# Feedback amplifier assignment

#### Rumination

Joined Mar 25, 2016
74

The figure shows a feedback amplifier comprising an op-amp and a number of resistors. The signal generator Vs has an output impedance Rs, and the circuit is unloaded i.e. no RL.

OP-AMP has the following data:
Gain from differential input to output A = 431400 [V / V]
Differential input impedance Rid = 6175 [K Ω]
Differential output impedance Ro = 416 [Ω]

1) Divide the circuit in an A (gain) circuit and a B (feedbacknetwork) circuit and determine the feedback amplifier gain Af.

2) Determine the circuit input impedance Ri_f with negative feedback.

3) Determine the circuit's output impedance Ro_f with negative feedback.

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The given values:
Rs = 131 ohm.
R1 = 87 Kohm.
R2 = 1 Kohm.
R3 = 17,4 Kohm.
R4 = 1,7 Kohm.

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#### DGElder

Joined Apr 3, 2016
351
And?

Are you expecting someone to do your homework for you? You need to work the problem yourself so you can learn something. Make a good effort and someone will help if you get stuck and have a specific question.

#### Rumination

Joined Mar 25, 2016
74
And?

Are you expecting someone to do your homework for you? You need to work the problem yourself so you can learn something. Make a good effort and someone will help if you get stuck and have a specific question.
NO, I don't want anybody to do my assignment. I was going to post my solution, but you were too quick!

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#### Rumination

Joined Mar 25, 2016
74
My solution ( I think, I have made some mistakes with the resistors - Parallel or in series)

B circuit:
R11 = R2 + (R3 II R4) = 1 KΩ + (17,4 KΩ II 1,7 KΩ) = 2,54 KΩ
R22 = R4 + R3 = 1,7 KΩ + 17,4 KΩ = 19,1 KΩ

A circuit:
RiA = Ri + R11 + Rs + R1 = 6175 KΩ + 2,54 KΩ + 0,131 KΩ + 0,087 KΩ = 6177 KΩ
RoA = Ro II R22 = 416 Ω II 19,1 KΩ = 407 Ω

Aa = (Ri / (Ri + R11)) * A * (R22 / (R22 + Ro))
= (6175 KΩ / (6175 KΩ + 2,54 KΩ)) * 431400 * (19,1 KΩ / (19,1 KΩ + 0,416 KΩ))
= 0,9996 * 431400 * 0,979
= 422171

β = (R2 / (R2 + R3)) = (1 KΩ / (1 KΩ + 17,4 KΩ)) = 0,0543

Af = (Aa / (1+ β * Aa)) = (422171 / (1+0,0543 * 422171)) = 18,41

Ri_f = RiA * (1+ β * Aa) = 6177 KΩ * (1+0,0543 * 422171) = 141,61 GΩ

Ro_f = (RoA / (1+ β * Aa)) = (407 Ω / (1+ 0,0543 * 422171)) = 0,0177 Ω

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#### MrAl

Joined Jun 17, 2014
7,891
Hello,

I got a different gain for Af, but a couple questions first...

The value of differential input resistance is 6175k. Is that taken to be from the inverting input to the non inverting input or from each input to ground?

The value of differential output resistance is 416 ohms. Is that in series with the output of the op amp or from output to ground or some other connection?

I took the input resistance to be from non inverting to inverting terminals.
I took the output resistance to be in series with the output of the (internal) op amp.
I got a voltage gain close to 18. I suppose we could sim this to check.

LATER:
Checked using a second method which involved calculating resistors in series and parallel, came out with the same result: slightly over 18 but i'll wait to post the exact solution. That is with my above assumptions about the input and output impedances.
Also note that the two main resistors are R3 and R2, and if they were the only feedback resistors the gain would be 17.4 which is close to 18 more or less.

Keep in mind that if you do it this way, you have to calculate the total resistance working from output to input, then from input to output, in order to get the proper total resistances for use in a voltage divider formula.
For example, to calculate the voltage at the inverting terminal due to the op amp very output voltage we have to do:
R51=R5||R1
R851=R51+R8
R2851=R851||R2
R32861=R2851+R3
R432851=R32581||R4

and finally, the output of the circuit is:
Vout*R432851/(R432851+Ro)
where Ro is the 416 ohm output resistance of the op amp.

So you see that there are seven resistors involved in the calculation of the voltage at the inverting terminal and that is just due to the output of the very internal op amp. We would then have to calculate forward in the same manner to get the voltage at the inverting input due to the input voltage itself. The sum of these two is the total voltage at the inverting input.
You can also see that to calculate the voltages at both the inverting and non inverting terminals requires four calculations when doing it this way.
An alternative is to use nodal analysis and using a voltage controlled voltage source for the op amp.

Since the circuit is different if we take the output resistance 416 to be from the op amp output to ground, so this has to be clearly specified. The gain however will still come out close to 18 though.

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#### The Electrician

Joined Oct 9, 2007
2,814
Rumination, in your circuit of post #1, you show the three resistors R2, R3 and R4 in a PI network topology, but in post #4 they have become a TEE network. Which one is correct?

If they are in a PI network topology, then β will be approximately R2/(R2+R3) rather than R4/(R4+R3).

Is the feedback amplifier gain specified as Vout/Vs, or is it Vout/Vin?

Also, why is the output impedance of the opamp specified as a differential impedance when the output of the opamp as shown in post #1 is single ended, not differential?

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#### Rumination

Joined Mar 25, 2016
74
Rumination, in your circuit of post #1, you show the three resistors R2, R3 and R4 in a PI network topology, but in post #4 they have become a TEE network. Which one is correct?

If they are in a PI network topology, then β will be approximately R2/(R2+R3) rather than R4/(R4+R3).
Yes you are right. It's a PI network topology as shows in post #1. My mistake.

Is the feedback amplifier gain specified as Vout/Vs, or is it Vout/Vin?
It's Vout/Vin

Also, why is the output impedance of the opamp specified as a differential impedance when the output of the opamp as shown in post #1 is single ended, not differential?
It's also a mistake. It's of course only a output impedance and NOT a differential.

#### The Electrician

Joined Oct 9, 2007
2,814
Yes you are right. It's a PI network topology as shows in post #1. My mistake.

It's Vout/Vin

It's also a mistake. It's of course only a output impedance and NOT a differential.
You should redo your calculations, leaving out Rs in the Af calculations, and with the PI topology. Then we can see if there are any more mistakes.

#### Rumination

Joined Mar 25, 2016
74
I'm just confused about if this is correct or not:

B circuit: (feedback network)
R11 = R2 + (R3 II R4)
R22 = R4 + R3

A circuit:
RiA = Ri + R11 + Rs + R1

#### Rumination

Joined Mar 25, 2016
74
You should redo your calculations, leaving out Rs in the Af calculations, and with the PI topology. Then we can see if there are any more mistakes.
I already did it in post #4. I edited the post. Buy why leave Rs out?

#### The Electrician

Joined Oct 9, 2007
2,814
I already did it in post #4. I edited the post. Buy why leave Rs out?
Because if Af is Vout/Vin, then Vin is taken to the right of Rs, and Rs doesn't have any effect.

#### The Electrician

Joined Oct 9, 2007
2,814
Because if Af is Vout/Vin, then Vin is taken to the right of Rs, and Rs doesn't have any effect.
Also note that in post #1, Ri_f is shown as taken to the right of Rs, so Rs shouldn't be involved in the calculation of Ri_f.

#### MrAl

Joined Jun 17, 2014
7,891
My solution ( I think, I have made some mistakes with the resistors - Parallel or in series)

B circuit:
R11 = R2 + (R3 II R4) = 1 KΩ + (17,4 KΩ II 1,7 KΩ) = 2,54 KΩ
R22 = R4 + R3 = 1,7 KΩ + 17,4 KΩ = 19,1 KΩ

A circuit:
RiA = Ri + R11 + Rs + R1 = 6175 KΩ + 2,54 KΩ + 0,131 KΩ + 0,087 KΩ = 6177 KΩ
RoA = Ro II R22 = 416 Ω II 19,1 KΩ = 407 Ω

Aa = (Ri / (Ri + R11)) * A * (R22 / (R22 + Ro))
= (6175 KΩ / (6175 KΩ + 2,54 KΩ)) * 431400 * (19,1 KΩ / (19,1 KΩ + 0,416 KΩ))
= 0,9996 * 431400 * 0,979
= 422171

β = (R2 / (R2 + R3)) = (1 KΩ / (1 KΩ + 17,4 KΩ)) = 0,0543

Af = (Aa / (1+ β * Aa)) = (422171 / (1+0,0543 * 422171)) = 18,41

Ri_f = RiA * (1+ β * Aa) = 6177 KΩ * (1+0,0543 * 422171) = 141,61 GΩ

Ro_f = (RoA / (1+ β * Aa)) = (407 Ω / (1+ 0,0543 * 422171)) = 0,0177 Ω
Hello again,

I see you redid your calculation of the gain
That's much better. I got 18.37 from input generator to output.
You can adjust for Vin by knowing the ratio of the two resistors at the input, and then can calculate input resistance too if you like.

#### The Electrician

Joined Oct 9, 2007
2,814
I'm just confused about if this is correct or not:

B circuit: (feedback network)
R11 = R2 + (R3 II R4)
R22 = R4 + R3
If you're using the PI topology (as shown in post #1), I would think it should be R11 = R2 || (R3+R4).

And R22 = R4 || (R2+R3)

A circuit:
RiA = Ri + R11 + Rs + R1
I would think it should be R1 || (Rid + R11)

Note that Ri_f can't be more than R1, because R1 is in parallel with whatever the input impedance of the amplifier is.

#### Rumination

Joined Mar 25, 2016
74
Now I have calculated everything again ( I checked my assignment again and found out that Rid is 6175 OHM and NOT Kohm). I still get Ri_f to be larger than R1.

B circuit:
R11 = R2 II (R3 + R4) = 1 KΩ II (17,4 KΩ + 1,7 KΩ) = 0,9502 KΩ
R22 = R4 II (R2 + R3) = 1,7 KΩ II (1 KΩ + 17,4 KΩ) = 1,5562 KΩ

A circuit:
RiA = R1 II (Rid + R11) = 87 KΩ II (6,175 KΩ + 0,9502 KΩ) = 6,586 KΩ
RoA = Ro II R22 = 416 Ω II 1,5562 KΩ = 328,25 Ω

Aa = (Rid / (Rid + R11)) * A * (R22 / (R22 + Ro))
= (6,175 KΩ / (6,175 KΩ + 0,9502KΩ)) * 431400 * (1,5562KΩ / (1,5562KΩ + 0,416 KΩ))
= 0,8666 * 431400 * 0,78907
= 294994,8

β = (R2 / (R2 + R3)) = (1 KΩ / (1 KΩ + 17,4 KΩ)) = 0,0543

Af = (Aa / (1+ β * Aa)) = (294994,8 / (1+0,0543 * 294994,8)) = 18,41

Ri_f = RiA * (1+ β * Aa) = 6,586 KΩ * (1+0,0543 * 294994,8) = 105,50 MΩ

Ro_f = (RoA / (1+ β * Aa)) = (328,25 Ω / (1+ 0,0543 * 294994,8)) = 0,0205 Ω

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#### Rumination

Joined Mar 25, 2016
74
I was thinking, that maybe we should include Rs, because in question 2 and 3 we have to find the input impedance WITH feedback, and output impedance WITH feedback. Or am I getting it wrong? Why would my teacher give the value of Rs, if I don't use it anywhere?

#### The Electrician

Joined Oct 9, 2007
2,814
I was thinking, that maybe we should include Rs, because in question 2 and 3 we have to find the input impedance WITH feedback, and output impedance WITH feedback. Or am I getting it wrong? Why would my teacher give the value of Rs, if I don't use it anywhere?
If the problem doesn't explicitly say that Af is to be Vout/Vin rather than Vout/Vs, I would solve it for both, a triviality. That way you are guaranteed to have the answer your professor wants, plus one extra one.

R1 is outside the feedback network, and you shouldn't include it until the end. Use these expressions:

RiA =Rid + R11

Ri_f = R1 || (RiA * (1+ β * Aa))

#### MrAl

Joined Jun 17, 2014
7,891
I was thinking, that maybe we should include Rs, because in question 2 and 3 we have to find the input impedance WITH feedback, and output impedance WITH feedback. Or am I getting it wrong? Why would my teacher give the value of Rs, if I don't use it anywhere?
Hi,

If you had an amplifier sitting around and decided to test it, you would connect a wave generator to the input. The wave generator would have some series output resistance Rs. You would turn up the generator to maybe 1v peak, and measure it at the input of the amplifier which is Vin. Knowing Vin is 1v peak and the output was 18v peak would tell you that the gain of the AMPLIFIER itself was 18. If you could look inside the generator box and saw a zero output impedance generator connected to a 100 ohm resistor, then you would know the actual output of the internal generator, which would not help with calculating the gain.

However, if you know the true internal generator voltage and the true internal generator impedance then you can calculate the input resistance of the AMPLIFIER itself because that information gives you the input current in addition to the input voltage Vin. Hence, if you know that extra information then you can easily calculate the other requirement which is the input resistance of the amplifier itself where the input is at the node labeled Vin.

In short, the input is at Vin, but knowing other external things can help calculate the input resistance. So you dont actually skip Rs except for the calculation of the gain.

If the internal generator voltage is Vs and the amplfier input is Vin, and the generator Rs is 100, then the input current is:
Iin=(Vs-Vin)/100

Basically since the input is Vin and the current is Iin, then the input resistance at that level is Vin/Iin, and now you have the input resistance too.

#### Rumination

Joined Mar 25, 2016
74
Is this correct now?

A circuit:
RiA = Rid + R11+ Rs = 6,175 KΩ + 0,9502 KΩ + 0,131 KΩ = 7,2562 KΩ
RoA = Ro II R22 = 416 Ω II 1,5562 KΩ = 328,25 Ω

Aa = (Rid / (Rid + R11)) * A * (R22 / (R22 + Ro))
= (6,175 KΩ / (6,175 KΩ + 0,9502KΩ)) * 431400 * (1,5562KΩ / (1,5562KΩ + 0,416 KΩ))
= 0,8666 * 431400 * 0,78907
= 294994,8

β = (R2 / (R2 + R3)) = (1 KΩ / (1 KΩ + 17,4 KΩ)) = 0,0543

Af = (Aa / (1+ β * Aa)) = (294994,8 / (1+0,0543 * 294994,8)) = 18,41

Ri_f = R1 II (RiA * (1+ β * Aa)) = 87 KΩ II (7,2562 KΩ * (1+0,0543 * 294994,8) = 86,93 KΩ

Ro_f = (RoA / (1+ β * Aa)) = (328,25 Ω / (1+ 0,0543 * 294994,8)) = 0,0205 Ω

#### The Electrician

Joined Oct 9, 2007
2,814
Is this correct now?

A circuit:
RiA = Rid + R11+ Rs = 6,175 KΩ + 0,9502 KΩ + 0,131 KΩ = 7,2562 KΩ
RoA = Ro II R22 = 416 Ω II 1,5562 KΩ = 328,25 Ω

Aa = (Rid / (Rid + R11)) * A * (R22 / (R22 + Ro))
= (6,175 KΩ / (6,175 KΩ + 0,9502KΩ)) * 431400 * (1,5562KΩ / (1,5562KΩ + 0,416 KΩ))
= 0,8666 * 431400 * 0,78907
= 294994,8

β = (R2 / (R2 + R3)) = (1 KΩ / (1 KΩ + 17,4 KΩ)) = 0,0543

Af = (Aa / (1+ β * Aa)) = (294994,8 / (1+0,0543 * 294994,8)) = 18,41

Ri_f = R1 II (RiA * (1+ β * Aa)) = 87 KΩ II (7,2562 KΩ * (1+0,0543 * 294994,8) = 86,93 KΩ

Ro_f = (RoA / (1+ β * Aa)) = (328,25 Ω / (1+ 0,0543 * 294994,8)) = 0,0205 Ω
Everything is looking good except for Ro_f. I get a much different answer with a full nodal analysis than what the classical feedback analysis, (RoA / (1+ β * Aa)), gives.

Let me ask MrAl what he gets for Ro_f.