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

Discussion in 'Electronic Basics' started by Mr. J D, Aug 18, 2006.

1. ### Mr. J DGuest

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?

2. ### kfelGuest

T=RC, where R=Resistance in Ohms and C=capacitance in farads and T in
secs. Note that this is the time it would take for the capacitor to
reach 63.2% of the supplied voltage.

3. ### John FieldsGuest

I've arranged your post as if you had bottom posted.

If this is the circuit Mr. J D had in mind: (View in Courier)

|
C
IN>----[R]---+----B
| E
[C] |
| |
GND>---------+------+

then T = RC won't work

Note that the base-to-emitter junction is a diode with its anode
connected to the base and its cathode connected to the emitter,
which won't allow the cap to charge any higher than Vbe.

4. ### William HightowerGuest

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