# Calculating Coupling Capacitor Values

Joined Jun 16, 2023
128
Hi:

I am experimenting with an emitter-follower circuit. The input impedance is 14.236 KΩ
and the output impedance is 32 Ω. The corner frequency is 20 Hz. I have calculated the
input and output coupling capacitor values as follows.

 C1 = 1 / (2 * π * fmin * Rin(total)) = 0.559 μF C2 = 1 / (2 * π * fmin * Rout) = 249 μF

Do these values make sense?

RS

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#### BobTPH

Joined Jun 5, 2013
8,068
The output coupling capacitor needs to be based on the load impedance. Did you do that? I get 7.8uF using 1000 as the R.

Edited to add: Actually both resistors have an effect, but the the 1000 dominates.

#### AnalogKid

Joined Aug 1, 2013
10,776
Assuming you mean a 32 ohm load, such as an earphone, your math is correct. To build it with real-world parts, round up to the next available values. Unless you are trying to match a specific sonic profile, larger coupling caps rarely hurt anything.

A true 32 ohm output impedance is not a good idea. Modern headphones and earbuds are designed for amplifiers wht a low output impedance. Also, your circuit has an asymmetrical output impedance, meaning that the impedance when sourcing current into the load is difference from the impedance sinking current out of the load.

Note that for "flat" response down to some corner frequency, the response actually is 3 dB down at the corner freq. IOW, the output voltage will be only 70% of the "flat" value. One octave away it is down only 1 dB (89%). So if you want the amplifier to be "flat" down to 20 Hz, and if the two corner frequencies are 10 Hz, the response will be down 2 dB at 20 Hz, one dB per highpass filter.

ak

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Joined Jun 16, 2023
128
Just to clarify regarding a Common-Emitter Amplifier.

Zout = RC

Zout = RC

And the ouput coupling capacitor value is calculated using Rload?

Is this correct?

RS

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#### AnalogKid

Joined Aug 1, 2013
10,776
There is no RC on your schematic.

Separate from that - incorrect.

The output impedance of an emitter-follower circuit is not symmetrical. That is, it is a lower impedance when sourcing current through the emitter than when it is sinking current through Re. Most materials assume that the load impedance is much greater than Re, and does not affect the calculations. If this case, both the source and sink output impedances are considered equal.

https://inst.eecs.berkeley.edu/~ee105/sp08/lectures/lecture10_6.pdf

ak

Pedantic tidbit: BTW, the technical name for an emitter follower circuit is a common collector circuit.

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Joined Jun 16, 2023
128
There is no RC on your schematic.

Separate from that - incorrect.

The output impedance of an emitter-follower circuit is not symmetrical. That is, it is a lower impedance when sourcing current through the emitter than when it is sinking current through Re. Most materials assume that the load impedance is much greater than Re, and does not affect the calculations. If this case, both the source and sink output impedances are considered equal.

https://inst.eecs.berkeley.edu/~ee105/sp08/lectures/lecture10_6.pdf

ak
correct me. Thanks.

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#### AnalogKid

Joined Aug 1, 2013
10,776
As I said, your equation for Zout is incorrect. The correct one is in the link I posted.

ak

#### Audioguru again

Joined Oct 21, 2019
6,432
Your schematic shows an emitter-follower. A common emitter transistor is different with its output from the collector.
I increased the load resistance then it works much better, even at 20Hz.

Here are my simulations:

#### Attachments

Joined Jun 16, 2023
128
correct me. Thanks.