Electric OU: Supplement: Driving a MOSFET by LOWERING SOURCE VOLTAGE Part 1

Usually we think of turning on a MOSFET by raising its Gate voltage to 10 or 12 volts above the negative rail/source voltage. But what if the Gate is pinned to circuit ground or the negative rail itself?

Vgs is a relative voltage difference. If the Gate is at zero voltage, how can we arrive at a +10 Vgs in order to turn on the transistor? By lowering the Source voltage. A Gate of +10 and a source of 0 produces a Vgs of +10 volts, duh. But a Gate of 0 volts and a Source of -10 volts also produces a Vgs of +10 volts, and the mosfet will indeed turn on.

There are other considerations; the circuit I show in the video actually puts the main and bias supplies in series, and the current through the transistor also flows through the R1 potentiometer. The bias battery and potentiometer could be replaced by a Function Generator, clearly, and then it’s easy to see that the transistor drain-source current must also flow through the Function Generator, and that the FG is acting as a power supply in series with the main battery.

Now please go on to Part 2!

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