What is the value of R and C if we want 2 mSec Dealy from 4.2V Input. IC will get enable if Voltage is more than 1.5V.

 

What is the value of R and C if we want 2 mSec Dealy from 4.2V Input. IC will get enable if Voltage is more than 1.5V.

Check out the RC formulas and explanations here: http://www.electronics-tutorials.ws/rc/rc_1.html

Specifically, I think you'll want to use this formula and solve for RC.

 

What is the value of R and C if we want 2 mSec Dealy from 4.2V Input. IC will get enable if Voltage is more than 1.5V.

What is the amount of current you want delivered to the load?
What is the voltage at the load at steady state?
Knowing these two you can establish the maximum value of R.
Then you use the formula for RC charging circuit to determine C.

 

HI

You can use this formula:

T=-ln((V-Vc/V))*R*C

V is input voltage
Vc is voltage at Time T

So for V=4.2 and Vc = 1.5, I believe it works out to R=47K, C=0.1uF, T=2ms delay

eT

 

HI

You can use this formula:

T=-ln((V-Vc/V))*R*C

V is input voltage
Vc is voltage at Time T

So for V=4.2 and Vc = 1.5, I believe it works out to R=47K, C=0.1uF, T=2ms delay

eT

I came to almost the same conclusion. Solving for RC, I got:
RC= -t/Ln(1-Vc/Vs)
RC = 0.00452

I think in your formula you need parentheses around "V-Vc" so that it's (V-Vc)/V instead of (V-Vc/V)

 

A time-constant of 4.42ms looks correct.

 

I came to almost the same conclusion. Solving for RC, I got:
RC= -t/Ln(1-Vc/Vs)
RC = 0.00452

I think in your formula you need parentheses around "V-Vc" so that it's (V-Vc)/V instead of (V-Vc/V)

Yes...needs the parens. Should be:

T=-ln((V-Vc)/V)*R*C

Still resolves to 2ms, but if use your formula, I get 4.7ms.

When I simulate to check, Vc is about 1.45v at 2ms with 4.2v supply