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

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?

Reply to
Mr. J D
Loading thread data ...

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.
Reply to
Rheilly Phoull

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 =^.^=

Reply to
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...

Reply to
William Hightower

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

Cheers! Rich

Reply to
Rich Grise

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

Reply to
legg

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
Reply to
jasen

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.