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I'm designing an active bandpass filter (using only BJTs, voltage sources, resistors and capacitors) with the following properties:

  • Corner frequencies \$8 ~ \text{kHz}\$ and \$800 ~ \text{kHz}\$ (bandwidth of about \$790 ~ \text{kHz}\$)

  • \$40 ~ \mathrm{dB/dec}\$ roll-off of the LPF (second order filter)

  • \$20 ~ \mathrm{dB/dec}\$ roll-off of the HPF (first order filter)

  • Gain of about \$42 ~ \text{dB}\$

  • Low input impedance: \$\sim 60 ~ \Omega\$ (over a wide range of frequencies \$80 ~ \text{Hz} - 80 ~ \text{MHz}\$)

  • High output impedance: \$\sim 50 ~ \text{k}\Omega\$ (over a wide range of frequencies \$80 ~ \text{Hz} - 80 ~ \text{MHz}\$)

I thought about the following simple input stage: let the input be applied across a resistor (\$R'\$) with the desired input impedance followed by a unity-gain buffer (which has a high impedance and therefore wouldn't affect much the input impedance). Then, take the output of the buffer and process it through an HPF filter (and later through LPF). However when I tried to implement the buffer using emitter-follower I quickly ran into some problems.

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In order to establish a proper biasing of the transistor we have to have the resistors \$R_1\$ and \$R_2\$. Crucially, a coupling capacitor \$C_{in}\$ is required (otherwise the circuit just does't work properly - SPICE simulation yields extreme attenuation in the frequency response). But the introduction of this capacitor creates two serious problems:

  • It affects the input impedance. The input impedance now changes with frequency, and it's hard to maintain a constant low input impedance over such a wide range of frequencies

  • It acts as an HPF filter and therefore adds another pole. In other words, if I want to have \$20 ~ \mathrm{dB/dec}\$ roll-off I must remove the HPF filtering of the buffer output (which in turn completely changes the design of the circuit).

What are some possible ways to alleviate these issues without over-complicating the circuit?

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answer(s). \$\endgroup\$ – Dave Tweed Feb 2 at 12:34
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The high-pass filter will introduce a zero at \$f=0\$ and a pole at \$f=f_c\$ (the corner frequency of the high-pass filter). A pole and a zero will compensate each other, leading to a flat frequency response inside the frequency band.

The corner frequency \$f_c\$ is determined by \$C_{in}\$, and the parallel resistance of \$R_1\$ and \$R_2\$ (and for a small part the resistance seen at the base of the BJT). It can be found by analyzing the AC-equivalent circuit.

  • It affects the input impedance. The input impedance now changes with frequency, and it's hard to maintain a constant low input impedance over such a wide range of frequencies

"Inside the frequency band" means for \$f > f_c\$, in which case the signal frequency is so high that the capacitor will act like a short-circuit in the AC-equivalent circuit. The input impedance is therefore constant in the bandpass region, and equal to the parallel resistance of \$R'\$, \$R_1\$, \$R_2\$ and the base resistance of the BJT (usually very large because it is bootstrapped).

  • It acts as an HPF filter and therefore adds another pole. In other words, if I want to have 20 dB/dec roll-off I must remove the HPF filtering of the buffer output (which in turn completely changes the design of the circuit).

As explained, the zero at \$f=0\$ will introduce an increase of 20dB/dec, and then the pole at \$f=f_c\$ will decrease it again with 20dB/dec, so the net effect is 0dB/dec.

  • Low input impedance: ∼60 Ω (over a wide range of frequencies 80 Hz−80 MHz)
  • High output impedance: ∼50 kΩ (over a wide range of frequencies 80 Hz−80 MHz)

I doubt that the required specifications want you to design for a small input impedance and a large output impedance though. Typically, a common-collector/emitter follower/buffer circuit tries to achieve the opposite: high input impedance and low output impedance.

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  • \$\begingroup\$ Thanks. 1) "The input impedance is therefore constant in the bandpass region, and equal to the parallel resistance" - indeed, but notice that the input impedance is required to be constant at much wider range than the bandwidth. So it's tricky. 2) "I doubt that the required specifications want you to design for a small input impedance and a large output impedance though" - I know, but I don't see any other way of meeting the requirements (I know that usually the input impedance should be high and the output impedance low, but here they purposefully demand the opposite). \$\endgroup\$ – grjj3 Feb 1 at 20:40
  • \$\begingroup\$ "The high-pass filter will introduce a zero at f=0" - I'm not sure how this comes about, but I'll check it. Thank you anyway. \$\endgroup\$ – grjj3 Feb 1 at 20:41
  • \$\begingroup\$ I'd try to make the capacitor negligible compared to \$R'\$. If \$R'\$ is small enough (\$\approx60\Omega\$), then a large parallel resistance will not affect the input impedance by much. \$\endgroup\$ – Sven B Feb 1 at 21:40
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The transistor is an emitter-follower with a gain of about only 1. You will need a second transistor to provide the gain of 42dB which is a little more than 100 times.

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