1
\$\begingroup\$

My beginner’s electronics book has now started to teach me about microphones, specifically Electret microphones.

While searching the web for more info, I stumbled upon an article where the author mentioned that when a microphone has high impedance it is critical to keep the cables connected to the microphone as short as possible or else the signal would degrade significantly.

I have been trying to make sense of such statement but so far I have failed miserably. No matter how I think about this I can’t imagine why a cable with such small impedance compared to a microphone’s huge impedance would matter at all.

Could someone please help me undestand this phenomenon?

In case it helps, here is a link to the YouTube video where this is mentioned.

Thanks.

\$\endgroup\$
6
\$\begingroup\$

The problem is the cable's capacitance, which is proportional to its length. This capacitance works in conjunction with the high source impedance of the microphone to create a low-pass filter that can severely attenuate the higher frequencies that you're interested in.

For example, if your microphone has a 100K output impedance, and your cable has 1000 pF of capacitance, together they create an R-C time constant of 100 µs, which corresponds to a cutoff frequency of about 15 kHz. Anything above this frequency will be attenuated at a rate of 6 dB/octave. Longer cables mean more capacitance, reducing the cutoff frequency proportionally.

\$\endgroup\$
  • 2
    \$\begingroup\$ I expect noise coupling is another reason to keep the leads on a high-impedance microphone short. \$\endgroup\$ – akohlsmith Aug 6 '15 at 21:49
  • \$\begingroup\$ @Dave Tweed How do you infer the time constant from the impedance values ; and subsequently the cutoff frequency and attenuation from the value of the time constant? \$\endgroup\$ – K. Rmth Aug 7 '15 at 9:22
  • 1
    \$\begingroup\$ @K.Rmth $$Z_0=\sqrt\frac{L}{C}$$ This is ofcoruce for losless line where L and C are both expressed as F/m and H/m - distributed capacitance/inductance. \$\endgroup\$ – Golaž Aug 7 '15 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.