# Bootstrap capacitor

Discussion in 'General Electronics Chat' started by sean900911, Feb 14, 2013.

1. ### sean900911 Thread Starter Member

Jan 28, 2013
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0
Hi, I had some question about the bootstrap capacitor on gate driver.
first , the bootstrap capacitor is used because the voltage on the high side drivers gate need to be about 10-15 volts higher than the voltage on its drain. However , if my input supply is about 20v and the gate voltage is not higher than the source voltage as well . Is it possible to 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 cannot supply more than 15v right? if my input supply also contribute about 20v, can it be turn on?

Thank you so much for your reply ... Really appreciate for your all helping.
Thank you

2. ### praondevou AAC Fanatic!

Jul 9, 2011
2,939
490
Did you have a look at the application note AN978?

The bootstrap capacitor itself provides the high current for the gate. It acts as a temporary power supply which is being charged when the lower MOSFET is ON.

The path it is charged is from the 15V IR2110 power supply through a high voltage diode to the + side of the capacitor, then from the - side through the conducting lower MOSFET back to the 15V power supply.

What do you mean by "input power supply"?

Let's determine two terms:

VDD: power supply of the IR2110 = 15V
Vpower: power supply for the load (connected to the H-bridge) <500V

3. ### sean900911 Thread Starter Member

Jan 28, 2013
35
0

Hi , thanks for your reply . First , im connecting 4 quadrant full bridge power mosfet as well. I found that , a lot of theory state that " to turn on an n-channel FET , a gate voltage must higher that the source voltage" . So, if Vpower =12 and VDD=15 and it surely can turn on the mosfet as well. But how about if Vpower=50 and VDD=15 and it sure cannot turn on the mosfet right? because the gate voltage is not higher that the source voltage right? I also attach the schematic diagram of my circuit and please point my mistake if i was wrong. Thank you for your patient with me . Thank you so much.

Jan 28, 2013
35
0
5. ### shortbus AAC Fanatic!

Sep 30, 2009
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This is what the bootstrap part of the circuit does. It allows the gate to go to the higher voltage needed.

When the source of the high side is not conducting it allows the bootstrap cap to charge to Vdd level. When the source is conducting, the plate of the boot cap connected to the source then charges to Vpower, displacing the Vdd charge into the gate.

6. ### sean900911 Thread Starter Member

Jan 28, 2013
35
0

Thank you for your reply, so this means that the voltage value charge to the gate is depend on the Vpower voltage value? Thank you .

Jul 9, 2011
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8. ### sean900911 Thread Starter Member

Jan 28, 2013
35
0
Hi ,if Vpower =12 and VDD=15 and it surely can turn on the mosfet as well. But how about if Vpower=50 and VDD=15 and it sure cannot turn on the mosfet right? because the gate voltage is not higher that the source voltage right? Although your gate voltage VDD =15 and Vpower is greater than Vgate right ? Because a lot of theory state that " to turn on an n-channel FET , a gate voltage must higher that the source voltage" but now my source voltage is higher than the gate voltage, how it can be turn on? Thank you for your patient with me , please point my mistake if i was wrong.Thank you

9. ### praondevou AAC Fanatic!

Jul 9, 2011
2,939
490
Vpower can be up to 500V.

The bootstrap capacitor voltage is floating with respect to -DC when the lower MOSFET is OFF.

And yes, the upper MOSFET can be driven by the IR2110 even if Vpower (+DC in the picture) is 50V, or 200V, or 500V....

Try to understand how the bootstrap capacitor is charged. I think it is pretty much well explained in the link I sent.

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