Here is the image, it is a really basic bandpass filter, BUT imagine a 5Mohm load resistor where Vout isI have to find the components of a bandpass filter given only the two corner frequencies. The only components are \$C_1\$ and \$R_1\$ for the high pass, and \$R_2\$, \$C_2\$, and Load resistance (\$5\$M\$\Omega\$) on the low pass.

I set \$R_1\$ to \$10\$k\$\Omega\$ and got \$C_1 = 53.05\$nF when corner frequency for high pass is \$300\$Hz using \$C=1/2\pi fR_1\$.

However, for the low pass... I can't figure it out. I combined \$C_2\$ and the load into equivalent impedance \$Z\$ (they are in parallel), then plugged \$Z\$ into the transfer equation for a low pass filter where ever a \$C\$ would have normally appeared, and get \$C_2 = 1.588\$nF while \$R_2 = 10\$k\$\Omega\$ (setting transfer of low pass filter equal to \$1/\sqrt{2}\$). Also corner frequency of low pass filter is 10kHz. But the graph of the output voltage looks like a high pass filter only, and is in the micro volts and the -3db frequencies are nowhere even close... I don't get what is going wrong with the low pass filter

So was wondering if anyone can show me how to calculate the \$R\$ and \$C\$ values of the low pass filter part of a bandpass filter when it is connected to a load?

It is quite a basic bandpass filter, but imagine a 5Mohm load where Vout is.

  • 2
    \$\begingroup\$ You should really post a schematic of the circuit you are trying to analyze/design. \$\endgroup\$ Feb 17, 2015 at 2:29

1 Answer 1


Your low pass filter is loading your high pass filter. Think of a simple voltage divider: enter image description here. In general, if you have 2 circuits connected together, the first one's output to the second one's input, you do not want the second one to attenuate the signal that passes through the first one. Therefore, you must make sure that the second circuit's input impedance (represented by R2 in the voltage divider) is larger than your first circuit's output impedance. How much larger? The rule of thumb is 10x larger.

The output impedance of your high pass is roughly the value of R1, or 10 kOhm. The input impedance of your second circuit is also roughly R2 = 10 kOhm. So you need to decrease R1 by a factor of 10 or increase R2 by a factor of 10 (or some other combination such that R2 = 10R1) and adjust C1 and C2 accordingly to obtain the same cutoff frequencies. Once you do, you will see that your bandpass filter behaves much better.

Along the same lines, you don't want to increase R2 beyond 500 kOhm because your load is 5 MOhm. The same rule applies!

  • \$\begingroup\$ Oooh I see. So once I get R2 10x larger than R1, I can set up: 1/square-root(2) = transfer function of low pass with Zeq of load resistor and C2, and I should see a desired bandpass output? \$\endgroup\$ Feb 17, 2015 at 5:56
  • \$\begingroup\$ Yes, that's right, you will get the cutoff frequency of the low pass if you do that. As an exercise, I would try to get the transfer function of the entire circuit and set it equal to \$ \frac{1}{\sqrt{2}} \$ to get both cutoff frequencies. If you need help with deriving the transfer function, see this question: electronics.stackexchange.com/questions/152159/… \$\endgroup\$ Feb 17, 2015 at 15:14
  • \$\begingroup\$ Eh, I still don't think it is working out. I attempted to find Vx (Vout of high pass), then take Vout/Vin = LPF transfer function * Vx, since Vx = HPF * Vin. Except I equivalent impedanced everything to the right of C1 including R1 to get Vx. This has turned into 30+ minutes of scribbling and solving, which really shouldn't be the case for something so basic... And finding the entire transfer function doesn't really help finding the component values I need, I feel I need each individual part to do that. \$\endgroup\$ Feb 17, 2015 at 20:43
  • \$\begingroup\$ Yes, you don't need to write the entire transfer function to find your values. You can find your values for each filter individually. Just make sure that R2 > 10*R1, and Rload > 10*R2. That's why I said "as an exercise"...it's not required. Sorry if I confused you. \$\endgroup\$ Feb 17, 2015 at 22:20
  • \$\begingroup\$ To clarify, if the input impedance of the second stage is approximately 10 times the output impedance of the first stage, the total transfer function \$H(\omega)\$ is approximately \$H_2(\omega) \times H_2(\omega)\$, the product of the individual transfer functions of both stages. Therefore, you don't have to calculate \$H(\omega)\$ in order to construct your BPF - you just need \$H_1(\omega)\$ and \$H_2(\omega)\$. \$\endgroup\$ Feb 18, 2015 at 3:16

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