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Problem for making the excersise (given by the professor)

On the picture above, you'll see a circuit. The exercise is to make a transfer function and draw a bode plot. I've tried to do that, but because there's no solution, I don't know if I'm right.

Specifications: C1 and C2 are infinity, there's no value given for the two resistors (R), R1=180 ohm, R2=180k ohm, C3=22 microF

My attempt

This is an non-inverting opamp-circuit, so the attenuation is: $$\frac{V_{out}}{V_{in}}=1+\frac{R_1+Z_{C_3}}{R_1+R_2+Z_{C_3}}$$

Converting this to a transfer function: $$H(s)=\frac{R_1*C_3s+1}{C_3s(R_1+R_2)+1}$$ The zero and pole point is: $$z_1=-\frac{1}{R_1*C_3}\quad and \quad p_1=-\frac{1}{(R_1+R_2)*C_3}$$ The magnetude and break-frequencies are: $$|H(j \omega)|=\frac{\sqrt{(R_1 C_3 \omega)^2+1}}{\sqrt{(C_3 \omega (R_1+R_2))^2+1}}\\ f_{break1}=\frac{1}{2 \pi R_1 C_3}=40Hz\\ f_{break2}=\frac{1}{2 \pi (R_1+R_2) C_3}=0,04Hz$$

Can someone check if this is correct, or improve me if I'm wrong, please?

EDIT by the correction from the answers

This is an non-inverting opamp-circuit, so the attenuation is: $$\frac{V_{out}}{V_{in}}=\frac{R_1+R_2+Z_{C_3}}{R_1+Z_{C_3}}$$

Converting this to a transfer function: $$H(s)=\frac{C_3s(R_1+R_2)+1}{R_1*C_3s+1}$$ The pole and zero point is: $$p_1=-\frac{1}{R_1*C_3}\quad and \quad z_1=-\frac{1}{(R_1+R_2)*C_3}$$ The magnetude and break-frequencies are: $$|H(j \omega)|=\frac{\sqrt{(C_3 \omega (R_1+R_2))^2+1}}{\sqrt{(R_1 C_3 \omega)^2+1}}\\ f_{break1}=\frac{1}{2 \pi R_1 C_3}=40Hz\\ f_{break2}=\frac{1}{2 \pi (R_1+R_2) C_3}=0,04Hz$$

Bode Plot, created in Matlab

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  • \$\begingroup\$ Three answers, one upvote. \$\endgroup\$ – pipe Aug 28 '17 at 22:07
  • \$\begingroup\$ You've now changed the question, using the answers posted on the site. So all the correct answers are now worthless. I'm deleting mine + downvote. \$\endgroup\$ – Chu Aug 28 '17 at 22:08
  • \$\begingroup\$ Chu, don't be butthurt, the OP is just a guy worried about his homework, we should help not bitch. \$\endgroup\$ – peufeu Aug 28 '17 at 22:15
  • \$\begingroup\$ It's the second time I post something here. I've now made an extra section with the corrections from the answers and above the original post. Now, everyone can see the problem and the solution \$\endgroup\$ – JonathanC Aug 28 '17 at 22:32
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The equation for the gain of your non-inverting amplifier is incorrect, it should be:

$$\frac{V_{out}}{V_{in^+}}=1+\frac{R_2}{R_1+Z_{C_3}}$$

Since this is homework it is more interesting to have you guess a little bit, so I suggest you have a look at the non-inverting amp equations from your course and check this.

Note that the high-frequency pole is not present here. If a perfect opamp model is used, the circuit is strictly a highpass.

EDIT

You should change the x-axis of your plots to the bandwidth of interest. Since this seems to be an audio preamp, use 1Hz-100kHz, which will let you see the main bandwidth (20Hz-20kHz) plus what happens on the edges.

The purpose of the cap C3 in the feedback network is to reduce the gain of the opamp at DC to unity gain. This is very useful, since your opamp will usually have some DC offset due to mismatch between the input transistors, and without the cap, this offset would be amplified by the circuits' gain, which is 1001 here. So an opamp will 2mV offset, which is not uncommon, would result in a 2V offset at the output, which considering the 0-5V rails, would compromise then available voltage swing. So keep in mind that the use of C3 is to compensate for defects which exist in real opamps, but not in perfect opamp models.

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  • \$\begingroup\$ I've edited my post by following your corrections. When I see the bode plot, it's more logical with the goal of the given circuit in mind. \$\endgroup\$ – JonathanC Aug 28 '17 at 21:47
  • \$\begingroup\$ Answer edited............ \$\endgroup\$ – peufeu Aug 28 '17 at 22:14
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Almost .. Looks good.

Since the breakpoints are 3 decades apart, you can simplify it to basically R1C3, R2C3 for low reject and high reject breakpoints at -3dB f2,f1 from a gain of 60dB = 20log R2/R1. neglect the rounding error of actual gain = 1+R2/R1=1001

your initial Av- is wrong . but 1+ negative Feedback ratio is the correct gain as above answer.

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  • \$\begingroup\$ Thank you for your quick answer. I've plotted this in Matlab (you can see it here), but the plot starts on 0dB and go down. When I read your answer I noticed that you have another beginning point then I've in Matlab. How can I calculate that, or can I determine that from the formulas above. link to plot \$\endgroup\$ – JonathanC Aug 28 '17 at 20:27
  • \$\begingroup\$ I was too quick in reading just your conclusion. The inverting gain shud be impedance ratio of Av-=-Zf/Zs and non-inv gain is 1 + |Av-| \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 28 '17 at 23:06

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