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I'm trying to convert an MXL 603s circuit, which has a non-electret capsule, to an electret version. I'm pretty certain that the only real difference between the two would be the fact that the non-electret capsule needs a polarizing voltage, whereas the electret is already polarized. You can find the schematic for the 603s here.

I'm pretty sure what would need to be removed is the section of the circuit starting from the connection to the capsule, then moving counter-clockwise until you hit R10 (1.5k). I think all of the components in that entire section are for polarizing the capsule properly (although I definitely don't understand the purpose of the 2N5551 transistor and the inductors).

If someone could tell me:

  • Am I correct in thinking that those components are for polarizing voltage, and can be removed?

  • How would I need to change the preamplifier components (the ones that will stay) so that they will work with an electret capsule that has an internal FET? This is the capsule I'll use.

Thanks!

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The 2N5551 and its components are an oscillator and a charge pump that takes the 12 Vdc from the zener supply and boosts it to 40-60 Vdc. You are correct, none of this is necessary with an electret cartridge; 12 V is more than enough to excite the capsule. Keep R10, D4, and C6. Note that D1 and D2 are backwards on the schematic.

The mic needs a bias current. Connect the left side of R13 directly to the zener diode +12 V, and reduce the value to around 15K-22K.

Using the numbers on the datasheet:

Mic current = 0.5 mA

Mic voltage = 1.5 V

Source voltage = 12 V

Voltage across the R13 bias resistor = (12 - 1.5) = 10.5 V

R13 = E / I = 10.5V / 0.0005 A = 21,000 ohms

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  • \$\begingroup\$ Great answer. I realized that I'll be using the 48VDC phantom power supply that comes from the console to power this mic. So I guess I'll just use that same equation but with 48V? Comes out to 93kOhm. Also, how do you know that the diodes are flipped, and what is their purpose/how do they affect signal flow? \$\endgroup\$ – Jack Ritter Feb 2 at 17:39
  • \$\begingroup\$ The 48 V lines also have audio on them. Better to keep the 12 V regulator to cilp out interference and noise. There is +48 V on both audio lines. The diodes OR them together to a common voltage source for the circuit. This attempts to place the current load equally on both lines, minimizing another form of interference and/or feedback. The benefits of a balanced signal pair happen only if everything about the pair is - wait for it - balanced. \$\endgroup\$ – AnalogKid Feb 2 at 17:46
  • \$\begingroup\$ If this answers your original question, please vote up the response and flag it as the answer. \$\endgroup\$ – AnalogKid Feb 2 at 17:49
  • \$\begingroup\$ Ok. Just one last question - once the 12V Zener becomes reverse-biased at at least 12VDC, wouldn't all that DC voltage just go to ground? How does it work to "regulate" the voltage at 12VDC? \$\endgroup\$ – Jack Ritter Feb 2 at 18:14
  • \$\begingroup\$ Even though it is caused by reverse-bias rather than forward-bias, a zener diode functions like a regular diode. If there is enough voltage to start conduction (and a certain minimum amount of current), the voltage across the p-n junction is (relatively) constant regardless of how much current (not voltage) is going through it. This is why there must be an external impedance of some kind to limit the current to a safe level. The excess 36 V is dropped across the resistor feeding the zener. \$\endgroup\$ – AnalogKid Feb 2 at 19:07

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