Actually any MOSFET will have a switching losses, every time you turn it fully on or off, so the low-side MOSFET as well.

Apart from the higher current it can deliver the gate driver is really just there to make the upper gate drive signal independent from the zero reference. That's how you can use a much higher voltage on your H-bridge than would otherwise be possible.

Hi , erm , can u explain a bit that what did u mean by "the gate driver is really just there to make the upper gate drive signal independent from the zero reference."I am a bit confuse on this. Thank you.

 

The gate of the upper N-fet needs to be 15V above the high voltage rail, and the maximum voltage between gate and source is typically +/-20V otherwise the mosfet will become dead very soon.
This means you need a completely separate supply to power the top driver. This could be a standalone transformer generating 15V supply for the top rail, or it can be the bootstrap capacitor, which gets charged when the bootom fet is on, and after that it provides the voltage required to turn the upper fet on.

 

The gate of the upper N-fet needs to be 15V above the high voltage rail, and the maximum voltage between gate and source is typically +/-20V otherwise the mosfet will become dead very soon.
This means you need a completely separate supply to power the top driver. This could be a standalone transformer generating 15V supply for the top rail, or it can be the bootstrap capacitor, which gets charged when the bootom fet is on, and after that it provides the voltage required to turn the upper fet on.

Hi, actually, what do u mean by gate of the upper N-fet needs to be 15V above the high voltage rail ? what us voltage rail ? Vsupply ? because a lot a people say that the gate must be higher than the source voltage in order to turn on the voltage . But now my Vsupply is 100V and it is impossible for the gate voltage to reach above 100V right ? This is because the gate mosfet voltage maximum value is just about 15V. Am I correct ?

 

switching sequence:

left up: bootstrap cap charges, FET does not conduct
left down: FET conducts
right up: bootstrap discharges, FET conducts
right down: FET does not conduct

left up: bootstrap cap discharges, FET conducts
left down: FET does not conduct
right up: bootstrap charges, FET does not conduct
right down: FET conducts

 

Hi, actually, what do u mean by gate of the upper N-fet needs to be 15V above the high voltage rail ? what us voltage rail ? Vsupply ? because a lot a people say that the gate must be higher than the source voltage in order to turn on the voltage . But now my Vsupply is 100V and it is impossible for the gate voltage to reach above 100V right ? This is because the gate mosfet voltage maximum value is just about 15V. Am I correct ?

Yes I mean the 100V Vsupply.

That is exactly what the high side driver and bootstrap capacitor achieve. When the bottom transistor is on, the source of the top transistor is basically at 0V and the lower side of the bootstrap cap as well, so the cap charges to 15V through that diode. Then the bottom transistor turns off.

This leaves the source of the top transistor be at "some" potential anywhere between 0V and 100V, but there is still 15V stored in the bootstrap cap, so the driver can apply 15V to the gate relataive to its source no matter where the transistor is floating right now. As soon as it turns on, the source is at 100V and the gate is at 115V.

 

Yes I mean the 100V Vsupply.

That is exactly what the high side driver and bootstrap capacitor achieve. When the bottom transistor is on, the source of the top transistor is basically at 0V and the lower side of the bootstrap cap as well, so the cap charges to 15V through that diode. Then the bottom transistor turns off.

This leaves the source of the top transistor be at "some" potential anywhere between 0V and 100V, but there is still 15V stored in the bootstrap cap, so the driver can apply 15V to the gate relataive to its source no matter where the transistor is floating right now. As soon as it turns on, the source is at 100V and the gate is at 115V.

Hi, so u means that the gate mosfet can support up to 115V ? I thought on mosfet gate maximum voltage is only 15-20 volts? thank you

 

switching sequence:

left up: bootstrap cap charges, FET does not conduct
left down: FET conducts
right up: bootstrap discharges, FET conducts
right down: FET does not conduct

left up: bootstrap cap discharges, FET conducts
left down: FET does not conduct
right up: bootstrap charges, FET does not conduct
right down: FET conducts

Hi , Thank you for the diagram. Really appreciate ... thank you....

 

Hi, so u means that the gate mosfet can support up to 115V ? I thought on mosfet gate maximum voltage is only 15-20 volts? thank you

No. The transistor can only support ~20V difference between source and gate. The fact that the source is at 100V above the circuit ground doesn´t change anything in that, because the gate voltage is nowhere referenced to that ground, but it moves along with the source voltage.
Imagine you had a 12V battery instead of the bootstrap capacitor, and left out the bootstrap diode, the situation would be the same.

 

No. The transistor can only support ~20V difference between source and gate. The fact that the source is at 100V above the circuit ground doesn´t change anything in that, because the gate voltage is nowhere referenced to that ground, but it moves along with the source voltage.
Imagine you had a 12V battery instead of the bootstrap capacitor, and left out the bootstrap diode, the situation would be the same.

Hi, actually, i am confuse on this part, it take me about 2 week and i still did not understand well on it.

 

so, which part exactly confuses you?

 

Hi, actually, i am confuse on this part, it take me about 2 week and i still did not understand well on it.

You could write a sequence on how you understand the bootstrap function and we tell you where you are getting it wrong.

1.
The n-channel FET is turned on with 15V from gate to source. It doesn't matter where this voltage is with respect to ground.

2.
Once the bootstrap capacitor charged it acts like a battery. You use this charge to transfer it to the gate capacitance. This turns ON the FET.

 

so, which part exactly confuses you?

Actually, mosfet driver is only use to provide additional peak current to the mosfet gate to perform fast switching. This is because the high side of the mosfet having a high voltage and the source is not connect toward to ground. If we are using micro controller without mosfet driver to provide signal to the gate, it might ruin the mosfet(more heat) because of slow charging time between the gate and source due to low current provided from micro-controller. This is what I understand on why the mosfet driver is needed on high and low side mosfet. About the bootstrap circuitry, it only perform as a additional battery to charge to the gate driver. But, there is a lot of explanation say that,

1. "In order to fully turn on an N-channel MOSFET, you generally need to get the gate in the vicinity of 10V higher than the source. "

2. "Say your V+ is +12V. And your bootstrap diode has a Vf of 0.6v. In that case, when the capacitor is being charged, it is charged to (12-0.6) = 11.4V. When being discharged, there is 11.4v across the cap, and there is 12v on the line. So, voltage at source = 12V on load, and therefore, voltage on gate = (12+11.4) = 23.4V
This is more than 10V greater than your source voltage and therefore, you can successfully turn the MOSFET on in high side configuration.


See the number 2 problem, i know that mosfet gate only can support maximum 15-20V and it will be burn as the voltage exceed above 20V. But now it say that the voltage on gate is 23.4v which had exceed the requirement for gate driver. Is it possible to turn on the mosfet? It will no ruin the mosfet?

Thank you.

 

You could write a sequence on how you understand the bootstrap function and we tell you where you are getting it wrong.

1.
The n-channel FET is turned on with 15V from gate to source. It doesn't matter where this voltage is with respect to ground.

2.
Once the bootstrap capacitor charged it acts like a battery. You use this charge to transfer it to the gate capacitance. This turns ON the FET.

Hi,
First , the bootstrap circuitry is only used to provide peak current to the gate of MOSFET. This is because microcontroller only gives a very low current and it will not completely fully charge on the capacitor between the gate and source. Besides, if we do not use mosfet driver on the high side mosfet, it will ruin the mosfet because of slow switching and produce heat on the mosfet. That why we need the bootstrap circuitry to provide current and perform fast switching to reduce the heat toward mosfet.

If we say provide current to the gate of mosfet by using capacitor , I only know that capacitor is voltage charge and mosfet also is a voltage control device right? Then, why it is not using voltage to provide to the gate instead of current? I am confuse in this part as well. Thank you so much.

 

See the number 2 problem, i know that mosfet gate only can support maximum 15-20V and it will be burn as the voltage exceed above 20V. But now it say that the voltage on gate is 23.4v which had exceed the requirement for gate driver. Is it possible to turn on the mosfet? It will no ruin the mosfet? Thank you.

There´s your problem. The voltage on gate is 23.4V higher than the ground potential. But the mosfet doesn´t know what the ground potential is, it only cares about what is the difference between gate and source, and that is set only by the voltage coming from the high side driver and bootstrap cap.
If the source of terminal of the mosfet is at 12V, then the gate will be at 23.4V. If the source is at 500V, the gate will be at 511.4V, but still there is only 11.4V between gate and source, so the mosfet is safe.

 

There´s your problem. The voltage on gate is 23.4V higher than the ground potential. But the mosfet doesn´t know what the ground potential is, it only cares about what is the difference between gate and source, and that is set only by the voltage coming from the high side driver and bootstrap cap.
If the source of terminal of the mosfet is at 12V, then the gate will be at 23.4V. If the source is at 500V, the gate will be at 511.4V, but still there is only 11.4V between gate and source, so the mosfet is safe.

Hi, what do u mean by "the gate will be at 511.4V " i am confuse on this part.
Thank you

 

Hi, what do u mean by "the gate will be at 511.4V " i am confuse on this part.
Thank you

Look at the following picture. The bootstrap cap has been charged while the lower FET was ON. So now the lower FET is turned offand we turn the upper FET ON. The diode does not conduct, the charge from the bootstrap cap is being transfered to the gate. The voltage from gate to source of the upper FET is 15V, the FET conducts.

Now digest this:
- The voltage from drain to source of the lower FET is now 500V because the upper FET conducts.
- The source of the upper FET is connected to the drain of the lower FET
- therefore the source of the upper FET is also at 500V with respect to ground
- the gate of the upper FET is 15V higher than the source of the upper FET
- the gate of the upper FET is 515V with respect to ground and 15V with respect to the source of the upper FET.

(The voltages indicated are ideal voltages without any losses in the driver or power FETs)

 

Look at the following picture. The bootstrap cap has been charged while the lower FET was ON. So now the lower FET is turned offand we turn the upper FET ON. The diode does not conduct, the charge from the bootstrap cap is being transfered to the gate. The voltage from gate to source of the upper FET is 15V, the FET conducts.

Now digest this:
- The voltage from drain to source of the lower FET is now 500V because the upper FET conducts.
- The source of the upper FET is connected to the drain of the lower FET
- therefore the source of the upper FET is also at 500V with respect to ground
- the gate of the upper FET is 15V higher than the source of the upper FET
- the gate of the upper FET is 515V with respect to ground and 15V with respect to the source of the upper FET.

(The voltages indicated are ideal voltages without any losses in the driver or power FETs)

Hi, Thank you for your explanation. So, the conclusion is the driver is only use to provide 15v to the gate in order to turn on the mosfet as fast as possible. That all. Am I right ?
Thank you for drawing and explain to me along the way. Thank you.