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.

 

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

 

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"

 

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"

Not even that "bit of algebra" available Mathew?

 

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?

 

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

 

\(\frac{v}{5} = 1-e^{\frac{-t}{RC}}
\frac{v}{5}-1=-e^{\frac{-t}{RC}}
1-\frac{v}{5}=e^{\frac{-t}{RC}}
\)
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.

 

Was just looking for the equation and didn't want to have to spell out my limitations. Sorry for coming off badly.

Thank you.

 

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.

 

Doesn't matter. He only has enough fish for a day -- not enough to feed himself long enough to get a degree.

 

Doesn't matter. He only has enough fish for a day -- not enough to feed himself long enough to get a degree.

Too true, my friend, all too true.