# what is the time delay for a 10M and a 10uf?

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#### shweeeet

##### New Member
???

anyone???

?

Oh dear......whats happened?........its all gone......Arghhh

The time CONSTANT, not the time delay, is C in microfarad x R in megs,
= 10 x 10 = 100secs.

The time 'delay' depends on the circuit it is used in.

I misread the orignal post...... :?

The time delay i.e. Charging Time (t)= 0.693 * R * C

therefore , t = 0.693 * 10 * 10 = 69.3 Seconds

kinjalgp said:
The time delay i.e. Charging Time (t)= 0.693 * R * C

therefore , t = 0.693 * 10 * 10 = 69.3 Seconds
You're talking about the time constant - not the time delay.

A time delay assumes there is a delay before something is actioned. The delay would depend on how far up the charging curve the charge on the capacitor has to be before the delayed device is triggered. For example, if a CMOS gate is being used for the trigger threshold, then it may be anywhere between 30% and 70% of its final value, resulting in about a 4:1 variation in delay time

Ok I got it but what about your time constant i.e. R*C
Which one is correct 0.693*R*C or just R*C ?

kinjalgp said:
Ok I got it but what about your time constant i.e. R*C
Which one is correct 0.693*R*C or just R*C ?

For a cap charging from zero volts to Vf,

v=Vf*(1-e^(-t/R*C)), where v is the voltage across the cap at time t, and e is Napier's constant, e=2.71828......, which is the base for natural logarithms.

Solving for t, t=-R*C*ln(1-v/Vf), where ln is the natural log operator.

When v/Vf=0.5, t=0.693*R*C

In other words, it takes 0.693RC seconds for the capacitor to charge to half the supply voltage (0.5*Vf).
Using the first equation, when t=RC (one time constant), the cap will have charged to 0.632*Vf.

Ron

kinjalgp said:
Ok I got it but what about your time constant i.e. R*C
Which one is correct 0.693*R*C or just R*C ?
The time constant is the product of C and R. I had not seen a reference to 0.693 before and did not realise its significance until Ron's reply.
Ron H said:
....In other words, it takes 0.693RC seconds for the capacitor to charge to half the supply voltage (0.5*Vf).
Using the first equation, when t=RC (one time constant), the cap will have charged to 0.632*Vf
Ron's second sentence is the application of time constant that I remember.

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