# Quiescent voltage of op-amps

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The Electrician
Gold Member
peasngravy, do you want any more help on this, or will post #22 be your last post here?

peasngravy, do you want any more help on this, or will post #22 be your last post here?
Hi - I am based in Europe so it’s just early in the morning here now. I will spend some time today digesting some of the info you guys have given and have another go at an answer when my daughter allows me to spend the time :). Thanks to everyone for your time so far.

I have not seen anything like the stuff that has been posted so far in my notes so it seems like a bit of a jump to just throw this stuff in without really teaching us about it - this is the first op-amp circuit in the course to use capacitors so far.

This is a distance learning course so it’s difficult to understand some of it without speaking to someone.

The reason designers like to use op-amp is their (nearly) ideal characteristics. These are infinite input impedance, infinite gain, zero output impedance (for voltage amplifiers). This allows you to solve feedback networks by splitting them easily into parts that don't interact.

So, if you aren't familiar with this yet. First study the simple op-amp configurations like on this site (there are about a million places on the web for this stuff, I chose this one at random). You will want to first understand and memorize the simplest negative feedback results (inverting and non-inverting circuits). These sites will all initially show you the answers for simple resistors, but in filter circuits like the one you posted you can combine reactive elements into a complex, frequency dependent, impedance. So, for example, R9||C7 can be treated as an impedance Z=R9/(1+jω⋅R9⋅C7). This example is an impedance whose magnitude |Z|=R9/√(1-ω2⋅R92⋅C72) is Z≅R9 at low frequencies, but around the frequency fc=1/(2π⋅R9⋅C7) will start to decrease, at high frequencies it will be dominated by the capacitor C7 with Z≅1/(jω⋅C7).

This approach will allow you to split up your analysis into pieces like I've shown below. Except for their inclusion in the basic formulas, you don't have to worry about interactions between the groups.

View attachment 272841
this is very helpful to see it broken down this way, I didn’t realise they could almost be looked at as independent circuits

The Electrician
Gold Member
Suppose C3 and C7 were removed, and C9, C2, C5 and C8 were replaced with a short (each replaced with a wire in other words), could you calculate the voltage gain Vout/Vin? Then the circuit would be all resistors, no capacitors; opamps still there, of course.

The Electrician
Gold Member
The problem statement says:

"Estimate ... the overall voltage gain (in decibels) of the circuit at the
frequency of operation. State any assumptions made. "

Notice that it says "estimate". You made some calculations and found that the operating frequency is probably in the neighborhood of a few hertz. Assume that it's one hertz for easy calculations. State that you have made this assumption and explain that it's based on your calculations of the high and low pass frequencies.

Now calculate the reactance of each capacitor at one hertz; you get a value in ohms. Replace each capacitor with a resistor whose value is equal to the reactance in ohms. Now you have a circuit without any capacitors, just resistors. Calculate the gain Vout/Vin for this modified circuit. This can be your "estimate", and it's not completely unreasonable.

So I estimated the voltage at the non-inverting terminal as 3.707V

(R1*R3)/(R1+R3) = 7674
5V*(7674/(7674+R12)) = 3.707V

Ideal op-amp so this voltage applies across both input terminal

Gain of non-inverting op amp:
Av = 1+ (Rf/Rin)
Rin being R4 here, 10k
Av = 1+ (2.2meg/10k) = 221
Av = Vout/Vin
Vout = Vin*Av = 3.307*221
Vout = 819V

Seems rather high but hopefully I am the right track here?

The Electrician
Gold Member
So I estimated the voltage at the non-inverting terminal as 3.707V

(R1*R3)/(R1+R3) = 7674
5V*(7674/(7674+R12)) = 3.707V
The 5V is a DC voltage and it's blocked by capacitors.

It looks like you're calculating the quiescent voltage on the non-inverting terminal of the first opamp. You already did that correctly in post #5. Now you're doing it incorrectly.

Ideal op-amp so this voltage applies across both input terminal

Gain of non-inverting op amp:
Av = 1+ (Rf/Rin)
Rin being R4 here, 10k
Av = 1+ (2.2meg/10k) = 221
Av = Vout/Vin
Vout = Vin*Av = 3.307*221
Vout = 819V

Seems rather high but hopefully I am the right track here?
Now you're calculating the gain of the opamp for the DC on the "+" terminal. The capacitor C2 prevents the opamp from having any gain (other than unity) for DC because for DC the 10k resistor R4 is not Rin. Rin is the series combination of R4 and C2 and the resistance of C2 for DC is infinite. The resistor R5 connects the "-" terminal to the output for DC so the gain of the opamp for DC is just 1. The voltage at the "-" terminal is the same as the opamp output voltage.

The problem statement wants you to estimate the overall voltage gain. They are referring to the small signal gain: https://en.wikipedia.org/wiki/Small-signal_model

The AC signal at Vin is at a frequency of around 1 or 2 hertz, but it is a small signal, probably only millivolts.

The incoming small AC signal passes through C9 and R3. Those two components appear to be high pass but their output is loaded by R1 and R12 which changes the frequency determined by the C9 R3 pair.

Also, the small signal which has passed through the C9 R3 pair is decreased in amplitude because of the loading from R1 and R12. Note that for small signal purposes, the 5 volt line is a ground for small signals so that R1 and R12 are effectively in parallel as a load on the C9 R3 pair.

For small signals the series combination of R4 and C2 serves as Rin in the formula for non-inverting gain, but you must calculate the impedance magnitude |Z| of the R4 C2 series connected pair to use in the formula.

The formula also needs a value for Rf. This value will be the impedance magnitude |Z| of the parallel combination of C3 and R5.

peasngravy
The 5V is a DC voltage and it's blocked by capacitors.

It looks like you're calculating the quiescent voltage on the non-inverting terminal of the first opamp. You already did that correctly in post #5. Now you're doing it incorrectly.

Now you're calculating the gain of the opamp for the DC on the "+" terminal. The capacitor C2 prevents the opamp from having any gain (other than unity) for DC because for DC the 10k resistor R4 is not Rin. Rin is the series combination of R4 and C2 and the resistance of C2 for DC is infinite. The resistor R5 connects the "-" terminal to the output for DC so the gain of the opamp for DC is just 1. The voltage at the "-" terminal is the same as the opamp output voltage.

The problem statement wants you to estimate the overall voltage gain. They are referring to the small signal gain: https://en.wikipedia.org/wiki/Small-signal_model

The AC signal at Vin is at a frequency of around 1 or 2 hertz, but it is a small signal, probably only millivolts.

The incoming small AC signal passes through C9 and R3. Those two components appear to be high pass but their output is loaded by R1 and R12 which changes the frequency determined by the C9 R3 pair.

Also, the small signal which has passed through the C9 R3 pair is decreased in amplitude because of the loading from R1 and R12. Note that for small signal purposes, the 5 volt line is a ground for small signals so that R1 and R12 are effectively in parallel as a load on the C9 R3 pair.

For small signals the series combination of R4 and C2 serves as Rin in the formula for non-inverting gain, but you must calculate the impedance magnitude |Z| of the R4 C2 series connected pair to use in the formula.

The formula also needs a value for Rf. This value will be the impedance magnitude |Z| of the parallel combination of C3 and R5.
this is really helpful. Thank you. As soon as I get a chance I’ll have another shot at this

The Electrician
Gold Member
Hi, peasngravy, any luck on your latest efforts?

Hi, peasngravy, any luck on your latest efforts?
Hi - I actually tried building this circuit in pspice to help me understand it a bit better yesterday but I didn't have much luck.

Most of my learning material was in my workplace so I didn't have access to a lot of it over the weekend so I am looking through this now.

I was confused on what the AC input signal (vin) is though, the one you said will be in mV - i am not sure how to calculate that

The Electrician
Gold Member
You don't calculate Vin--it's a given. You assume a sine wave of voltage at a frequency of 1 Hz is applied to Vin. I like to assume that the voltage of the input is 1 volt AC, but any voltage is OK. Then you calculate the way this voltage signal is amplified as it passes through the circuit. The voltage at Vout is your final result. The voltage gain is then given by Av = Vout/Vin.

In the real world the voltage at Vin due to the sensors will probably be on the order of millivolts, but for your problem since it's just math, you can assume any Vin. Since the circuit is being treated as if it's perfectly linear with unlimited output voltage capability from the opamps for the purpose of doing the math, any Vin will do.

DaveE
Gold Member
Concerning C4 and C6, it is NEVER a good idea to hang a capacitor directly on the output of an op-amp. I will defend your original statement about them being put there by an idiot.
-
Now, that being said, many years ago I was likely guilty of doing that exact same thing in some homebrew project before I knew better. If something like this were to ever help prevent an oscillation it is likely an example of 2 wrongs making a right. And that in and of itself is still wrong.
Yes, however, in this case it doesn't matter for stability. That pole is around 1MHz and these circuits cross-over at 1-2KHz. Still, I have no idea what they think it's supposed to do.

The Electrician
Gold Member
Yes, however, in this case it doesn't matter for stability. That pole is around 1MHz and these circuits cross-over at 1-2KHz. Still, I have no idea what they think it's supposed to do.
Based on the frequency response of the circuit, it appears that the operating frequency is about 1-2 Hz, not 1-2 kHz.

You don't calculate Vin--it's a given. You assume a sine wave of voltage at a frequency of 1 Hz is applied to Vin. I like to assume that the voltage of the input is 1 volt AC, but any voltage is OK. Then you calculate the way this voltage signal is amplified as it passes through the circuit. The voltage at Vout is your final result. The voltage gain is then given by Av = Vout/Vin.

In the real world the voltage at Vin due to the sensors will probably be on the order of millivolts, but for your problem since it's just math, you can assume any Vin. Since the circuit is being treated as if it's perfectly linear with unlimited output voltage capability from the opamps for the purpose of doing the math, any Vin will do.
Well I wish I asked you that much sooner as I spent a long time on google trying to work it out :D

The Electrician
Gold Member
In pspice you simply connect an AC voltage source to Vin. Set the parameters of the source for your desired frequency and voltage amplitude.

Given that it's an ideal op-amp and the gain can't be anything other than unity, then all inputs and outputs of the op-amps are 2vdc?

And I just need to perform a small signal ac analysis to work out the overall voltage gain?

berkeman
The Electrician
Gold Member
Given that it's an ideal op-amp and the gain can't be anything other than unity, then all inputs and outputs of the op-amps are 2vdc?

And I just need to perform a small signal ac analysis to work out the overall voltage gain?
Yes and Yes.

peasngravy and berkeman
The Electrician
Gold Member
Have you tried to derive the non-inverting gain of just the first opamp stage using equivalent impedance of the C3 R5 pair as Rf, and the C2 R4 pair equivalent as Rin? Ignore the effect of C9, R2, R1 and R12; they are just a voltage divider in front of the "+" input of the first opamp.

Have you tried to derive the non-inverting gain of just the first opamp stage using equivalent impedance of the C3 R5 pair as Rf, and the C2 R4 pair equivalent as Rin? Ignore the effect of C9, R2, R1 and R12; they are just a voltage divider in front of the "+" input of the first opamp.
I did calculate those pairs - using 1hz frequency, I got Z values of 16.06Meg and 12.34K for those which gave me a gain of 1301 (1+(rf/rin))

I checked this on circuitlab (with an AC source at 1V and 1 hz) and found the value did not match up so I am a bit lost now. This gave me a gain of around 135. These are the values

The Electrician
Gold Member
I did calculate those pairs - using 1hz frequency, I got Z values of 16.06Meg and 12.34K for those which gave me a gain of 1301 (1+(rf/rin))
If you'll show your work, I can see where you went wrong and suggest a fix.

Sure - so using 1hz as the frequency, ω = 2πf = 6.283
C2 = 22uF
R4 =10k
|Z| = √ R² +(1/(ωC²))
|Z| = √ 10k² +(1/(6.283*0.000022²))
|Z| = 12342 for Rin

I just noticed where I went wrong actually, and I used this equation for both pairs, but one is in series and the other is parallel.

So the other pair using this formula
|Z|=1√(1R)2+(ωC)2

Gives me 2.179Meg for Rf

Av = 1+(Rf/Rin)
= 1+(2179000/12342)
=176

Which is a bit closer but still slightly out

The Electrician
Gold Member
You need to carry more digits in your arithmetic.

When I carry out this division: (2179000/12342) I get 176.55161238
After adding 1, I get 177.55161238

But carrying more digits in all the calculations, I get the desired quotient as: 2179278.18/12342.4196 = 176.568148
The after adding 1 I get: 177.568148

What do you think the correct result is, and where did you get the value you believe is correct?

You will get a tiny error in your calculations because you aren't using complex arithmetic. When you add 1 to the quotient of the two impedances, the impedances should be kept in complex form for the exactly correct final result which is: 177.4516107

I don't think spice will give you that many digits, but that is the exact correct result.

Your problem statement asks you for an estimate. I would say you have a very good estimate, but do carry a few more digits so your result is 177.55, not 176.

Next, what is the ratio of the voltage divider made of C9, R3, R1 and R12.

Then, show your work for the second opamp stage.

peasngravy
You need to carry more digits in your arithmetic.

When I carry out this division: (2179000/12342) I get 176.55161238
After adding 1, I get 177.55161238

But carrying more digits in all the calculations, I get the desired quotient as: 2179278.18/12342.4196 = 176.568148
The after adding 1 I get: 177.568148

What do you think the correct result is, and where did you get the value you believe is correct?

You will get a tiny error in your calculations because you aren't using complex arithmetic. When you add 1 to the quotient of the two impedances, the impedances should be kept in complex form for the exactly correct final result which is: 177.4516107

I don't think spice will give you that many digits, but that is the exact correct result.

Your problem statement asks you for an estimate. I would say you have a very good estimate, but do carry a few more digits so your result is 177.55, not 176.

Next, what is the ratio of the voltage divider made of C9, R3, R1 and R12.

Then, show your work for the second opamp stage.
I think the number around 177 is mathematically correct so it has to be the correct value - the pspice was simply to use a as a bit of a guide and I was hoping they'd be a bit closer, that's all.

For the voltage divider you mentioned, can I use the Z value of C9 and R3 in there?

The Electrician
Gold Member
The signal coming in at Vin passes through the series combination of C9 and R3. Calculate the |Z| of those two and that forms a voltage divider followed by a resistance to ground equal to the parallel combination of R1 and R12, with the output of the divider applied to the "+" input of the opamp. You will multiply the voltage divider ratio (which will be less than unity) times the gain of the first opamp. That will give the gain from Vin to the first opamp output.

Then you will calculate the gain of the second opamp and multiply that factor in to get the overall gain.

The signal coming in at Vin passes through the series combination of C9 and R3. Calculate the |Z| of those two and that forms a voltage divider followed by a resistance to ground equal to the parallel combination of R1 and R12, with the output of the divider applied to the "+" input of the opamp. You will multiply the voltage divider ratio (which will be less than unity) times the gain of the first opamp. That will give the gain from Vin to the first opamp output.

Then you will calculate the gain of the second opamp and multiply that factor in to get the overall gain.
So R1 and R12 have a ratio of 0.4 - I am not quite sure how to factor in the impedance of c9 and r3.

The ratio of 0.4v is quite close to what my pspice model suggests as the input (430mv)

After that - I think I have to calculate the series capacitance of c5+c8 and then use that figure to calculate the impedance of the rc series along with r6 which should be my Rin for the second op amp?