Bootstrap circuit function

Question

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?

Answer 1

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.

The IC in the picture is FAN7842. The next picture is the block diagram of the FAN7842 itself.

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:

2. Initial conditions: Q1 is turned off. Q2 is turned on. The Gate of Q2 is at V_cc.
3. The bootstrap capacitor C_boot is charged when the lower MOSFET Q2 is conducting and the source of the upper MOSFET Q1 is at a low potential ( V_S1≈0 ). C_boot is charged from V_cc through D_boot.
4. 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, V_S1>V_cc. C_boot remains charged for the time being. D_boot prevents it from discharging into V_cc. C_boot haven't been used for driving the gate of Q1, yet.
5. 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 C_boot. Also, the value of C_boot is chosen such that it's bigger than the gate capacitance of Q1 ( C_boot>>C_gate ). Now, the Q1 is turned on by connecting its gate to the charged C_boot. The gate capacitance is charged from C_boot, and the gate voltage goes up.
6. Finally, Q1 is turned off by connecting its gate to its source. Q2 is turned on by driving it's gate to V_cc. 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. V_cc=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 D_boot diode).

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.

Answer 2

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 C_BOOT 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.

Answer 3

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