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I have this question in Linear Circuit Analysis course where I need to find the values of R1, R2, C1, C2, of the given bandpass filter circuit in order to get a peak frequency wp = 1.5 MHz and a bandwidth Bw = 300 kHz.

the first step is to find the transfer function of the circuit, to compare it to the general TF of the bandpass filter:

\$H(s) = \frac{Ks}{s^{2} + 2\sigma s +\omega p^{2}}\$ (1)

and this is what I get :

\$H(s) = \frac{\frac{-1}{C2R1} s}{S^{2} + (\frac{1}{R2C2} + \frac{1}{R1C1})s + \frac{1}{R1R2C1C2}}\$ (2)

by comparing the two functions, we get these 2 equations (Bw and wp) with 4 unknowns:

\$\frac{1}{R1C1} + \frac{1}{R2C2} = 6\pi \ast 10^{5}\$ (I)

\$ \frac{1}{R1R2C1C2} = 9\pi ^{2}\ast 10^{12}\$ (II).

I am allowed to assume 2 unknowns out of the four, so, I went with C1 = C2 = 1uF.

now I'm supposed to solve a system of 2 equations with 2 unknowns, but for some reason, it is not working with me; I tried solving for R1 in eq 2, and the substitute in eq 1 and I got a negative R2 resistance, what could be wrong?? is there another way to solve it that I am not aware of??

Edit 1: fixed the equation format.

enter image description here

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  • \$\begingroup\$ whoever draws capacitors filled like in educational material should be ashamed, to be honest. Obada, totally not your fault, but this is indicative of lower-quality material. Which textbook are you using in your course? \$\endgroup\$ – Marcus Müller Jun 2 at 16:49
  • \$\begingroup\$ Linear Circuit Analysis, Decarlo-Lin. but for this question, in particular, it is not from the textbook. it is a different schematic style. \$\endgroup\$ – Obada Jun 2 at 17:21
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It won't make a very sharp or precise filter using the configuration shown but, to make life simpler (given that it is a really sloppy filter) you can regard C1 and R1 as forming a high-pass network and C2 and R2 as forming a low-pass network.

The two filter points need to be different to obtain a 300 kHz bandwidth so,

  • make the low-pass network cut-off at 1.5 MHz + 150 kHz (1.65 MHz) and
  • make the high-pass network cut-off at 1.5 MHz - 150 kHz (1.35 MHz).

That allows you to calculate both resistors based on both capacitors being 1 uF. However, what you will find is that 1 uF is too high in value for an op-amp to work with at circa 1.5 MHz and you will need to choose a much lower default value.

For instance, using a 100 pF capacitor, it will form a 3 dB point at 1.65 MHz when R = 964 ohms.

If you'd have used 1 uF, R would be 0.096 ohms and that isn't practical for any op-amp.

However, you will never get a centre frequency to bandwidth ratio as required by the question using a simple op-amp and high-pass and low-pass networks. To get a centre frequency if 1.5 MHz and a bandwidth of 300 kHz implies a Q factor of 1500/300 = 5. This cannot be achieved by the op-amp configuration shown because the maximum Q factor is limited to 0.5. You can try it of course, but you'll get just a sloppy band pass filter like this: -

enter image description here

It just won't deliver what you want. Bandwidth is approximately 3.53 MHz - 610 kHz = 2.92 MHz and centre frequency is 1.47 MHz hence, Q = 0.503.

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  • \$\begingroup\$ what you mean is to consider C1, R1 as a high pass filter with C2, R2 as a low pass filter, but in order to do that, I have to find the transfer function for each filter right?? what would be the Vin and vout for each filter then??? besides that, I'm pretty sure that the answer the instructor expects is easier than this because we haven't been taught such a method. \$\endgroup\$ – Obada Jun 2 at 14:25
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    \$\begingroup\$ "Network" is what I said not "filter"; regard C1 and R1 as forming a high-pass network and C2 and R2 as forming a low-pass network. Given your question about Vin and Vout, I assume you don't really understand op-amps. \$\endgroup\$ – Andy aka Jun 2 at 14:29
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    \$\begingroup\$ I am not complaining but, this is actually my second course in circuits and unfortunately, I haven't got the chance to go deeper into many topics, that is why there are some things that I don't understand well. \$\endgroup\$ – Obada Jun 2 at 14:45
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To achieve a controlled bandwidth (here Q = 5), you'll need a different topology.

The inverting opamp topology ensures there is NO INTERACTION between the input series RC network and the output parallel RC network.

By the way, for a successful 1.5MHz active filter, you'll need an opamp with 20Mhz or 50MHz or 100MHz Unity Gain Band Width.

Walt Jung discussed this, decades ago, explaining the stop_band attenuation requires accurate cancellation of charges. To achieve accuracy, the opamp must have lots of amplification at the high (stop_band) frequencies.

If you want -40dB at 100MHz, you'll need a gain of 100X (+40dB) at 100MHz inside the opamp.

Even the topology you are using will demonstrate this demand.

Change the opamp UGBW, and notice how the high_frequency attention is directly affected by that change.

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