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Why does the implementation of a buffer amplifier separating the high-pass and low-pass filter of a bandpass filter affect the transfer function of the bandpass filter?

I thought the point of a buffer amp was to keep the output voltage equal to the input voltage regardless of current draw at the output. Therefore I would expect there to be no change assuming no current is being drawn at the input or the output and all components are ideal.

Unfortunately it seems the transfer function of both filters and their behaviour is different and I don't know why.

  • Buffer amplifier present:

schematic

simulate this circuit – Schematic created using CircuitLab

  • Buffer amplifier not present:

schematic

simulate this circuit

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    \$\begingroup\$ Hint: (1) What is the load impedance seen by C1+R1 in each case? (2) What is the source impedance seen by R2+C2 in each case? \$\endgroup\$
    – Transistor
    Mar 18, 2021 at 9:53
  • \$\begingroup\$ "assuming no current is being drawn at the input or the output and all components are ideal" - yes this is the case for the circuit with the buffer, but do you think that this statement holds for the circuit without the buffer too? Spilt the circuit into v2 sections and think about what effect each section has on the other with no buffer to isolate them. \$\endgroup\$
    – brhans
    Mar 18, 2021 at 10:00
  • \$\begingroup\$ Also check out this question, it explains the difference: electronics.stackexchange.com/questions/220050/… \$\endgroup\$
    – jusaca
    Mar 18, 2021 at 10:20
  • \$\begingroup\$ When you have a simple voltage divider - will the divider ratio change when you connect an aditional resistor? \$\endgroup\$
    – LvW
    Mar 18, 2021 at 16:59

2 Answers 2

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One of the most important uses of buffers (being them emitter followers, voltage followers or whatever) is that the isolate the stages presenting a stable impedance on both input and output.

Since a passive filter is influenced by these impedences (RF people fight with this all the day) simply chaining them has the first one seeing the input of the second one and the second one seeing at the input the output of the first one. These are complex impedances and the result is… well, not immediately obvious. That being said ladder filters are designed with that in mind, but as a single unit.

A buffer in the middle present to the first stage a (usually high) fixed impedance and to the second stage a (usually low) impedance. In this way you can design both stages independently and 'glue' them with a buffer.

This is valid not only for filters but for signal stages in general. There is a whole two port network theory based on this chaining.

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The TL081 has a GBP (Gain Bandwidth Product) of 4 MHz.

Its transfer function, which has 2 poles, do participate to the overall transfer function of your circuit.

The TL081 is used in your circuit is a unity gain buffer. That means that the first pole is at 4 MHz and that means that starting from 400 KHz the overall transfer function starts "feeling" the presence of that pole.

To answer your question:

The TL081 is ideal up to the frequency of 400 KHz.

That means that the overall transfer function is the product of the RC and the CR networks transfer functions only if the poles and the zeros of the RC and CR networks are below 400 KHz.

Try R = 1 kOhm and see if you see the same effect that you mentioned in your answer.

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  • \$\begingroup\$ good catch, didn't notice the circuit was bandwidth starved \$\endgroup\$ Mar 18, 2021 at 12:20
  • \$\begingroup\$ TL081 is the 'default' opamp in the Circuit-Lab schematic editor. I wouldn't assume that the OP is actually talking about this specific device, particularly since they're referring to 'ideal' components. \$\endgroup\$
    – brhans
    Mar 18, 2021 at 12:43

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