# NMOS bootstrapping without an IC

Like the title says, I was trying to bootstrap a high-side NMOS. After some googling I found this video https://youtu.be/zcQV_ZpK1W8 which helped me a lot except for one thing which I cannot understand.

Now, my proposed circuit is this It is pretty much similar to his but with the capacitor connected directly to ground.

When the NPN transistor is ON, the capacitor charges up to Vcc, which is the same for my and his circuit (I think). The problem is when it is OFF, when the load should be ON. What I previously thought was that the capacitor “pushes” Vcc into the MOSFET’s gate, and so does the battery (or whatever voltage source we’re using) through the diode to the gate. Thus, we have 2Vcc at the gate. I now see that this is wrong because my circuit has it as if the battery and the capacitor were connected in parallel, which will NOT give the sum of the emfs.

However, I can’t see why his circuit works. He says at the end of the video that since the load has 12V across it and so does the capacitor which is in series with it, then the end result is that the gate has 24V. How does electricity work “backwards”? If “potential” is basically the energy per unit charge, how does the energy flow backwards from the load to the capacitor to the gate? I am just not convinced. What am I missing?

• It is pretty much similar to his but with the capacitor connected directly to ground That translates to 'it's pretty much similar to his but I've changed the bit that makes it work' – Neil_UK Sep 9 '19 at 13:14