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Old 16th Jun 2006, 1:57 pm   #21
YC-156
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Default Re: Noise Gate

Here is a quick suggestion for a circuit, drawn using Notepad(TM) technology.

Will be back later with comments and explanations if needed.

Frank N.
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Old 16th Jun 2006, 9:14 pm   #22
YC-156
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Default Re: Noise Gate

Mkay, no comments?

OK, here is how it works:

V1 is the preamp for the full wave rectifier V4, which will determine when we switch the output on and off. Supply voltages are just examples, and needs adjustment for a particular situation.

V2 is the combined noise gate and simple class A audio amp.

V3 is an inverting DC amplifier and "X" is the mythical Aditron device.

Imagine no signal enters at Line In. Then the rectifier has no signal to rectify, and its output is zero. Note that the rectifier is configured to generate a negative(!) voltage proportional to the input signal.

With V4 not generating any (negative) voltage, V3 is essentially free to do as it sees fit. Due to the negative voltage rail R5 is in series with V3's positive supply and is part of the anode load.

If we imagine the current through "X" to be initially zero, it will suffer the full 100V supply voltage and 'break down', Ie. it starts to conduct. V3 will then limit the current to some sane level due to its cathode resistor generaing a suitable negative grid bias voltage.

With V3 conducting, the intersection between R4, R5 and "X" will be dragged down to some negative voltage below zero, since the supply current for V3 runs from the ground rail through R5, "X" and down to V3.

The trick now is to balance the resistance of R5, the voltage drop across "X" and V3 and R10 (the cathode resistor), such that the R4 / R5 / "X" intersection is dragged down to a suitably low level, say minus 25ish volts, when V3 and "X" is conducting.

V2 never draws any grid current. So the voltage, if any, at the bottom end of R4 is felt in full force by V2. With V3 conducting V2 is completely cut off with -25V or so on the grid.

Ie. the gate is closed with no or a low level input signal present at Line In.

As the Line In signal increases, the rise of which can be adjusted via VR1, V4 starts to generate an increasingly negative voltage onto the grid of V3. This in turn causes V3 to move ever closer to its particular anode current cut off point. As the anode current decreases, the voltage at the anode increases in proportion, leaving an ever smaller voltage surplus to drive a current through R5 and "X".

(Note that V4 generates a negative voltage with respect to the negative supply rail, here at -100V. Thus the grid of V3 goes below -100V. )

At some particular point the voltage across "X" becomes too small, and it turns off. The V3 anode current drops to zero as long as V4 keeps generating a solid negative voltage.

When "X" turns off, it and V3 magically disappears from the grid circuit of V2, since neither draws any current. The net effect is that suddently V2 feels it has a grid leak resistor of resistance R4 + R5 connected straight to the signal ground rail. (Since the current through R5 is zero, no voltage drop is generated across it.)

V2 now works as an ordinary class A buffer stage with cathode bias generated by R6. The gate is open, the music plays.

Phew!

There are a few tricky details with the C2/R9 and R4 / V2 grid cap time constants, plus getting everything to balance. But if a suitable device "X" can be located, I think this ought to work.

Or perhaps not?

Best regards

Frank N.

NB: V3 and V4 must share a separate, floating heater supply. If they were allowed to run off the same supply, which powers V1, V2 and the rest of the amp, there would in essence be a voltage of around 100V or possibly even more between the heaters and the cathode sleeves of V3 and V4. That is almost certainly beyond the permissible limits.
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