Design of High-pass Passive Filter

  • #1

Homework Statement


Design a filter that will pass frequencies greater than 20 Hz and set the -3dB point at 10 Hz. Assume zero source impedance and 10k load impedance.

Homework Equations


##f=\frac{1}{2\pi RC}## which is the equation used to determine the 3 dB frequency of an RC filter.

The Attempt at a Solution


Select a high-pass RC filter. Active filters have no been presented yet in the text, so I am using passive components.
As the input impedence of the load is assumed to be 10k, then I would like the output impedence of my filter to be about 1k: a tenth of the load.
Deriving the output impedence of the high-pass filter:
[itex]
Z_{input}=\frac{Z_{C} R}{R+Z_{C}}=\frac{\frac{R}{j\omega C}}{R+\frac{1}{j\omega C}}=\frac{R}{j\omega R C +1}
[/itex]
The next step is where my thinking is unclear. While I know that I want this output impedence to be about 1k, what kind of value should I select R to be and why? Does the 3 dB frequency I am designing for affect this in any way or is it independent?
Arbitrarily assuming an R value of 1k to match the impedence I want yields the following:
## \frac{1}{C}=2\pi fR## so ##C=\frac{1}{20\pi R}=16 pF##

The design is complete.

The information I am looking for is how to select an R value and why.

Thank you in advance.
 

Answers and Replies

  • #2
berkeman
Mentor
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Homework Statement


Design a filter that will pass frequencies greater than 20 Hz and set the -3dB point at 10 Hz. Assume zero source impedance and 10k load impedance.

Homework Equations


##f=\frac{1}{2\pi RC}## which is the equation used to determine the 3 dB frequency of an RC filter.

The Attempt at a Solution


Select a high-pass RC filter. Active filters have no been presented yet in the text, so I am using passive components.
As the input impedence of the load is assumed to be 10k, then I would like the output impedence of my filter to be about 1k: a tenth of the load.
Deriving the output impedence of the high-pass filter:
[itex]
Z_{input}=\frac{Z_{C} R}{R+Z_{C}}=\frac{\frac{R}{j\omega C}}{R+\frac{1}{j\omega C}}=\frac{R}{j\omega R C +1}
[/itex]
The next step is where my thinking is unclear. While I know that I want this output impedence to be about 1k, what kind of value should I select R to be and why? Does the 3 dB frequency I am designing for affect this in any way or is it independent?
Arbitrarily assuming an R value of 1k to match the impedence I want yields the following:
## \frac{1}{C}=2\pi fR## so ##C=\frac{1}{20\pi R}=16 pF##

The design is complete.

The information I am looking for is how to select an R value and why.

Thank you in advance.

For a single pole HPF, you have an input capacitor and an output resistor in series. Just put that 10k load resistor in parallel with your output resistor and use the parallel combination value for your R of the filter.

And 16pF is *way* too small to be used at this low frequency. You pick your combination of R & C to be some practical values. The biggest non-polar capacitor that is not too physically big and expensive is around 1uF. Start with that value to see what value of R you need to get that -3dB point, and go from there... :smile:
 
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  • #3
And 16pF is *way* too small to be used at this low frequency. You pick your combination of R & C to be some practical values. The biggest non-polar capacitor that is not too physically big and expensive is around 1uF. Start with that value to see what value of R you need to get that -3dB point, and go from there.

Thank you very much for your reply. I assume that the problem with having selected a 16pF capacitor is that they do not come in a non-polar variety and a signal may require going into negative voltages?

For a single pole HPF, you have an input capacitor and an output resistor in series. Just put that 10k load resistor in parallel with your output resistor and use the parallel combination value for your R of the filter.

Starting with a 1uF capacitor and a 10Hz point for the 3dB frequency yields an equivalent resistance of 16kΩ. Due to the load resistance being 10kΩ and the network being a parallel configuration, the 16kΩ value is impossible to attain without changing the network topology.

If I cannot go larger than 1.0uF, then this is more complicated than I was thinking. Knowing that the load resistance is 10kΩ, I can neglect the extra resistor entirely. Using a capacitance of 1.5uF would yield a 3dB frequency of 10.6 Hz, which may be close enough?
 
  • #4
berkeman
Mentor
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I assume that the problem with having selected a 16pF capacitor is that they do not come in a non-polar variety
No, 16pF is a very small capacitance, and only comes in non-polar versions. It's just that using 16pF at 10Hz would require a huge resistance value (impractical).

Knowing that the load resistance is 10kΩ, I can neglect the extra resistor entirely. Using a capacitance of 1.5uF would yield a 3dB frequency of 10.6 Hz, which may be close enough?
That works. Also, you can make a non-polar capacitor of a larger value by putting two polar caps back-to-back. This is done commonly in audio power circuits. For example, you can place two electrolytic 22uF caps back-to-back to make an 11uF non-polar capacitor. :smile:
 
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  • #5
you can place two electrolytic 22uF caps back-to-back to make an 11uF non-polar capacitor. :smile:

This is really awesome to know! It makes sense, but I'd never be sure enough of how the caps work to actually try it. Thanks!
 
  • #6
CWatters
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Just for info....

In your first post you made an error in the calculation.

1/(20*π*1,000) = 16uF not 16pF
 
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  • #7
Just for info....

In your first post you made an error in the calculation.

1/(20*π*1,000) = 16uF not 16pF
Very right, thank you!
 

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