I have some question about the bootstrap capacitor on a gate driver.

First, the bootstrap capacitor is used because the voltage on the high-side driver's gate needs to be about 10-15 V higher than the voltage on its drain. However, if my input supply is about 20 V and the gate voltage is not higher than the source voltage as well, can it turn on?

Second, to turn on an N-channel FET, we need a gate voltage which is higher than the source voltage. How can this be? The gate voltage can't supply more than 15 V, right? If my input supply also contributes about 20 V, can it turn on?

  • \$\begingroup\$ @sean900911 For your first question, you haven't specified what type of MOSFET you're trying to control - N-channel or P-channel. The answer depends on the type. For the second question, the maximum gate-source voltage again depends on the specific N-channel MOSFET you're trying to control. Please add more details. \$\endgroup\$ Commented Feb 22, 2013 at 20:49

3 Answers 3


To make the explanation easier, here's the diagram for a typical bootstrap gate driver. Perhaps, the O.P. could post his actual circuit diagram.

enter image description here
The IC in the picture is FAN7842. The next picture is the block diagram of the FAN7842 itself.

enter image description here

Bootstrap gate drive circuits are used with H-bridge and half-bridge MOSFET topologies. The overall idea of the bootstrap gate drive circuits is this:

  1. Initial conditions: Q1 is turned off. Q2 is turned on. The Gate of Q2 is at Vcc.
  2. The bootstrap capacitor Cboot is charged when the lower MOSFET Q2 is conducting and the source of the upper MOSFET Q1 is at a low potential ( VS1≈0 ). Cboot is charged from Vcc through Dboot.
  3. Now, the direction of current through the bridge needs to change. Q2 is turned off by driving it's gate low. The source of Q1 is no longer tied to ground and it floats up. As a result, VS1>Vcc. Cboot remains charged for the time being. Dboot prevents it from discharging into Vcc. Cboot haven't been used for driving the gate of Q1, yet.
  4. The gate drive circuit for Q1 is inside the IC. This special gate drive circuit is not connected to Vcc. It's powered exclusively by Cboot. Also, the value of Cboot is chosen such that it's bigger than the gate capacitance of Q1 ( Cboot>>Cgate ). Now, the Q1 is turned on by connecting its gate to the charged Cboot. The gate capacitance is charged from Cboot, and the gate voltage goes up.
  5. Finally, Q1 is turned off by connecting its gate to its source. Q2 is turned on by driving it's gate to Vcc. This cycle can repeat again.

Below is an oscilloscope screenshot of a gate drive waveform. It was taken with one of my own circuits, not with the FAN7842 circuit above. The principles are the same, though.

The gate drive signals go above the H-bridge supply voltage. Vcc=12V in this circuit. In the waveform, it's the difference between the high state of the gate signal and the H-bridge supply voltage (minus the drop across the Dboot diode).

enter image description here

An important thing about bootstrap gate drive circuits is that the duty cycle has to be D < 100%. It doesn't work at 100%.

In case you already know how charge pump voltage doublers work, you would recognize that the bootstrap gate drive circuit is somewhat similar.

  • \$\begingroup\$ @Kortuk Idea behind capacitor charge pump circuits. A capacitor is charged to a (relatively low) voltage V1. Then the negative side of the capacitor is connected to another voltage V2. As a result, the circuit can produce V1+V2. This action is present in the bootstrap circuit too. \$\endgroup\$ Commented Feb 25, 2013 at 16:34
  • \$\begingroup\$ @Kortuk Buck circuit has an inductor (as well as boost). We are also trying to generate a gate drive signal, which is higher than the supply voltage. On a slightly different note: sometimes a bootstrap gate drive circuit is used to drive the gate of N-channel MOSFET in a buck. \$\endgroup\$ Commented Feb 25, 2013 at 16:39
  • \$\begingroup\$ Ohh crap, I just saw which connection I misread. I blame jippie. Removing my comments. I thought the lo side was somewhere else, very awesome indeed. \$\endgroup\$
    – Kortuk
    Commented Feb 25, 2013 at 21:21
  • \$\begingroup\$ Point 3) Vs1 almost equal to zero means that no current goes through the load, correct? Point 4) Based on what Vs1 > Vcc? At point 3) it was about 0 and now all of a sudden it became so high? How that happened? Just asking! :) \$\endgroup\$
    – Barzi2001
    Commented Aug 18, 2023 at 16:08
  • 1
    \$\begingroup\$ @Barzi2001 For an example let's say that the bridge supply Vsupply=34V (or it can be up to 200V), and Vcc=12V (or 15V). Vsupply > Vcc. Steps 4 and 5 happen in quick succession. There's only a very short dead time when both Q1 and Q2 are off. When Q1 turns on during step 5, Vs1 becomes connected to Vsupply through Q1. That's how Vs1 becomes greater than Vcc when Q1 conducts. \$\endgroup\$ Commented Aug 18, 2023 at 18:59

Your preoccupation is well justified: How to turn on the high side N-MOS if we need a very high voltage at the gate?

At some point someone had the brilliant idea of first charging a capacitor on a separate circuit (with enough Vgs to turn on the transistor, in this case around 15V), then disconnect it from the "charging" circuit (note that the capacitor retains its charge even though it got disconnected), and then place it between Gate and Source of the transistor that is to be turned on. When it is time to turn the transistor off, the capacitor is removed from the gate (leaving maybe a resistor that discharges the gate capacitance) and the process can be repeated when it is time to turn it back on.

This is in essence what the driver circuit does, and for the specifics on how exactly to do the charging/disconnection/connection of this bootstrap capacitor you can refer to Nick's answer.

The reason that the bootstrap capacitance needs to be larger than the transistor's gate capacitance is that CBOOT is charging the gate capacitance, so it needs to have enough charge so that it does not drop too much voltage in doing so, otherwise the transistor would not turn on.

The reason that this does not work with 100% duty cycle is that Cboot will eventually get discharged because of the R2 and any other leakages involved.


this mosfet driver is much better it have a very less rise time compared to fan http://datasheets.maximintegrated.com/en/ds/MAX15018-MAX15019.pdf

the supply voltage to this ic is mosfet max gate voltage (stay 2 volt less than that)

and this also have internal diode

just use a boot capacitor

while choosing a mosfet drivet things to consider 1)n channel or p channel mosfet(n channel for lower side of power supply p channel for positive side)

2)vgs(voltage required to turn on gate for n channel it is positive for p channel it is negetive)

3)resistance of output(this is necessary cause mosfet internal resistance should be 10 time less than output resistance otherwise mosfet will consume a lot of power)

4)switching frequency it depends on driver rise time and gate capacitance of mosfet. usually all mosfet drivers give some data about rise time vs gate capacitance

5)for higher voltages bootstrap n channel is used (more than 25v usually) cause for p channel we may blow up gate while switching every thing else needed is given in data sheet wire it up it would work


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