BJT Amp Frequency Response

  • Thread starter Elektron
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I would like to know how to design an amplifier similar to the attached circuit with set -3db points, the lower on the order of 10Hz, and the higher on the order of 10kHz.

I have designed amplifiers to work with a set gain, but have not dealt with frequency response before.

I designed an amp on PSPICE with a theoretical gain of over 200, but when I tacked on a low-pass filter, the output pretty much went to 0.

Therefore, how would you go about designing for gain in addition to -3db points?
 

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  • #2
berkeman
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Welcome to the PF.

Ignoring the emitter degeration, what RC time constants set the LPF and HPF breakpoints in that circuit?

EDIT -- Actually I'm not seeing the LPF part of this circuit...
 
  • #3
Baluncore
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In the simple analysis, gain is decided by the ratio of Rc to Re. But for a 10Hz to 10kHz band all sub-modules will effect each other to some extent.

Place a small value capacitor, Cbd, from node B to node D. That makes the BJT a miller integrator which is a low-pass filter. Select Cbd to give –3dB at 10kHz.

Both C1 and C2 set the 10Hz low frequency point. But their surrounding resistive circuit is difficult to analyse. Simulate it with C1 as a very high value capacitance, adjust C2 for –1.5dB at 10Hz. Remember that value of C2 for later. Then simulate it with C2 as a very high capacitance, adjust C1 for –1.5dB at 10Hz. Replace C2 with the value remembered earlier. The values of C1 and C2 that give –1.5dB each will together give –3dB at 10Hz.

Because the 10Hz HPF and the 10kHz LPF both give some in-band loss, you will need to adjust Rc/Re to increase the DC gain to compensate. Notice that Ce will give a step change to in-band gain if the value of Ce is too small.
 
  • #4
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Thank you both for your replies. I was able to implement your suggestions on PSPICE and am observing the -3db points.
 

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