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I found this formula used in a circuit to calculate the RC delay time. Can someone please explain what it means?

*Delay time = - 14.7K * 4.7uF * ln(1 - (1.2/5))= 18.96ms

Thank youenter image description here

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The standard formula for charging a capacitor via a resistor is this: -

$$V_C = V_S(1-e^{\frac{-t}{RC}})$$

Where \$V_C\$ is the voltage on the capacitor,

and \$V_S\$ is the supply voltage (5 volts in your circuit example).

So, rearranging you get: -

$$t = -RC\cdot \ln(1 - \frac{V_C}{V_S})$$

So the delay time is how long it takes the capacitor voltage to reach 1.2 volts (the enable threshold nominal value it appears to be) when supplied from a 5 volt input voltage. This dictates how long it takes the 3.3 volt rail on the G571195T (unknown to me and a bit of a guess) to activate.

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A better equation is :-

t = RC.ln(Vi/Vf)

Where:-

t = Capacitor charging or discharging time.

Vi = The initial voltage across the charging resistor.

Vf = The final voltage across the charging resistor.

This equation, which can be used to find both charge and discharge times of the capacitor, makes life more simple than using the two separate more conventional equations relating to capacitor charge and discharge times.

This equation, unlike the equation in your question, can be used to calculate the capacitor charge time between two different arbitrary voltage levels. For example, deriving the Tmark and Tspace equations for a 555 timer in astable mode for initial and final charging resistor voltages of 2Vcc/3 and Vcc/3 respectively.

Note that Vf cannot be 0V as division by zero is not allowed. Vf equal to 0 would mean that the capacitor has charged 100% (right up to the supply) which in theory (and practice) can’t happen. This limitation is also present in your equation.

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