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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?

Thanks in advance!

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    \$\begingroup\$ Bootstrapping can be used to generate a voltage that is higher than VCC. I do not see that happening in this circuit. Have a look at figure 6 on this page: nutsvolts.com/magazine/article/power_mosfets \$\endgroup\$ – Bimpelrekkie Sep 9 '19 at 10:54
  • \$\begingroup\$ Which circuit? I know my circuit doesn’t work, but why does his work? I know that his works because it works in the video. \$\endgroup\$ – user401445 Sep 9 '19 at 10:55
  • \$\begingroup\$ In a video anything can be faked. Or your circuit is actually different. Check carefully and also read the page I linked above. \$\endgroup\$ – Bimpelrekkie Sep 9 '19 at 10:57
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    \$\begingroup\$ 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' \$\endgroup\$ – Neil_UK Sep 9 '19 at 13:14
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The circuit in the linked video works because the source of the MOSFET attains a potential of about 12 volts and this lifts the lower end of the bootstrap capacitor to around 12 volts and, because the top terminal of the capacitor was previously at about 12 volts, the new voltage at the top terminal of the bootstrap capacitor is raised to about 24 volts.

The capacitor retains pretty much all of its charge (and hence voltage) because the diode that previously allowed it charge is now reverse biased and cannot extract current/energy from the capacitor.

Your circuit won't work because the capacitor you use is firmly tied to ground and is therefore un-bootstrappable.

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