I am studying a buck converter using an NMOS for the high-side switching.

The inverter is supplied by a 3.7 V battery and should power a 0.6 Ω load. The switching is done by a 100 kHz 0-3.3 V PWM generated by a microcontroller. The duty cycle is around 50%.

I suspect a bootstrap is used to drive the high-side MOSFET ON but I don't fully understand it. Can someone tell me if this circuit works and if so, how? (I suspect the inductor value is also too low, causing strong current ripples (around 0.9 A))

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2 Answers 2


Your MOSFET is wired as a source follower and this means that without boot-strapping it will have poor efficiency. The gate has to rise several volts higher (usually 10 volts higher) than the source to properly turn on the MOSFET and allow full conduction in other words.

Given that your circuit can only drive the gate to 3.7 volts (at best) and, given that the threshold voltage for very light current conduction (1 μA) of the NDS7002A is 2.1 volts typically, you are likely to get very poor results.

The inductor does appear to be a little low in value but, concentrate on getting the MOSFET boot-strapped. That's your big issue.

Maybe this will help (a synchronous buck converter): -

enter image description here

  • The lower MOSFET M1 is easily turned on because its source is connected to ground.
  • So, when M1 activates, capacitor CB is charged to Vcc by connecting CB's negative terminal to ground and, the positive terminal of CB is connected to Vcc via the diode.
  • After M1 deactivates and M2 initially activates, M2's gate voltage is at Vcc and is enough to start activating it.
  • As M2 turns on, it rapidly lifts its source and, in doing so, it will lift the bootstrap capacitor's negative terminal.
  • In lifting CB's negative terminal, CBs positive terminal (which is Vcc higher) becomes even higher than Vcc and starts activating M2 even more.
  • As M2 fully turns on, CBs, positive terminal will be at "High voltage" plus Vcc and, of course, this satisfies the requirements to fully turn on M2 (its source voltage equals its drain voltage).

It's all about storing enough charge in CB so that it can power M2's driver circuit inside the chip.


Does it have to be N-MOS?

Note that Vgs determines on/off state. Not Vg to GND. When on, Vs = Vin, so Vg = Vin + Vgs(on).

So we use a bootstrap circuit to supply the gate voltage while it's on. Fine.

But this can only happen if there is something to charge the bootstrap in the first place. Not fine.

In a synchronous buck converter, we can have the low side fire first (or disable the high side until this has happened), ensuring charge. In the diode version, we have no such luck. Perhaps only on first startup, the output is at zero (the load has discharged the capacitor), and the bootstrap charges through the load. (This requires Cboot ≪ Cout, which is normally the case.) But that's not all. Sufficient load current must always be flowing, so that the catch diode is active, drawing more than enough current to keep the bootstrap charged. If load current drops low enough, for long enough, the bootstrap runs out and never recovers.

We can partially solve this with a pull-up resistor, and optionally an output pull-down in case there isn't always a load attached. This works for Vout near zero, but Vout subtracts from Vin into the bootstrap and eventually we'll run into the same problem as before. (This is essentially immediate for the single cell battery case, but can be feasible for higher input voltages, like say 100V in, 12V bootstrap, and output up to, maybe 75V?)

Better is to use a charge pump, or fully isolated supply, so the gate drive is always powered. In the charge pump case, an AC source powers a capacitor, the far end of which goes to clamp diodes at the gate drive circuit (the basic half-wave voltage doubler arrangement). This loses voltage from the diodes, again not very battery friendly, but is reasonable for say 10-15V supplies and Vin up to 200V or so (notice that the swing of the switch node itself, injects much more charge through the capacitor -- putting a huge burden in terms of peak current flow on both the AC source and clamp diode load; this makes it unsuitable for high voltages). With an isolator, just any DC-DC supply, flyback or whatever, will do; Vgs(on) can then be other than battery voltage, allowing for example non-logic-level parts (10V drive could be an option; not that it will be necessary here).

Best, however, will simply be using a P-ch MOSFET. While performance is poorer than N-MOS, this is a fairly small penalty at these low voltages. Gate drive can be direct from a controller, driver IC, or even MCU if you're doing it that way, and the driver can be direct powered by the battery (assuming battery or 5V compatible supply range). (If using 3.3V logic, a simple e.g. 74HC3G04 can be used as level translator + driver, using one as inverter and two in parallel as driver. Or all three in parallel and not minding the inversion.)


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