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I have a question regarding using an n-channel mosfet in the high side with a charge pump bootstrap.

Lets assume the following:

• Vgs max is 15v
• Vgs(th) is 2v (logic level)
• voltage pressent at source pin is 10v
• charge pump is charged to 20 volts
• all grounds are common

Now, the voltage at the gate must be higher than the source to turn on. Since the source pin is at 10v, the gate must be at least 2 volts (Vgs(th)) higher. So you have to apply at least 12 volt to the gate, so that the fet can turn on. Now if i apply the output of my charge pump to the gate with the help of a transistor ( and here is the real question ) what voltage does the gate of the fet see ?

Is it 20 volts ? Bad because Vgs max is 15v. Or does it see 10 volts ? (20v pump - 10v source = 10v gate ?)

And if it sees only 10v, why or how ? I ask why, because i dont understand why its 10v even though im applying 20v.

Second question: Would it be still 10v at the gate if i use an external 20v battery with no common ground. + of the battery to the gate and - to the source ( which is at 10 volts )

Thanks

Edit: Added schematics for this example.

Edit2: Added second schematics for my second question. (ofc there isnt a common gnd, but just saying)

I have a question regarding using an n-channel mosfet in the high side with a charge pump bootstrap.

Lets assume the following:

• Vgs max is 15v
• Vgs(th) is 2v (logic level)
• voltage pressent at source pin is 10v
• charge pump is charged to 20 volts
• all grounds are common

Now, the voltage at the gate must be higher than the source to turn on. Since the source pin is at 10v, the gate must be at least 2 volts (Vgs(th)) higher. So you have to apply at least 12 volt to the gate, so that the fet can turn on. Now if i apply the output of my charge pump to the gate with the help of a transistor ( and here is the real question ) what voltage does the gate of the fet see ?

Is it 20 volts ? Bad because Vgs max is 15v. Or does it see 10 volts ? (20v pump - 10v source = 10v gate ?)

And if it sees only 10v, why or how ? I ask why, because i dont understand why its 10v even though im applying 20v.

Second question: Would it be still 10v at the gate if i use an external 20v battery with no common ground. + of the battery to the gate and - to the source ( which is at 10 volts )

Thanks

Edit: Added schematics for this example.

Edit2: Added second schematics for my second question. (ofc there isnt a common gnd, but just saying) Edit3:( Only the +15 volts gnd is the same gnd as the 10 volts battery )

I have a question regarding using an n-channel mosfet in the high side with a charge pump bootstrap.

Lets assume the following:

• Vgs max is 15v
• Vgs(th) is 2v (logic level)
• voltage pressent at source pin is 10v
• charge pump is charged to 20 volts
• all grounds are common

Now, the voltage at the gate must be higher than the source to turn on. Since the source pin is at 10v, the gate must be at least 2 volts (Vgs(th)) higher. So you have to apply at least 12 volt to the gate, so that the fet can turn on. Now if i apply the output of my charge pump to the gate with the help of a transistor ( and here is the real question ) what voltage does the gate of the fet see ?

Is it 20 volts ? Bad because Vgs max is 15v. Or does it see 10 volts ? (20v pump - 10v source = 10v gate ?)

And if it sees only 10v, why or how ? I ask why, because i dont understand why its 10v even though im applying 20v.

Second question: Would it be still 10v at the gate if i use an external 20v battery with no common ground. + of the battery to the gate and - to the source ( which is at 10 volts )

Thanks

Edit: Added schematics for this example.

I have a question regarding using an n-channel mosfet in the high side with a charge pump bootstrap.

Lets assume the following:

• Vgs max is 15v
• Vgs(th) is 2v (logic level)
• voltage pressent at source pin is 10v
• charge pump is charged to 20 volts
• all grounds are common

Now, the voltage at the gate must be higher than the source to turn on. Since the source pin is at 10v, the gate must be at least 2 volts (Vgs(th)) higher. So you have to apply at least 12 volt to the gate, so that the fet can turn on. Now if i apply the output of my charge pump to the gate with the help of a transistor ( and here is the real question ) what voltage does the gate of the fet see ?

Is it 20 volts ? Bad because Vgs max is 15v. Or does it see 10 volts ? (20v pump - 10v source = 10v gate ?)

And if it sees only 10v, why or how ? I ask why, because i dont understand why its 10v even though im applying 20v.

Second question: Would it be still 10v at the gate if i use an external 20v battery with no common ground. + of the battery to the gate and - to the source ( which is at 10 volts )

Thanks

Edit: Added schematics for this example.

Edit2: Added second schematics for my second question. (ofc there isnt a common gnd, but just saying)

I have a question regarding using an n-channel mosfet in the high side with a charge pump bootstrap.

Lets assume the following:

• Vgs max is 15v
• Vgs(th) is 2v (logic level)
• voltage pressent at source pin is 10v
• charge pump is charged to 20 volts
• all grounds are common

Now, the voltage at the gate must be higher than the source to turn on. Since the source pin is at 10v, the gate must be at least 2 volts (Vgs(th)) higher. So you have to apply at least 12 volt to the gate, so that the fet can turn on. Now if i apply the output of my charge pump to the gate with the help of a transistor ( and here is the real question ) what voltage does the gate of the fet see ?

Is it 20 volts ? Bad because Vgs max is 15v. Or does it see 10 volts ? (20v pump - 10v source = 10v gate ?)

And if it sees only 10v, why or how ? I ask why, because i dont understand why its 10v even though im applying 20v.

Second question: Would it be still 10v at the gate if i use an external 20v battery with no common ground. + of the battery to the gate and - to the source ( which is at 10 volts )

Thanks

I have a question regarding using an n-channel mosfet in the high side with a charge pump bootstrap.

Lets assume the following:

• Vgs max is 15v
• Vgs(th) is 2v (logic level)
• voltage pressent at source pin is 10v
• charge pump is charged to 20 volts
• all grounds are common

Now, the voltage at the gate must be higher than the source to turn on. Since the source pin is at 10v, the gate must be at least 2 volts (Vgs(th)) higher. So you have to apply at least 12 volt to the gate, so that the fet can turn on. Now if i apply the output of my charge pump to the gate with the help of a transistor ( and here is the real question ) what voltage does the gate of the fet see ?

Is it 20 volts ? Bad because Vgs max is 15v. Or does it see 10 volts ? (20v pump - 10v source = 10v gate ?)

And if it sees only 10v, why or how ? I ask why, because i dont understand why its 10v even though im applying 20v.

Second question: Would it be still 10v at the gate if i use an external 20v battery with no common ground. + of the battery to the gate and - to the source ( which is at 10 volts )

Thanks

Edit: Added schematics for this example.