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I'm trying to work out how to bias a dual gate MOSFET that I'm working with. Its behavior is halfway between depletion and enhancement modes. That is, its ideal VG range is about -1.5V up to about 0.5V. It looks like it needs VG-S to be biased to about -0.7V to work best (linearity/gain). In particular it seems that the modulation effect (multiplying, rather than adding, the signals) happens best at pretty specific bias voltages, especially on G1.

I have an idea that if I take the D.C. current from the source through a diode, I might be able to "jack the source up" by 0.7 volts, and then by leaving the gate lightly tied to ground, I would have the bias that I need, but I have a nagging feeling that I don't think I've ever seen this done, and I'm suspecting that I'm having a moment of gross foolishness. Can this work, or am I being very silly?

I realize it's more traditional to do this with a resistor, but as I understand it, the quiescent current seems to vary pretty widely with these devices, so that would be hard to choose, and it seems like I need to be particularly picky with the bias voltages in this particular device for the particular use I'm working on. The quiescent current itself seems far less important in this case.

The general idea is sketched below (and no, that's not the device, yes, I know that device symbol represents an enhancement mode device, but no, there doesn't appear to be a depletion mode device in the library, and heck, this thing's kinda halfway anyway :)

Thanks for any input!

schematic

simulate this circuit – Schematic created using CircuitLab

Toby Eggitt
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1 Answers1

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because a resistor reacts more to changes in current than a diode does, a resistor will give you a more predictable quiescent current

this is the priciple of negative feedback the wronger it is the stronger is the correcting force/

Because the higher the quescent current through the resistor the more that the gate is biased towards off, this resulting in less variation in quescent current because the gate bias automagically adjusts.

If you build it with a diode instead you get a thermometer and RF detector that also varies more between different MOSFETS.

Jasen Слава Україні
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  • Interesting. Can you elaborate on the RF detector concern? Does not the shunt capacitor kill that? Also, I edited my question to clarify that, based on data sheet and usage, it *appears* that the bias voltage is much more significant than the actual quiescent current. Does this change anything? Using a resistor, won't the voltage shift pretty significantly between devices? Perhaps I misunderstand the device of course (dual-gate, as a modulator). – Toby Eggitt Jan 16 '22 at 14:58
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    it probably does (as it will most-likely be a low-ESR type) , and it's probably not a diode with good RF performance. but on the other hand most people have a 2W transmitter in their pocket. – Jasen Слава Україні Jan 16 '22 at 21:13
  • Aaaand.. this sub-project is a 28 MHz AM modulator, so there's that. I guess I could go with multiple power rails, though I was hoping to avoid that complexity. – Toby Eggitt Jan 17 '22 at 18:16