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?
- posted
17 years ago
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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