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I've been staring at this problem for two days now and can't get the correct transfer function. I've modeled it in Simulink (Matlab) and find the magnitude of the transfer function to be ~ $$H(s) = \frac{s+20}{s+0.028}$$

I'm getting huge numbers when trying to solve it so I must be doing something wrong. I've applied KCL to solve for the currents going into the parallel combo. Can anyone offer guidance? My work is shown below.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

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  • \$\begingroup\$ It's usual to sum currents away from the node of interest and equate to zero. You seem to have mixed the directions. You also appear to have five currents at the Vn node, but there are only three. Vn is zero, btw \$\endgroup\$
    – Chu
    Mar 10 '17 at 17:12
  • \$\begingroup\$ Thanks for your response! So was I on the correct path to finding H(s) just summing the currents in the wrong direction? Would it be: (vx-vi)/R1 + (vx-vn)/R2 + (vx-0)/R3 + (vo-vx)/R4 + (vo-vn)/(R5+1/(C*s))? \$\endgroup\$ Mar 10 '17 at 17:30
  • \$\begingroup\$ No, there are only three currents at the Vx node \$\endgroup\$
    – Chu
    Mar 10 '17 at 18:10
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Can anyone offer guidance?

I'm thinking the big issue you have is R3 in that it goes to ground half way along the feedback path. If so then my advice is convert Vo, R3 and R4 to a much smaller voltage source (Vx) in series with one resistor.

  • Vx will be Vo.R3/(R3+R4)
  • Series resistance will be R3||R4

It's just nortons and thevenins theorums I'm using.

Now you have a voltage source Vx in series with Rx where Rx is R2 + R3||R4.

Now go about applying the feedback in your equations but using the above.

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  • \$\begingroup\$ The inverting input to opamp is also 0V so when calculating Vx you have to assume R2 and R3 are in parallel. Though to be fair R2 is much bigger than R3 so it probably does not make much difference \$\endgroup\$ Mar 10 '17 at 17:52
  • \$\begingroup\$ @WarrenHill that comes out in the wash when R2 is added to R3||R4 (in series). My Vx isn't the same position is your Vx. \$\endgroup\$
    – Andy aka
    Mar 10 '17 at 17:55
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    \$\begingroup\$ Sorry Andy, I misread your answer I chose a real point you could measure, you chose a Thevenin equivalent: Both approaches are equally valid. \$\endgroup\$ Mar 10 '17 at 18:01
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What is the gain at DC? Assuming virtualGND (pin-), the output comes into a severe voltage divider (200K/1.13K); that feedback voltage then is amplified by 200K/50K. Your DC gain is 160*4 = 640x.

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  • \$\begingroup\$ I'm actually looking for how to derive the transfer function of this op-amp. Thank you for the description about DC gain though! \$\endgroup\$ Mar 10 '17 at 17:31
  • \$\begingroup\$ Since I could not get 640X from your derived equation, I figured the derivation had errors. Sometimes I use easy-to-find equations (DC gain) as sanity checks on my own suspicious derived equations. Congrats on making the effort. Kirchoff would be delighted to see how we use his concepts. \$\endgroup\$ Mar 10 '17 at 17:36
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Lets start by calling the junction of R2, R3 and R4 x. We know the inverting input of the amplifier is at 0V so what's the voltage at x?

$$V_x = V_o \cdot \dfrac{\dfrac{R_2R_3}{R_2+R_3}}{\dfrac{R_2R_3}{R_2+R_3}+R_4} = V_o \cdot \dfrac{R_2R_3}{R_2R_3+R_2R_4+R_3R_4}$$

Next stage is to note that the current into ether the inverting or non-inverting input is 0 we have

$$\dfrac{V_i}{R_1} = \dfrac{V_x}{R_2} + \dfrac{V_o}{R_5 + \dfrac{1}{s \cdot C_1}}$$

Substitute for \$ V_x \$ and re-arrange to get an equation of the form \$ \dfrac{V_o}{V_i} = \$ something and job done.

I'll leave this as an exercise as I do not want to do your homework for you but I hope I've shown enough for you to see how to approach this kind of problem.

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  • \$\begingroup\$ I see your approach now and it makes sense to have only three terms in the KCL equation as the node on VN has 1 input and 2 outputs. I am a bit confused about how you solved for Vx at the x junction. Isn't it a voltage divider where Vx = vo*R3/(R3+R4)? I tried computing Vo/Vi with both Vx equations and I still am not getting H(s) = (s+20)/(s+0.028). What I'm evaluating is of much greater magnitude. (499 s + 10000) 40226000 -------------------------- (40282387 s + 1130000) 499 \$\endgroup\$ Mar 10 '17 at 18:10
  • \$\begingroup\$ Isn't it a voltage divider where Vx = vo*R3/(R3+R4)? @Andy's answer takes this as a virtual Vx which works by adding in the Thevinin series impedance. I am calculating the actual voltage at the junction and since one end of R2 and one end of R3 are at 0V you can think of R2 and R3 being in parallel. \$\endgroup\$ Mar 10 '17 at 18:16
  • \$\begingroup\$ Why not just recognize that R2 is in Parallel w/ R3, and replace that whole feedback path with (R2R3/(R2+R3)) + R4 ? \$\endgroup\$ Mar 13 '17 at 18:51
  • \$\begingroup\$ @ScottSeidman If you do that you get a different and wrong answer just considering R2 and R3 to be in parallel suggests the entire current in R4 is going into the inverting input node. It isn't and this significantly effects gain. \$\endgroup\$ Mar 14 '17 at 19:11
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At the \$\small V_x\$ node ( \$\small R_2;R_3;R_4\$ node): $$\frac{V_x}{R_2}+\frac{V_x}{R_3}+\frac{V_x-V_o}{R_4}=0$$

At the \$\small V_n\$ node, noting that \$\small V_n=0\$ (virtual ground): $$-\frac{V_i}{R_1}-\frac{V_o}{R_5+\frac{1}{sC_1}}-\frac{V_x}{R_2}=0 $$

Then eliminate \$\small V_x\$ to give the TF: $$G(s)=\frac{V_o(s)}{V_i(s)}=\frac{s+200}{s+0.28}$$

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You can obtain this transfer function using Fast Analytical Circuits Techniques or FACTs. You start by determining the gain for \$s=0\$: remove capacitor \$C_1\$ and determine the quasi-static gain \$H_0\$. Several options are available to get there. The most straightforward is superposition. Determine \$V_{(-)}\$ by setting \$V_1\$ to 0 V then by setting \$V_o\$ to 0 V. If you solve \$V_o/V_1\$ you should obtain the dc gain:

\$H_0=-\frac{R_2+R_4}{R_1}+\frac{R_2R_4}{R_1R_3}\$

If you compute this dc gain with the components values you gave, you find \$H_0=-717.399\$ or 57.115 dB.

The second option uses the Extra-Element Theorem or EET (https://en.wikipedia.org/wiki/Extra_element_theorem). The principle is the following. Identify an element in the circuit under study that is bothering you. Here, \$R_3\$ is clearly the one. Determine the gain when this element is either removed (set to an infinite value) or replaced by a short circuit (set to 0). This is what we call the reference gain or \$H_{ref}\$. In this example, we will determine \$H_{ref}\$ when \$R_3\$ is removed. The dc gain in this mode is straightforward:

\$H_{ref}=-\frac{R_2+R_4}{R_1}\$

The second step is to reduce the excitation source, \$V_1\$ to 0 V (\$R_1\$ is thus grounded). Now, you have to calculate the resistance \$R_d\$ offered by \$C_3\$ terminals when it is removed. You can install a current source \$I_T\$ and calculate the voltage \$V_T\$ across its terminals. If you do that, you find that \$R_d=0\$. The second exercise is to calculate the resistance \$R_n\$ seen from \$C_3\$ terminals when it is removed while \$V_1\$ is back in place and \$V_o\$ nulled. This is what is called a null double injection (NDI). Basically, from your schematic, thanks to the op amp and its virtual ground, we have 0 V on \$V_{(-)}\$ and \$V_o\$ being nulled, the right terminal of \$R_4\$ is also grounded. As such, the resistance \$R_n\$ seen from \$C_3\$ terminals is simply \$R_2||R_4\$. We can now apply the EET:

\$H_0=H_{ref}\frac{1+R_n/R_3}{1+R_d/R_3}=-\frac{R_2+R_4}{R_1}(1+\frac{R_2||R_4}{R_3})\$

If you compute this gain, you obtain exactly -717.399 as in the above.

Ok, we have the dc gain but what about the pole? Same as before, reduce the excitation to 0 V and compute the resistance seen from \$C_1\$ terminals in this mode. If you do well, you should get:

\$\tau_1=C_1(R_4(1+\frac{R_2}{R_3})+R_5+R_2)\$ from this value, you have:

\$\omega_p=\frac{1}{C_1(R_4(1+\frac{R_2}{R_3})+R_5+R_2)}\$

The zero is determined by a possible impedance combination in the transformed circuit (meaning \$C_1\$ is replaced by \$1/sC_1)\$ which could prevent the excitation to reach the output thus creating a null in \$V_o\$. Obviously, this is the series combination of \$R_5\$ and \$1/sC_1)\$ which can become a transformed short. The zero is thus located at

\$\omega_z=\frac{1}{R_5C_1}\$

This is it, we have the complete transfer function written in a low-entropy format:

\$H(s)=H_0\frac{1+s/\omega_z}{1+s/\omega_p}\$

with your values, the dc gain is -717.399, the zero is located at 3.189 Hz and the pole is placed at 4.44 mHz.

This low-entropy format is the one you should adopt versus the expression you first wrote. A low-entropy format lets you see a dc gain (if any), a zero and a pole. From your first formula, I cannot immediately see a dc gain and where the pole and zero are.

You can find more about these fast analytical circuits techniques in an APEC presentation taught in 2016: http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

Good luck!

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