Need Formula Concerning RC Timing

Thread Starter

Mathew Lewis McGill

Joined Nov 5, 2015
4
I have a circuit that generates a pulse width. What I need is a formula to determine the time it will take for the output to switch given a voltage at the noninverting input to the comparator.

upload_2015-11-5_16-16-12.png
 

Papabravo

Joined Feb 24, 2006
12,514
Let's see a +5V source
R=1.62 MΩ
C=10nF

The response of an RC circuit is

\(v=5(1-e^{\frac{-t}{RC}})
\)
You should be able to answer your own question with a bit of algebra
 

Thread Starter

Mathew Lewis McGill

Joined Nov 5, 2015
4
Yippy, a quiz.
OK, was you hoping that I'd show you? Cause I found that equation; It's all over the place.
I was kind'a hoping to get an equation to solve "t"
 

Papabravo

Joined Feb 24, 2006
12,514
Let's see a +5V source
R=1.62 MΩ
C=10nF

The response of an RC circuit is

\(v=5(1-e^{\frac{-t}{RC}})
\)
You should be able to answer your own question with a bit of algebra
\(\frac{v}{5} = 1-e^{\frac{-t}{RC}}
\frac{v}{5}-1=-e^{\frac{-t}{RC}}
1-\frac{v}{5}=e^{\frac{-t}{RC}}
\)
Take the natural logarithm of both sides
\(ln(1-\frac{v}{5})=ln(e^{\frac{-t}{RC}})
\)
Can you take it from there?
 

Thread Starter

Mathew Lewis McGill

Joined Nov 5, 2015
4
for a voltage of 125mv, the delay is 410.2ms

You're good with algebra, but seriously struggle with providing a simple straight answer.

By the way, I plugged the original equation in Excel and massaged the numbers.

If it's not too painful, I'd appreciate that equation now.

Thanks
 

Papabravo

Joined Feb 24, 2006
12,514
\(\frac{v}{5} = 1-e^{\frac{-t}{RC}}
\frac{v}{5}-1=-e^{\frac{-t}{RC}}
1-\frac{v}{5}=e^{\frac{-t}{RC}}
\)
Take the natural logarithm of both sides
\(ln(1-\frac{v}{5})=ln(e^{\frac{-t}{RC}})
\)
Can you take it from there?
Guess not.
\(ln(1-\frac{v}{5})=\frac{-t}{RC}
-RCln(1-\frac{v}{5})=t
\)
Are you really sure you could not have done that?
Remember ln is the natural logarithm, or log to the base e (≈2.718281828...), and log is logarithm to the base 10. Seriously struggle with providing a straight answer? If I show you, but you don't understand how it is done, you'll just keep having to ask other people to solve your problems for you. Struggle a bit on you own and you'll remember how to do it next time it comes up. We're here to help not spoon feed answers to you.

Also -- try not to bite the hand that feeds you next time. Petulance is unbecoming.
 
Last edited:

WBahn

Joined Mar 31, 2012
24,854
Congratulations on tricking someone into doing your homework for you. Doesn't happen too often around here, but it does happen from time to time.
 
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