WHy BJT Drive Mosfet

[ycbq999]

I an just a new grad. This motor controller typical diagram is quite confusing for me. Why does it use two bjts to drive top two mosfets? And to be honest, I do not see how do these two bjts work to turn on the mosfet. I hope someone can explain to me with patient. Thank you.

 

[chuckey]

The BJTs are buffers they work as emitter followers on the AOUT1 pin, when its low the bottom transistor has a b-c short so its emitter, sits at Vb + .8V and when VOUT1 is high its the other way around and the top transistors emitter is at Vb - .8V. The gate input capacitance of a FET is very high and its important that it is discharged rapidly, to stop a "top" FET being held ON by the voltage on its gate capacitance when the "bottom" FET is switched on.
Frank

 

[ycbq999]

HI Chuckey, thank you for your reply. It is really helpful. There is a particular part I am still not sure of. It is the zener diode and diode in parallel with a resistor connected between Base and Collector of the top BJTs. Would that make the Vb constant no matter AOUT1 on or off?

 

[chuckey]

The best I can do is that the zener + diode sets the junction of the bases to 1/2 of Vcc. The diode is for temperature compensation. In the absence of pulses from Aout, both the BJTs are off, the top one conducts when its base is at 1/2Vcc +.8, the bottom when its base is at 1/2Vcc -.8. AS Aout has a finite rise and fall times to its edges, this gives a dead band around 1/2Vcc where both BJTs are off. The effect is that the output at the emitters is a pulse that starts later and finishes earlier then Aout. Its more protection against the top FETs being on when the bottom FETs are on.
Frank