What is the Capacitor and Resistor formula to delay the turn-on, saturation, of a transistor?

Aug 18, 2006 6 Replies

First reply

Rheilly Phoull

There is a secret way of looking for stuff like that it's called "Google". Amazing really, you just enter a subject like "capacitor charge rate" and up comes all this stuff. Not as easy as asking here but worth the effort.

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An Electrolytic Capacitor in series with a Resistor attachced to the base of a transistor would provide a simple timing circuit, yes? What is the formula to derive the needed Farads and Ohms?



There is a secret way of looking for stuff like that it's called "Google". Amazing really, you just enter a subject like "capacitor charge rate" and up comes all this stuff. Not as easy as asking here but worth the effort.

Cheers ......... Rheilly P Where theres a will, I want to be in it.

IIRC -> I=Io*e^(-t/rc)

Where Io is the initial current (through the resistor), and I is the current after time t, (wich decays exponentialy)

Alternativly if you just want to get in the right ballpark the time constant is just R*C, where the voltage falls most of the way.

However if your circuit is as simple as you say I dont think it will work as wanted.

Colin =^.^=

R E >------/\\/\\/\\/\\/-------+----------> Ec Er | | ------- C ------- | | ----- --- - E = 5v t = 1m sec R = 1K C = 1uF

RC = R * C Example: 1K * 1uF = 1m sec note: One Time Constant, 63.2% TC = e-( t/RC ) Example: e (1m sec/1m sec) = 367.88m sec Ec = E * ( 1 - TC ) Example: 5v * ( 1 - 367.88m sec ) = 3.16v Er = E * TC Example: 5v * 367.88m sec = 1.84v

At 2m sec

TC = e( 2m sec / 1m sec ) = 135.34m sec note: Two Time Constants Ec = 5v * ( 1 - 135.34m sec ) = 4.32v Er = 5v * 135.34m sec = 676.67mv

. . . note: Three through five Time Constants

At 6m sec

TC = e-( 6m sec / 1m sec ) = 2.48m sec note: Six Time Constants Ec = 5v * ( 1 - 2.48m sec ) = 4.99v Er = 5v * 2.48m sec = 10mv

Hope this helps.

Regards,

Mr. Bill

"Mr. J D" wrote in message news: snipped-for-privacy@i42g2000cwa.googlegroups.com...

It's just as easy, but nowheres near as much fun! :-)

Cheers! Rich

If it's only turn-on delay you're interested in, and the input voltage is larger (x~6) than the VEBon, then you can do simple approximations.

i = C dv/dt

Vin/R = C dv/dt

dt = C x 0.65 x R / Vin. - units are seconds, farads, volts and ohms

If there's a charge on the timing capacitor (across the EB jn) before the pulse is applied, then dv becomes 0.65- Vinitial.

RL

there's a thing called "time constant" that's calculated by multiplying the resistance (in ohms) by the capacitance (Farads) it'll give a ballpark figure (in seconds).

The circuit you describe will perform differently with different supply voltages and loads, use the time constant as a staring point but be prepared to go up or down by a factor of 10 or more. don't expect a good degree of stability or repeatability.

Using something like a LM555 will give a more predictable result, still nowhere near perfect.

Bye. Jasen

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