In the previous RC Charging Circuit tutorial, we saw how a Capacitor, C charges up through the resistor until it reaches an amount of time equal to 5 time constants or 5T and then remains fully charged.

If this fully charged capacitor was now disconnected from its DC battery supply voltage it would store its energy built up during the charging process indefinitely (assuming an ideal capacitor and ignoring any internal losses), keeping the voltage across its terminals constant.

If the battery was now removed and replaced by a short circuit, when the switch was closed again the capacitor would discharge itself back through the resistor, R as we now have a **RC discharging circuit**. As the capacitor discharges its current through the series resistor the stored energy inside the capacitor is extracted with the voltage Vc across the capacitor decaying to zero as shown below.

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As with the previous RC charging circuit, in a **RC Discharging Circuit**, the time constant ( τ ) is still equal to the value of 63%. Then for a RC discharging circuit that is initially fully charged, the voltage across the capacitor after one time constant, 1T, has dropped to 63% of its initial value which is 1 – 0.63 = 0.37 or 37% of its final value.

So now this is given as the time taken for the capacitor to discharge down to within 37% of its fully discharged value which will be zero volts (fully discharged), and in our curve this is given as 0.37Vc.

As the capacitor discharges, it loses its charge at a declining rate. At the start of discharge the initial conditions of the circuit, are t = 0, i = 0 and q = Q. The voltage across the capacitors plates is equal to the supply voltage and Vc = Vs. As the voltage across the plates is at its highest value maximum discharge current flows around the circuit.

With the switch closed, the capacitor now starts to discharge as shown, with the decay in the RC discharging curve steeper at the beginning because the discharging rate is fastest at the start and then tapers off as the capacitor looses charge at a slower rate. As the discharge continues, Vc goes down and there is less discharge current.

As with the previous charging circuit the voltage across the capacitor, C is equal to 0.5Vc at 0.7T with the steady state fully discharged value being finally reached at 5T.

For a RC discharging circuit, the voltage across the capacitor ( Vc ) as a function of time during the discharge period is defined as:

- Where:
- Vc is the voltage across the capacitor
- Vs is the supply voltage
- t is the elapsed time since the removal of the supply voltage
- RC is the
*time constant*of the RC discharging circuit

Just like the previous RC Charging circuit, we can say that in a **RC Discharging Circuit** the time required for a capacitor to discharge itself down to one time constant is given as:

Where, R is in Ω’s and C in Farads.

Then we can show in the following table the percentage voltage and current values for the capacitor in a RC discharging circuit for a given time constant.

Time Constant |
RC Value | Percentage of Maximum | |

Voltage | Current | ||

0.5 time constant | 0.5T = 0.5RC | 60.7% | 39.3% |

0.7 time constant | 0.7T = 0.7RC | 49.7% | 50.3% |

1.0 time constant | 1T = 1RC | 36.6% | 63.4% |

2.0 time constants | 2T = 2RC | 13.5% | 86.5% |

3.0 time constants | 3T = 3RC | 5.0% | 95.0% |

4.0 time constants | 4T = 4RC | 1.8% | 98.2% |

5.0 time constants | 5T = 5RC | 0.7% | 99.3% |

Note that as the discharging curve for a RC discharging circuit is exponential, for all practical purposes, after five time constants a capacitor is considered to be fully discharged.

So a RC circuit’s time constant is a measure of how quickly it either charges or discharges.

Calculate the RC time constant, τ of the following RC discharging circuit.

The time constant, τ is found using the formula T = R x C in seconds.

Therefore the time constant τ is given as: T = R x C = 100k x 22uF = __2.2 Seconds__

a) **What value will be the voltage across the capacitor at 0.7 time constants?**

At 0.7 time constants ( 0.7T ) Vc = 0.5Vc. Therefore, Vc = 0.5 x 10V = __5V__

b) **What value will be the voltage across the capacitor after 1 time constant?**

At 1 time constant ( 1T ) Vc = 0.37Vc. Therefore, Vc = 0.37 x 10V = __3.7V__

c) **How long will it take for the capacitor to “fully discharge” itself, (equals 5 time constants)**

1 time constant ( 1T ) = 2.2 seconds. Therefore, 5T = 5 x 2.2 = __11 Seconds__

Error! Please fill all fields.

what is the minimum time required to charge a capacitor up to its maximum value?

If starting from zero, 5T

Why is the time constant for a discharging circuit higher than that of a charging circuit? I’m confused.

They are the same as the capacitor is fully charged after 5T, and fully discharged after 5T. The time for a capcitor to charge up from 0V to 63% of Vc is equal to 1T. The time taken for a capacitor to discharge down from max Vc to 37% of Vc is equal to 1T.

how to reduce the RC delay in layout after you run the simulation of the circuit, Thank you

what is the relation between the discharging and rc ?smaller value of rc rapid will discharging but how?

How do you move the discharge fomula around to find the discharge time that isn’t an exact time constant to a known voltage value?

-t / thor

VC = Vs x e

where Vs = supply voltage

t = time

thor = time constant (which is found by R x C)

In discharging cicruit – what is the relationship between Voltage on the resistor vs voltage on the capacitor?

(As VC goes up, VR goes down.) Sorry just noticed your question was “Discharging”, so as VC goes down, VR goes up for the same supply voltage. As e = 2.718, its base 10 antilog = 0.434, therefore Vr = antilog(log(Vs) – 0.434x(t/RC))

Is it not that they are the same only with different polarity?

If the half life was found to be 40s in a discharging experiment, what is the time required for the voltage to fall from 24.0 V to 3.0 V?

Why is the rate of the capacitor dicharge not constnat non-mathematically?

Hey Wayne Storr,

First of all, I would like to thank you for the fast reply I’ve got.

Secondly, I have two questions regarding this situation :

Q1. Why wouldn’t the Capacitor discharge till it reaches near 0 ?

Q2. I have seen someone comment in my schematic that, as he stated,” The forward beta is enough big to hold the maximum ic for a few seconds before it starts to fall, this time should be calculated first and added to the exponential equation.” If so, how can I calculate it.

And thanks.

Hey Wanye Storr,

I have built a simple Rc circuit with a transistor to determine the Time t it takes for the capacitor to discharge to it’s lowest Voltage value. Capacitor, resistor, then the base of the transistor. This is the order I have placed them.

Let me tell you my values :

Vs = 9 V

R = 4.7 K ohm

C = 220 uF

So I have used the equation : Vc = Vs *e^(-t/RC) to find t when Vc = 0.7, the voltage to keep the Transistor base working.

0.7 = 9 *e^(-t/1.034), and I get t = 2.641 seconds. Yet, when I try it on the app I have on my android, The Led, which is on the collector side of the transistor, stays led for about 3.6 seconds and so.

What did I do wrong in my calculations?

Hey Hussain, Taking your values of R = 4700 Ohms, C = 220uF, Vs = 9v, Vc = 0.7, then Tau = RC = 1.034s

For discharging: Vc = Vs e^(-t/RC) therefore Vc/Vs = e^(-t/RC)

Converting to natural logs (ln) as ln(e) = 1, gives time, t as: t = RC ln(Vs/Vc) = 1.034 x ln(9/0.7) = 2.64s

Then I guess your “app” isn’t very good.

Hey Wayne Storr,

First of all, I would like to thank you for the fast reply I’ve got.

Secondly, I have two questions regarding this situation :

Q1. Why wouldn’t the Capacitor discharge till it reaches near 0 ?

Q2. I have seen someone comment in my schematic that, as he stated,” The forward beta is enough big to hold the maximum ic for a few seconds before it starts to fall, this time should be calculated first and added to the exponential equation.” If so, how can I calculate it. (Sorry had to copy it because I forgot to press reply.)

And thanks.