jjagaimo

@jjagaimo@lemmy.ca

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jjagaimo,

You could use a single MOSFET on the high side of the divider and use a cap + diode to boost the voltage and fully turn on the FET:

https://lemmy.ca/pictrs/image/8b22113e-3ad3-40c1-80ee-e7b3e6be24cb.png

My gate driver is fairly crude but you could probably make something a bit better with a PNP transistor and either pull it down or leave it floating, or instead use a szaiklai pair

jjagaimo,

It is an N channel FET. The concept is called “bootstrapping” since Vgs needs to be greater than Vth for the MOSFET to be on. When the FET is on the high side and you want the full 9V on the output, you use the diode to charge the capacitor, and the other side of the cap is 0V. Then, when the other side of the cap is connected to 9V, the charge on the cap can’t go anywhere so the voltage on the other side jumps to 18V. This creates a Vgs of 9V. Ideally you would have something better to drive the gate to fully turn off the FET, but I just used a quick and dirty driver where the bootstrap capacitor directly feeds the gate instead of being the input to the driver. Because if this, the Vgs doesn’t drop completely to 0

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