What is the voltage at the top of the H? The supply should be 8 - 10 volts higher that this.

That voltage doesn't matter. It can be anything from 0 V up to the breakdown Vds of the FETs. The gate drive will be supplied from the 12~15V on the other side of the diode, and that will determine the Vgs of the high-side FET. The drain supply has no effect on this. I have personally designed and built H-bridges with the Vgg gate drive supply independent of the Vdd drain supply. The Vdd can be anything at all, so long as the FETs breakdown is not surpassed.

Regards

 

That voltage doesn't matter. It can be anything from 0 V up to the breakdown Vds of the FETs. The gate drive will be supplied from the 12~15V on the other side of the diode, and that will determine the Vgs of the high-side FET. The drain supply has no effect on this. I have personally designed and built H-bridges with the Vgg gate drive supply independent of the Vdd drain supply. The Vdd can be anything at all, so long as the FETs breakdown is not surpassed.

Regards

You can do that if the 12 to 15 volt supply in his schematic is floating with it's negative side tied to the source. I see no indication of that in his schematic.
As shown as the FET begins to turn on the source voltage will rise above the gate voltage and it can turn on no further. If you see something different maybe a picture would help me understand.

 

Most power MOSFETs have an ABS-MAX Vgs rating around 15 - 18V or so, its not a bad idea to clamp the bootstrap rail to protect the gate from excess voltage.

In the circuit I posted the IRF830 has Max Vgs = 20 V and the bootstrap rail is a regulated 12 V which supplies the control part of the circuit so everything is well.

 

That voltage doesn't matter. It can be anything from 0 V up to the breakdown Vds of the FETs. The gate drive will be supplied from the 12~15V on the other side of the diode, and that will determine the Vgs of the high-side FET. The drain supply has no effect on this. I have personally designed and built H-bridges with the Vgg gate drive supply independent of the Vdd drain supply. The Vdd can be anything at all, so long as the FETs breakdown is not surpassed.

Regards

All along I have the impression that maybe I am missing the gist of the question or maybe the reason I am not quite getting the question is because the questioner is somewhat confused. I have to say I am still not quite sure what it is. I may be missing something.

 

All along I have the impression that maybe I am missing the gist of the question or maybe the reason I am not quite getting the question is because the questioner is somewhat confused. I have to say I am still not quite sure what it is. I may be missing something.

Sorry, my bad. It does bootstrap.

 

I cannot upload any pictures from my work PC because of security rules, but the magic is in the diode.
A bootstrap circuit is a classic diode/cap clamping circuit.

The 12-15V supply, call it VGG, is not floating, its negative terminal is grounded. VGG is the supply for the low-side gate drive circuitry.

The bootstrap diode points from VGG to the top plate of the bootstrap cap. The cap's bottom plate is tied to the 1/2 bridge output. Call it OUT. Assume an ideal diode with no voltage drop. OUT goes to 0 volts when the low side driver is on, and the cap charges to VGG. The low-side driver turns off, the high-side driver turns on. (By "on" I don't necessarily mean channel current flows. I mean the channel resistance drops to its lowest value. The switch is "closed".) OUT goes to any VDD that is > 0 V. The voltage from the top plate to the bottom plate of the cap remains at VGG, and the diode reverse biases. The cap is now a floating VGG supply that powers the high side gate drive circuit. The bottom plate is OUT. The top plate is OUT+VGG.

VDD can be ANY voltage greater than ground, and the bootstrap still works. I know it will. I've seen it. I've personally turned the supply continuously down to 0V, and back up, while maintaining VGG at 12V. It works until you get down to a few millivolts.

Regards

 

I cannot upload any pictures from my work PC because of security rules, but the magic is in the diode.
A bootstrap circuit is a classic diode/cap clamping circuit.

The 12-15V supply, call it VGG, is not floating, its negative terminal is grounded. VGG is the supply for the low-side gate drive circuitry.

The bootstrap diode points from VGG to the top plate of the bootstrap cap. The cap's bottom plate is tied to the 1/2 bridge output. Call it OUT. Assume an ideal diode with no voltage drop. OUT goes to 0 volts when the low side driver is on, and the cap charges to VGG. The low-side driver turns off, the high-side driver turns on. (By "on" I don't necessarily mean channel current flows. I mean the channel resistance drops to its lowest value. The switch is "closed".) OUT goes to any VDD that is > 0 V. The voltage from the top plate to the bottom plate of the cap remains at VGG, and the diode reverse biases. The cap is now a floating VGG supply that powers the high side gate drive circuit. The bottom plate is OUT. The top plate is OUT+VGG.

VDD can be ANY voltage greater than ground, and the bootstrap still works. I know it will. I've seen it. I've personally turned the supply continuously down to 0V, and back up, while maintaining VGG at 12V. It works until you get down to a few millivolts.

Regards

It seemed to simple to me because of the opto. I'm used to seeing level shift stuff there.

 

It seemed to simple to me because of the opto. I'm used to seeing level shift stuff there.

An optocouple is a level shifter, heh.

 

It seemed to simple to me because of the opto. I'm used to seeing level shift stuff there.

An optocouple is a level shifter, heh.

Yes, of course. A little slow in this case, but yes.