Electronic – Floating charge pump for high side N-channel MOSFET bias

The trouble with the approach of dual-using the N channel MOSFET from your switching regulator is that the gate needs to be taken probably at least 5 volts higher than the source to switch it on reasonably.

This is achieved in normal operation because of the bootstrapping (a small amount of switching energy is stolen from the output to produce a power rail that is a few volts higher than your input power rail). The bootstrapping relies on the output regularly switching and charging a capacitor up. Once the output becomes DC, there is no possibility of this idea working.

However, I don't really see that the relay contact couldn't just be placed in parallel with the MOSFET - in effect you short the MOSFET out.

In this example only intrinsic capacitances are used. The capacitance from the gate to the channel is 1pF or 0.5pF from gate to drain and 0.5pF from gate to source.

The capacitors that are attached to the drain and source are 0.5pF as well. So basically we have a capacitive voltage divider with a 1pF gate-channel and a 1pF channel-ground capacitor.

Before all this can happen a channel is required, so we need a threshold voltage to invert the area under the gate. Then the excess voltage (Vin-VT) is split equally between the two 1pF caps, resulting in (Vin-VT)/2 at the drain (and the source) side.

Maybe I should add, that the shown circuit is NOT a charge pump. It is merely used to demonstrate that shortcomings of non-charge-conservative transistors models that exhibit a charge pumping effect. A charge conservative model (and physically correct model) will show no charge pumping effect whatsoever.

To better understand why VT has to be subracted a Q-V curve for the MOS capacitor is shown below.

Below the threshold voltage the charge stays almost constant, the transistor has only a very small capacitance (almost zero). Above VT the relationship between the voltage and the charge is linear. The transistor acts as a capacitor.