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Was reading the following: "In order to deliver its full frequency response, a ribbon microphone needs to see an input impedance of at least 4-5 times its output impedance. For example, if a microphone has an impedance of 300-Ohms, the preamplifier should have an input impedance of 1200 to 1500 Ohms, or greater"

Why would lower impedance input on the mic preamp attenuate frequencies? I'm assuming high frequencies are attenuated?

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Ribbon microphone: -

enter image description here

Firstly, if you load the transformer output too much the high frequency response will be reduced because of parasitic inductances in series with each winding. But there's a more fundamental thing that happens.

The ribbon is moved by changes in air pressure i.e. it is an electrical generator hence too much load current from the output mechanically dampens the ribbon and alters its frequency response.

Information: Ribbon Microphones

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Microphone impedance together with preamp impedance form a voltage divider, where the input signal is taken at preamp impedance. So, assuming that microphone signal is given by Vs, signal sought at preamp input is given by

$$ V = \frac{Vs.Z_{amp}}{Z_{amp}+Z_{micro}} $$

So, as you can see, if Zamp is greater than Zmicro, input signal is almost the same as the signal generated inside microphone

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The example equivalent circuit should give a good idea of what's going on. If you model the output impedance of the mic as having 300 ohms then the current the mic generates produces some voltage. Usually a preamp is a voltage amplifier (rather than a current amplifier). If the preamp is only say 300 ohms as well, then you've effectively halved the voltage you're trying to amplify. On the other hand, if you have a very high input impedance preamp of say infinite, then whatever current the mic outputs drops only over the 300 ohm internal impedance and results in the highest possible voltage thereby giving the largest signal to noise ratio to the pre-amp.

4-5 times the output impedance of the mic will give us a drop of (1/300+1/1200)^-1 or 240 total ohms which is 240/300 or 80% of the ideal output voltage. Doing the same thing for 5 times shows that we would get 83.3% of the ideal output voltage. The authors are basically saying, about 80-83% of the ideal output voltage will give a good enough signal to noise ratio for the preamp to work well without introducing excess noise.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ How does this relate to frequency response? \$\endgroup\$ – Chu Apr 6 '16 at 6:43
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This could be due to the fact that higher frequencies are attenuating faster through the cabling than lower freqs. That makes the fixed length of cable look like a higher resistance at higher freqs. So, considering that the mic output impedance, cable impedance, and amp input impedance at the end of the cable form a voltage divider, AND, assuming the length of cabling is within some standard range, you could come up with a standard rule ("the 4-5 times" for example) for ensuring that the measured voltage across the amp input impedance is high enough above the noise floor to satisfy the response requirement at the highest freq spec.

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That seems a bit dubious as a broad, general axiom. Remember also that no ribbon microphone actually has a 300 ohm output impedance. The ribbon itself may very well have an impedance of < 1 ohm. All ribbon mics have an impedance transformer (or an active circuit) that raises the ultra-low impedance (and ultra-low signal level) up to something useful in the Real World.

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