Design an active bandpass filter to get 1.5MHz peak frequency and 300KHz band

Question

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.

Answer 1

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: -

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.

Answer 2

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.