If I have a capacitor connected to the emitter of a transistor, with the base biased from a voltage divider and the collector connected to the supply, how can I go about calculating the effective time constant of the circuit? Basically I'm interested in figuring out how long it will take a capacitor of a certain value to charge. My guess is it would involve solving the differential equation \(Ies(e^{(Vb-Vc)/Vt}) = C\frac{dVc}{dt}\), since the emitter current is going to determine the charging current through the capacitor and making the assumption that there is a "stiff" voltage divider. Does that look about right?

 

Pure speculation: there should be no dVc if power supply is well regulated. Charging in two stages, first RC time constant, R = saturated C-E resistance, second when Ve reaches about 1 V of base V, then R starts to increase as base looses drive, so we have an expotential charge of an expotential charge. ??

 

Thank you for your reply. In the equation I have the dVc/dt term not because there's ripple on the supply, but because my intuition is the transistor is acting like a voltage controlled current source which is "pumping" charge into the capacitor. This charge going into the capacitor is causing the capacitor to charge up and the voltage across the capacitor to increase, which simultaneously decreases the base drive voltage. The thing about the two stage approach is that if the base voltage divider can supply enough drive current and the base voltage is lower than the supply voltage I don't think the transistor will necessarily go into saturation - in that case the only thing limiting current through the transistor is the voltage across the capacitor in the emitter circuit itself, right?

I think you're right though that there's going to be an exponential-exponential relationship somehow, though.

 

I doubt you can simply use the diode equation to model the transistor current - after all you are still injecting base current and the transient response will include all regions of transistor operation from saturation to cut-off.

You should try simulating the circuit to better appreciate the overall behaviour.

 

Charging a capacitor in such a way will blow the transistor if the power supply can supply enough current. You can charge the capacitor by connecting it between the collector and Vcc. Put a resistor between the emitter and ground to set up a constant current source. If you leave the circuit like that, the capacitor will charge up to Vcc and the base current will increase because the negative feedback is lost and the current source will be lost too. To avoid that you can put a resistor in parallel with the capacitor as to prevent the capacitor from charging to the full Vcc.