# If two BJT bases are shorted, where does the voltage come from?

While trying to understand the circuit below in this question I got the idea that PNP base current is going into NPN base terminal. This way, you drive 2 transistors with only one base current. With R3 and R2 you prevent the ringing from dis/charging base capacitance. Simulating this I could get voltages all across the circuit. But how to manually calculate the voltage at PNP and NPN h bridge's bases? My thought is, if transistor are voltage switches, although BJT are current driven, where does the voltage base to saturate they come from ?

• R1 can be shorted and so can D4 , I have seen your circuit in 1982. It does work and was more common before cheap low resistance powermosfets arrived. – Autistic Nov 14 '15 at 11:18
• Your statements doesn't solve my doubt. I'm not trying to optimize this circuit as you did, I'm using this as an example to explain my doubt. Do you have any thoughts about it exclusively? – GabrielRado Nov 14 '15 at 11:26
• R2 or R3 will have most of the 12V supply across them when the respective PWM inputs are activated. – Autistic Nov 14 '15 at 11:33
• The R2/R3 resistors have nothing to do with ringing, they limit (detremine) the base currents of the bridge transistors. – Wouter van Ooijen Nov 14 '15 at 11:45
• What Andy gave you is the way to do it on paper. But it won't be incredibly accurate. For power transistors like 2SA1129 and 2SC2654 you have to use the datasheet saturation curves. And unfortunaly you have mutual non-linear equation... so simulation with good models is pretty much required. But that still won't cover the motor's inductance variation etc. These things are hard to calculate precisely. – Fizz Nov 14 '15 at 12:18