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A common-emitter amplifier can include a bypass capacitor from the emitter to ground, resulting in considerable extra gain (1000uF in the picture). What happens if the bypass capacitor is switched "ON" and "OFF" using an N-channel MOSFET, as shown in the picture? Will the MOSFET produce any distortions to the audio signal?

I've read that MOSFETs alone are not good switches for audio signals, particularly when voltage goes above 0.6V and the FET is OFF. But voltage here going through the bypass capacitor would be max. 200mV.

PD: Spice simulations and breadboarding seem to work fine, but I'm still unsure.

enter image description here

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4 Answers 4

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This will work perfectly fine. As long as you don't make the MOSFET's body diode conduct with a reverse voltage greater than a few hundred mV, there won't be any difference between a simple NMOS transistor and a proper transmission gate / analog switch.

In other words: Just keep the peak-to-peak amplitude of the signal at the BC549's emitter below 600mV or so (300mV in either direction).

If you want the circuit to function with slightly larger signals as well, you could do something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

This makes the circuit's biasing more robust as the common-emitter amplifier's gain is now 1 at DC (instead of 10), and it also decreases the voltage across the MOSFET, giving you more headroom before the body diode begins to conduct.

If you can fit an additional MOSFET, you can also use this configuration to avoid the body diode problem entirely:

schematic

simulate this circuit

This switch configuration only requires the gate voltage to get significantly higher than the maximum signal voltage in order to turn it on, and lower than (or equal to) the signal voltage to turn it off. For example, if you have a signal that swings between 0V and 2V at the BJT's emitter, a gate voltage swing of 0V / 12V would be more than sufficient to cleanly turn the two-MOSFET switch on or off.

Note, however, that you might need a stronger gate driving signal to avoid charge coupling from the floating source node into the gate. You could, for example, drive the gates hard to GND / VCC with a SPDT switch. If you do this, you should still add a pull-down resistor (i.e. 100k) to prevent the gates from floating during the switch transition. Additionally, you'll need to use a break-before-make switch.

It's also quite important for this configuration that the position of the capacitor is swapped as shown, as this prevents negative voltages from developing at the two-MOSFET switch's terminals. Otherwise you'd need a negative gate voltage to turn it off reliably.

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  • \$\begingroup\$ With a gain of 10 and a supply of 12V, this is not guaranteed if given a large input signal. \$\endgroup\$ Commented Nov 21, 2023 at 8:29
  • \$\begingroup\$ I get the point and your example is inspiring @Jonathan S. It gave me the idea of having multiple gain stages along the bypass line. I wonder if ie. using two MOSFETs with their sources connected (back to back) would eliminate the effect of the diode, effectively allowing large signals. \$\endgroup\$
    – Domingo
    Commented Nov 22, 2023 at 0:07
  • \$\begingroup\$ @Domingo Yes, you can connect two MOSFETs in an anti-serial configuration (source-to-source, gate-to-gate) to make a bidirectional switch. This is how FET-based solid-state relays are built. The gate voltage must be lower than (or equal to) both drain voltages to turn it off, and significantly higher than both drain voltages to turn it on. \$\endgroup\$ Commented Nov 22, 2023 at 14:00
  • \$\begingroup\$ @Domingo I've updated the answer to include the alternative two-MOSFET switch configuration. Let me know if anything's unclear about it! \$\endgroup\$ Commented Nov 22, 2023 at 15:11
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    \$\begingroup\$ The 1M resistor is a bleeder that prevents negative voltage from building up on the floating source node due to capacitive charge pump effects when the drain voltages oscillate. It's probably fine (tm) without it, but adding it won't hurt and will improve reliability. \$\endgroup\$ Commented Nov 22, 2023 at 15:38
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It is not the capacitor that is causing the increase in circuit gain, it is the 51 ohm resistor. (Reference designators - ?)

At audio frequencies, the cap is pretty much a dead short. It allows the 51 ohm resistor to be connected in parallel with the 100 ohm resistor without upsetting the DC bias point of the circuit. Decreasing the emitter resistance increases the gain of the stage. In your circuit, but about 3x.

As long as you have a SPST switch, why not just do the switching with the switch and eliminate the FET. Replace the FET with the switch between the resistor and cap, and delete the 1 M resistor.

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    \$\begingroup\$ I should have added in the question that I need to switch 3 amps simultaneously, for a quite miniature application. Thanks for the clear explanation on how the 'bypass network' works. \$\endgroup\$
    – Domingo
    Commented Nov 21, 2023 at 0:08
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At the moment you flip the switch, your audio will instantaneously distort, since you are changing the operating point of the signal. Assuming you mean steady state distortion, there are a couple considerations: ideally, a MOSFET can be used as a switch in this type of application, but you have to remember a couple things. First of all, a FET has some drain-source and drain-gate capacitance, meaning even when it's off, it will act like a capacitor. If you get enough movement on your drain, it will couple to your gate and turn the FET on instantaneously. This is even worse at higher frequencies. You can solve this by driving the gate off strongly. I would recommend using an SPDT switch so that when you switch the FET off, it stays off.

The other big thing is that even when turned off, there is a diode from source to drain (more accurately body to drain and body to source, but in a discrete 3 terminal device, body is shorted to source internally). That means if your drain goes below your source voltage, it will pull current from the source. You always have to be careful when working with capacitors for this reason. Generally, you want your VGS to be as large as your FET allows for max conduction in triode, so you would put your source to ground and your drain to the bottom plate of the cap. If you do this, you may want to put something like a 10k resistor in parallel with the cap so that it's bottom plate is pulled high and it's not rectifying every cycle if the cap is charged and the bottom plate moves high. When the FET is on, it will just be in parallel with the 100 Ohm, which is to say it will be negligible.

What is the advantage of using a FET for this vs a normal switch in this application?

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    \$\begingroup\$ The advantage comes because I need to control three amplifiers simultaneously. The gate is switched ON at 9V by a SPDT, a 1M resistor to GND keeps it from floating. Appreciate your general remarks, I'll consider as you say placing the FET between cap and GND a better practice, with a parallel resistor for discharge. \$\endgroup\$
    – Domingo
    Commented Nov 21, 2023 at 0:01
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You should switch NMOS and capacitor places, so it will always conduct in both directions, otherwise your gate-source will float and may distort. Anyway 1K is way too much for 3A, that's 3kW of power. If you need power amplifiers you should look for AB class at the very least or use class D.

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    \$\begingroup\$ Welcome too EE.SE. How did you arrive at the conclusion that there would be 3A flowing in this circuit? \$\endgroup\$ Commented Nov 21, 2023 at 16:16
  • \$\begingroup\$ Thanks. Mistook comment author, my bad. Of cource 3A with 1K and 12V is impossible. First part still stands. \$\endgroup\$
    – lokkiser
    Commented Nov 23, 2023 at 6:08

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