# Finding component values of bandpass filter with load?

I 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.

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

• 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/… Feb 17, 2015 at 15:14
• 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 $H_2(\omega)$. Feb 18, 2015 at 3:16