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I've read a few other questions on here which helped me understand analog FET switches better but I still have a couple of questions/clarifications which I'd like to understand better.

Consider a supply of 9 VDC with an AC guitar signal. With reference to the below circuit:

  1. Is R37 necessary? I know that it is there for biasing, but I did an LT Spice simulation without it and the circuit worked the same. I understood that R38 is necessary to have a fixed DC voltage, else the NMOS would conduct on the negative portion of the AC wave.
  2. What is the purpose of D4 at the gate? I think that the RC combo adds some filtering, but I'm still unsure about the pof D4 in this case.

enter image description here

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    \$\begingroup\$ That's not NMOS, it is a JFET, which means it may be a depletion-mode device. \$\endgroup\$ Commented Aug 10, 2022 at 21:39
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    \$\begingroup\$ Just a heads up that LTspice automatically adds tiny conductances to (and across) certain nodes to help them converge. Look up gmin, gfloat, and topologycheck in the LTspice help (under .OPTIONS section) for more info. If you're dealing with very high impedances, you might have to adjust the settings to get a better representation of real life. \$\endgroup\$
    – Ste Kulov
    Commented Aug 11, 2022 at 4:44
  • \$\begingroup\$ This has little to do with LTspice, other than a simple simulation. Your question is more about the actual JFETs than the SPICE representations of them (or simulating). \$\endgroup\$ Commented Aug 11, 2022 at 6:29

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R37 and R38 provide a path for leakage currents to ground. Without them, when the JFET is off, C29 and C28 would be charged by these leakage currents and end up with random voltages on them. Then, turning on the FET would result in a loud THUMP as the capacitor voltage is now being transmitted to the output.

Q8 is a JFET so its gate is insulated by a reverse biased diode, which is reverse biased only when when Vgs is negative (unlike a MOSFET). So if Vgs becomes positive, the gate diode will conduct and the JFET will let gate current through into the signal, which would be undesirable. I guess the diode prevents that.

No diode: JFET gate conducts, letting control signal leak into output.

enter image description here

With diode: JFET gate does not conduct, and input signal is switched.

enter image description here

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  • \$\begingroup\$ 1. With your reasoning, can R37 and R38 be connected to ground then? I think not since R38 is needed to keep the Source at a fixed voltage, else the JFET will conduct on the negative portion of the AC wave. For example would R37 still be needed if the Input signal was already biased at 1/2 VCC? 2. After reading other sources as well, injecting a positive voltage at Vgs might damage the N-Channel JFET or as you explained (some switching voltage can "get through" and appear as an annoying "thump" mixed with signal). Can you provide more information about the diode at the MOSFET gate please? \$\endgroup\$
    – Luke Galea
    Commented Aug 11, 2022 at 9:56
  • \$\begingroup\$ 1. If they were connected to ground, then a negative voltage would have to be applied to the gate to turn the JFET off. If this is a single supply circuit, that's not possible. 2. Link about differences between MOSFET and JFET. 3. There's something weird about this circuit, I think it uses leakage through the diode to turn the FET on. Otherwise how would it ever turn on? \$\endgroup\$
    – bobflux
    Commented Aug 11, 2022 at 10:46
  • \$\begingroup\$ Re no.3 - yes and it also prevents Vgs from being way too over 0V which might damage the FET from what I read (feel free to confirm this). \$\endgroup\$
    – Luke Galea
    Commented Aug 11, 2022 at 11:56
  • \$\begingroup\$ It would only be damaged if gate current exceeded the maximum rating, which is not possible due to 1Mohm resistors. Otherwise the gate just conducts like a diode and dumps the gate current into the signal via drain and/or source. But when that happens it doesn't perform the intended switching function. \$\endgroup\$
    – bobflux
    Commented Aug 11, 2022 at 12:11
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    \$\begingroup\$ Discrete MOSFET has a body diode, so it can't be used alone to block current in both directions. You need two. Look up the MOSFET switches in BMS, and MOSFET solid state relay. In an IC the body diode goes to the substrate, so it can't switch voltage above its power supply or below ground. \$\endgroup\$
    – bobflux
    Commented Aug 11, 2022 at 14:35

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