Calculate RC time constant

[John0922]

Hi-  Can someone explain how to calculate the RC time constant for this circuit from EEV Blog on YouTube:

https://youtu.be/Foc9R0dC2iI

Multiplying R * C doesn't work. I also tried finding a Thevinin equivalent by loading the circuit into a simulator and measuring voltage and current. Dividing, I came up with a resistance of 101 K but this doesn't work either.   Thanks in advance for the help.

-John

[rstofer]

The time constant is still RC but you don't have to go very high in voltage to get the transistor to turn on.  About 0.7V up the curve ought to work.
No, I'm not going to solve the exponential equation to figure out how long it would take mostly because I don't know the input voltage.
Here's a neat calculator:
http://ladyada.net/library/rccalc.html

I used 12V for the voltage input, 0.7V for the voltage output and 1000k ohms plus 22 ufd.
The calculator gets 1.32 seconds.

ETA:

Dave points out that there is some current flow into the base of the transistor which limits how fast the capacitor can charge.  It's not clear to me how he got such fast timing, I guess I'd have to model it and see what is actually happening on the capacitor.

[rstofer]

Furthermore, one RC time constant only gets you 63% of the way to the input voltage.  It takes about 6 time constants to get very close to the input voltage. But the capacitor voltage will never get that high because of the base-emitter junction diode clampling it at 0.7V.

[John0922]

Thank you for your reply. So, do I begin by multiplying 1 M x 47u ? And then, realizing that the capacitor is only charging to about .6 V, this means the time will be probably less than one time constant?

Thanks again.

[John0922]

I just used the calculator you linked and the result was perfectly consistent with the simulator. Thanks so much!

[rstofer]

I'm always happy when the simulation comes close.

You could always graph the exponential function and then pick off the point where it goes to 0.7V.  That was a chore I wasn't up for.

[Ian.M]

One thing that simulators and fancy graphing calculators has cost us is a facility with rough 'back of the envelope' calculations.

When the supply voltage is much greater than the voltage the capacitor charges to, you can treat it as if it was being charged by a constant current.  Even with only 5V supply, so the final voltage (0.7V) is greater than 10% of the supply, if you treat the voltage rise as linear take the average voltage across the charging resistor as 5-(0.7/2) to get i, and from Q=C*V, substituting i*t for Q, solve for t, it comes out within a small fraction of a percent of the result obtained by solving or modelling the exponential equation.

When the final voltage is 1/3 of the charging voltage, the error is a bit over 1% and as 1% components are as precise as you are ever likely to see in a RC circuit, you should probably treat that as the limit for using the linear approximation.

[John0922]

Great point, Ian. I gave some thought to how I would approximate the answer, but had no idea how to approach it. Thanks.