BJT High Pass Filter

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KCHARROIS

Joined Jun 29, 2012
311
Hello,

I've been studying active filters using op amps in school for the past week so I decided to take a look at discrete active filters using a transistor but I'm not sure if I understand how the circuit works.

So here is what I think, when frequency goes up reactance of the C1 goes down and depending what the input impedance of the transistor say double the reactance of C1, acts as a High Pass Filter.

When frequency goes down C1 reactance goes up, so if input impedance of transistor is lower less signal goes through.

Am I missing something with the 5K feedback resistor or something else?

Thanks
 

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t_n_k

Joined Mar 6, 2009
5,455
The circuit interpretation is more complex than your initial one.

The 5K feedback resistor R4 plays a critical role.

You may note that the BJT part is a voltage follower.

If I reduce the circuit to a simplified version for AC purposes I "see" something like that shown in the attachment.

The analysis of this circuit requires some effort and turns out to have the form ...

\(\frac{v_{out}}{v_{in}}=\frac{s^2}{s^2+\frac{(C_1+C_2)}{RC_1C_2}s+\frac{1}{R^2C_1C_2}}\)

where

R=R4=R1||R2=5K in your example
C1=20nF
C2=10nF

The result is a 2nd order high pass filter with a -6dB frequency somewhere about

\(\omega=\frac{1}{\sqrt{R^2C_1C_2}} \ [radians \ sec^{-1}] \)
 

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t_n_k

Joined Mar 6, 2009
5,455
Yes - the variable 's' is the Laplace frequency domain operator - for which one may substitute the complex term from which it's possible to find the frequency domain transfer function explicitly in terms of angular frequency.

If you consider your previous investigations of op-amp based filters you may [with some careful consideration] see some similarities of the BJT version you posted with the op-amp Sallen-Key filter topology.

See http://en.wikipedia.org/wiki/Sallen–Key_topology
 
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