I'm designing a totem-pole by BJTs in order to drive a MOSFET. I studied on several online examples and built up my circuit according to what I understood from them. However, there is a detail which got stuck in my mind. I would like to know why doesn't shoot-through occur in this circuit during the transition time of the clock pulse (e.g.; when \$V_{clk} \tilde= 6V\$)? In other words, why doesn't the two BJTs become turned on at the same time during the transition?
simulate this circuit – Schematic created using CircuitLab
Simulation result:
(Vtp and Vgs overlap.)
Best Answer
These transistors don't conduct unless Vbe>0.6V for the NPN, Vbe<-0.6V for the PNP. And as the bases and emitters are tied together, it is impossible for both these conditions to be true at the same time. So when one transistor is turned on, the other is turned off.
HOWEVER
if R2 is too low, the transistor being turned on will "saturate". And when saturated, it will take a significant time to turn off after the base current is removed. This question and answers discuss one famous solution to that problem.
However the present value of R2 limits the base current, because the voltage across R2 will be relatively low, so the transistors will not saturate hard and will turn off relatively fast.