Can anybody help?
sizing a capacitor for a BJT base drive circuit and need to know if the voltage value used is the voltage over the base resistor (Vdrv - Vbe) or just Vdrv, the voltage driving the current at the base.
Tempted to assume the former?
Thanks

 

Tempted to ask for a drawing so I can understand exactly what you are talking about.
(A speed-up capacitor is not an often used term.)

 

Speed up capacitor?
placed across (parallel) the base resistor (controls the static current) at the base of a BJT to improve the transient response of the BJT (acting as a switch obviously). Helps speed up the charge and discharge of the transistors base capacitance to improve its switching times and reduce energy loss.
Sized using C=Q/V where Q is the base capacitance (to the emitter) but not sure on the value of the voltage.
Any ideas?

 

Well yeah. There is something driving the base with a voltage and that voltage will be present across the resistor that is in series with the base. If you're doing 5 volt digital switching, then nearly 5 volts is going to hit the base resistor and therefore 5 volts minus a Vbe is going to hit the capacitor that is in parallel with it.

(Other capacitors that might be called "speed up" capacitors are used in operational amplifier circuits. IIRC it's called, "feed forward compensation", and again, it's about getting the semiconductors to switch faster.)