Understanding High Entropy Expressions in Circuit Analysis

[peasngravy]

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]

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]

Hi, peasngravy, any luck on your latest efforts?

 

[peasngravy]

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]

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]

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]

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.

 

[peasngravy]

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]

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

 

[peasngravy]

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?

 

[The Electrician]

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.

 

[The Electrician]

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.

 

[peasngravy]

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]

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.

 

[peasngravy]

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]

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.

Thank you for that answer

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

 

[The Electrician]

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 into get the overall gain.

 

[peasngravy]

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 into 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?

 

[DaveE]

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?

The key problem for solving the 2nd amp is dealing with the C5/C8 network at the input. Because the virtual ground of the 2nd amp will keep the AC (signal) voltage = 0 at the input (BTW, don't confuse this with the non-zero DC voltage there), this problem amounts to given a signal voltage applied at the 1st amp output, what will the current flow be into the 2nd amp. That current will have to flow through the feedback components to create the output voltage. Something like this picture:

edit: note that you will end up with an equivalent impedance Zeq = Vi(jω)/Io(jω), although I phrased it as an admittance (i.e. find Io given Vi); whatever, there sort of the same, in a reciprocal fashion, LOL. Then you can pretend it is a two terminal complex impedance, if you like to think of things that way. BTW this Zeq simplification only works for Vi and Io, it won't be correct if you want to know other things in the network, like the current through R13, etc.

 

[The Electrician]

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?

The ratio of R1 and R12 is not relevant. Forget about the quiescient DC voltages in the circuit. Have a look here: https://en.wikipedia.org/wiki/Voltage_divider

In Figure 1 on that page let Z1 be the impedance magnitude of the series combination of C9 and R3, and let Z2 have the value of R1 and R12 in parallel. Then if 1 volt is applied to Vin in Figure 1, the output voltage will be applied to the "+" input of the first opamp. Multiply that value times the gain of the opamp and you have the gain from Vin of your circuit to the output of the first opamp. The voltage at the output of the first opamp is applied to the left side of C5. Then you calculate the gain from that point to the output of the second opamp.

To calculate the gain of the second opamp, follow the advice of DaveE and calculate the current in R6. Multiply that current times the impedance magnitude of the C7 R9 pair (remember that they're in parallel) and you have the voltage Vout at the output of the second opamp.

 

[peasngravy]

The key problem for solving the 2nd amp is dealing with the C5/C8 network at the input. Because the virtual ground of the 2nd amp will keep the AC (signal) voltage = 0 at the input (BTW, don't confuse this with the non-zero DC voltage there), this problem amounts to given a signal voltage applied at the 1st amp output, what will the current flow be into the 2nd amp. That current will have to flow through the feedback components to create the output voltage. Something like this picture:

edit: note that you will end up with an equivalent impedance Zeq = Vi(jω)/Io(jω), although I phrased it as an admittance (i.e. find Io given Vi); whatever, there sort of the same, in a reciprocal fashion, LOL. Then you can pretend it is a two terminal complex impedance, if you like to think of things that way. BTW this Zeq simplification only works for Vi and Io, it won't be correct if you want to know other things in the network, like the current through R13, etc.

As this is time dependent due to the capacitors, is there any particular point in time I should choose for this? Or does jω negate the need for this?
I can't find a way to use j, excel and my calculator do not like it

Edit - i will show my working up to this point. I am very close now, thank you allfor guiding me this far

Series combination of C9 and R3 forms ad RC series with the magnitude of Z, which forms a voltage divider with the parallel combination of R1 and R12

So AC voltage at output of first op amp = 515.79 * 177.658 = 91.765V

I had written this in word and for some reason the equations don't copy over so had to screenshot

 

[The Electrician]

Your procedure is correct, but there are some small arithmetic errors.

The value 515.79 should be .51579 in the following calculation, but even then the calculation you show is giving a slightly wrong result: .51579 * 177.658 = 91.765V should be as shown below:

But using .51679 instead of .51579 you should get:

.51679 * 177.658 = 91.812

Next calculate the C5 R13 C8 R6 impedance input to the second opamp. DaveE gave the procedure.

 

[berkeman]

I can't find a way to use j, excel and my calculator do not like it

What calculator are you using? My HP-42 (simulator on my smartphone) has "Complex" mode.

And Excel has a COMPLEX function. Just look it up in Excel Help:

I am very close now, thank you allfor guiding me this far

 

[peasngravy]

Are C5/R13 and C8/R6 just two impedances in parallel?

 

[The Electrician]

Are C5/R13 and C8/R6 just two impedances in parallel?

Calculate an equivalent for C8 R6 and then that equivalent is in parallel with R13 giving another equivalent. Then that equivalent forms a voltage divider with C5. The output of that voltage divider is a voltage which drives a current through C8 and R6. Divide the voltage out of the voltage divider by the equivalent impedance magnitude of C8 R6, giving you a current. Multiply that current times the equivalent impedance magnitude of C7 R9 and you have the gain of the second opamp.

 

[peasngravy]

C8 R6 = 33783.64566
C8 R6 in parallel with R13 = (33783.64566*10000)/(33783.64566+10000) = 7716.042178

Using this formula:

With Xc of C5 being 1/(2 pi fC) = 3386.2754

Vout = 91.812 * (3386.2754/7716.042178) = 0.4388617V

Dividing by Z of C8 R6, for current = 0.00001299A
Impedance of C7 R9 = 990580.99 ohm

Gain of 2nd op amp = 0.00001299 * 990580.99 = 12.86800311

Gain of 2nd op amp * Vin = 1170V

 

[peasngravy]

Thanks for that - i was trying to simply use a cell to calculate the square root of -1 which obviously is not possible, but I thought excel would recognise it, I didn't realize there was a specific function

 

[The Electrician]

C8 R6 = 33783.64566
C8 R6 in parallel with R13 = (33783.64566*10000)/(33783.64566+10000) = 7716.042178

Using this formula:
View attachment 273118

You are correct up to this point, but this formula is not what we want to use here.

With Xc of C5 being 1/(2 pi fC) = 3386.2754

Vout = 91.812 * (3386.2754/7716.042178) = 0.4388617V

I'll show corrections in red.

Use the voltage divider formula from Wikipedia with the appropriate impedance magnitudes:
Vout = 1 volt *(7716.042178)/ (3386.2754+7716.042178) = 1 volt * .6949968 = .6949968
We will multiply by 91.812 later.

Dividing by Z of C8 R6, for current = .6949968/33783.64566 = .00002057198
Impedance of C7 R9 = 990580.99 ohm

Gain of 2nd op amp = 0.00002057198 * 990580.99 = 20.3782

Overall circuit gain = Gain of 2nd op amp * Vin = 20.3782 * 91.812 = 1870.96
The Vin used just above isn't the Vin of the overall circuit; it's the output of the first opamp.

This is somewhat in error because impedance magnitudes were used in calculations rather than
the complex impedances. This was done because apparently your course work hasn't taught you
how to use complex arithmetic when capacitors are in the circuit, but using impedance magnitudes is better than doing nothing. I suppose the final result can be called a reasonable estimate given the use of impedance magnitudes.

If you're interested, I will show how I solved the complete network using nodal analysis.