2) and the equations can be simplified: replace R2+R6 by R26 = 2 × R and R7+R3 by R73 = 2 × R.
For AC you will have to take into account the complex impedance of C2. This will make the analysis considerably more complex, too.

 

[electronicsLearner77]

Thank you for the clarifications, but few clarifications before i attempt the problem, considering the impedance. Now i understand why i need to do the AC analysis since the input is not simple DC current they are sinusoidal phase current inputs. Few questions
1. If i do the AC analysis, do i still need to make the DC analysis or not required to find the gain etc?
2. If i referred the example differentiator circuit https://www.electronics-tutorials.ws/opamp/opamp_7.html
the input is still AC like pulse etc, but he has not considered the impedance of the capacitor. The equation is

Why did he not considered the impedance? I reviewed some of the active filters in that section they have considered, the impedance of the capacitor to find the pass band and other parameters of the filter. So, by seeing the circuit and analyzing, i need to see to include impedance or not in analysis. Am I correct? Please help.

 

[electronicsLearner77]

Ok, i included the equations which includes the impedance, can you please confirm if it is ok to solve the equations?

 

[Harald Kapp]

If i do the AC analysis, do i still need to make the DC analysis or not required to find the gain etc?

Gain will be frequency dependent.

What is the purpose of your analysis?
- do you want to make a complete analysis of the frequency behavior of the circuit?
- or do you want to analyze the gain within the low frequency range where the filter has little or almost no effect (see post #4, corner frequency is in the MHz range)? In this case you could simply ignore C

Your equations are set up o.k. (green).
You make assumptions that you should always state when solving such a task (red):
- label all variables used . Here you leave it to others to know what VO is. With respect to your equations it is the output of the op-amp, but with respect to the overall transfer characteristic you should move VO to the right and include the output filter in the analysis, too. Since the output of the op-amp has a low impedance, you can consider the output filter separately to simplify the math.
- The differential input voltage of the op-amp is assumed as 0 V. This is the usual proceeding in the analysis of such a circuit and therefore o.k. Just not explicitly stated. Note, however, that this assumption is (more or less) true only within a certain range of input voltages and output loads where the op-amp can achieve this situation by driving a high enough current throughthe feedback resistor ZF. If the input voltages rise above the range this circuit is designed for or the output load tries to draw more current than the op-amp can deliver, this assumption is no longer valid. But this is outside the nominal operating range of the circuit and you don't have to consider this case for your analysis. I just wanted to make you aware of this situation.