Half Bridge Gate Driver IC Circuit

[veryuniqueid]

I am looking at the Fairchild FAN7382 half bridge gate driver. On page 2 of their data sheet (attached) they show a typical application circuit. In this diagram, there is a resistor R2 from the gate of Q1 (the high side MOSFET) to the source of Q1. Similarly, there is a resistor R4 from the gate of Q2 (the low side MOSFET) to the source of Q2.

What is the purpose of these resistors? Are there any rules of thumb for determining values to use for these?

Ed

 

[itsmoked]

Purpose: To assure turnoff. If the gates are the same potential as the sources then the devices are OFF. These resistors assure that if no drive is present then the FETs are OFF. They are important!

Furthermore as the system is powered up they provide a 'brainless' solution to transient charge from briefly turning ON the devices.

Value: Generally you look at the leakage that can be expected from the design, in this case the driver chip. If there is a stated amount of leakage at the LOW output state, the value of the resistors must be low enough to preclude the FETs from turning ON. If no spec exists you start at something like 47k and go from there. Too low a value and you will affect the drive current negatively, and the resistors will dissipate too much power.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[veryuniqueid]

Thanks for the explanation. Does this also mean that if I want the initial condition to be with the high side FET (Q1) off and the low side FET (Q2) on that I could have R4 pull the gate of Q2 up to 15V?

Using a micro-controller to drive the gate driver IC, it could take tens or even hundreds of microseconds after power is applied to drive the gate output signals. Pulling Q2's gate high while tying Q1's gate to its source through the resistor R2 would put the half bridge into the desired initial state while the micro-controller was going through its power on sequence. If the micro-controller failed for some reason, the switch would then return to this initial state. Although it might take a while to charge up Q2's gate through R4 if R4 is something like 47K.

Ed

 

[MacGyverS2000]

Depending upon how the micro failed, it may short-circuit the switch, so don't count on it in a safety-critical application.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[LionelHutz]

I wouldn't. R3 and R4 make a voltage divider. If R4 connects to +15V then when the output (LO) is turned off (to ground) there will still be some gate voltage on Q2 and Q2 won't be fully off. This will cause shoot through current.

 

[veryuniqueid]

Thanks for the help. I think I'll handle the problem with logic on the input side of the half-bridge driver. It's a little more expensive than the pull-up resistor, but will work in all conditions.

 

[MacGyverS2000]

This design idea might prove useful...

http://www.edn.com/index.asp?layout=article&articleid=CA6558503

Dan - Owner

http://www.Hi-TecDesigns.com

 

[itsmoked]

Hit "skip ad" button.

Keith Cress
kcress -

http://www.flaminsystems.com