R C Circuit Transient Analysis

Transient Analysis 

Transients are generated in electrical circuits having energy storage elements like inductors or capacitors when sudden changes occur in the operating conditions. This occurs when the circuit is powered up for the first time. A first order circuit has a single energy storage element for example either a capacitor or an inductor. A second order circuit has two storage elements for example both an inductor and a capacitor.

R C Circuit Transient Analysis
Consider the circuit shown in FIG. 1 below:

FIG. 1: Circuit with Series Resistor and Capacitor

Using to Kirchhoff's Voltage Law (KVL) which states the voltage around a loop equals the sum of 
every voltage drop in the same loop. Referring to FIG. 1 it can be said:
V = VR(t) + Vc(t)...........equation (1)

For a capacitor:
Vc(t) = 1/C∫i(t)dt or
dVc(t)/dt = 1/C i(t)
1/C i(t) = dVc(t)/dt
i(t) = C(dVc(t)/dt)..........equation (2)

In this circuit the current that flows through capacitor C also flows through resistor R.
So VR(t) = R i(t) or
VR(t) = RC(dVc(t)/dt)
Equation (1) can be rewritten as:
V = RC(dVc(t)/dt) + Vc(t)
RC(dVc(t)/dt) = V – Vc(t)
RC dVc(t) = (V – Vc(t))dt
RC dVc(t)/(V – Vc(t)) = dt
-RC ln (V – Vc(t)) = t + k.......equation (3)

To determine k a starting condition must be considered. Initially the switch is open and the voltage across the capacitor is zero. Thus at t=0, Vc(t)=0. Replacing these values in equation (3) it can be said that:
-RC ln V = k or
k = -RC ln V............equation (4)

Now substitute the value of k in equation (4) in equation (3):
-RC ln (V – Vc(t)) = t -RC ln V
-RC ln (V – Vc(t)) + RC ln V = t
-RC ((ln (V – Vc(t)) - ln V) = t
ln (V – Vc(t)) – ln V = -t/RC
ln ((V – Vc(t))/V) = -t/RC
Taking the anti-log on both sides
(V – Vc(t))/V = EXP(-t/RC)
V – Vc(t) = V(EXP(-t/RC))
– Vc(t) = V(EXP(-t/RC) – V
Vc(t) = V – V(EXP(-t/RC)
Vc(t) = V(1 – EXP(-t/RC))............equation (5)
Equation (5) is the voltage across the capacitor in relation to time.

From equation (1) that is V = VR(t) + Vc(t) it can be said that:
VR(t) = V – Vc(t)................equation (6)

From equation (5) that is Vc(t) = V(1 – EXP(-t/RC)) or Vc(t) = V – V(EXP(-t/RC))
Equation (6) can be re-written as:
VR(t) = V – V + V(EXP(-t/RC))
VR(t) = V(EXP(-t/RC))..................equation (7)

The current i(t) can be written as:
i(t) = (VR(t))/R

But VR(t) = V(EXP(-t/RC))
i(t) = (V(EXP(-t/RC)))/R
i(t) = (V/R)EXP(-t/RC).............equation (8)

The RC time constant, also called tau or τ, the time constant (in seconds) of an RC circuit, is equal to the product of the circuit resistance (in ohms) and the circuit capacitance (in farads).
τ =RC (seconds)
So after one time constant τ = 1xRC = RC
Vc(RC) = V(1 – EXP(-RC/RC)) = V(1 – EXP(-1)) = V (1 – 0.3679) = V(0.6321) = 0.6321V
VR(RC) = V(EXP(-RC/RC)) = V(EXP(-1) = V(0.3679) = 0.3679V
i(RC) = (V/R)EXP(-t/RC) = (V/R)EXP(-RC/RC) = (V/R)EXP(-1) = (V/R)(0.3679) = 0.3679(V/R).

After five time constants τ = 5xRC = 5RC
Vc(5RC) = V(1 – EXP(-5RC/RC)) = V(1 – EXP(-5)) = V (1 – 0.0067) = V(0.9933) = 0.9933V
VR(5RC) = V(EXP(-5RC/RC)) = V(EXP(-5)) = V(0.0067) = 0.0067V
i(5RC) = (V/R)EXP(-t/RC) = (V/R)EXP(-5RC/RC) = (V/R)EXP(-5) = (V/R)(0.0067) = 0.0067(V/R).

FIG. 2: VC(t) and VR(t) as a Function of Time in Time Constants

Shown below is the circuit current i(t) as a function of time in time constants.

FIG. 3: I(t) as a Function of Time in Time Constants

A simulation was done in LTspice. The resistor R1 was chosen as 10 000Ω. The capacitor C1 was chosen as 100uF. This gives a time constant of τ= R x C = 10 000Ω x 100uF = 1 second. The input pulse starts at 0V, rises to 1V, has a delay of 0.1s, a rise time of 0.01s, a fall time of 0.01s, an on time of 5s, a period of 10s and the cycle is repeated 3 times. The simulation was run for 20 seconds. The circuit is shown in figure 4 below.

FIG. 4: Circuit Simulated in LTspice.

Measurements were made which are shown in figure 5 below. The read trace is the input pulse. The green trace is the voltage across resistor R1. The blue trace is the voltage across the capacitor C1.

FIG. 5: Traces Input Voltage-Red, R1-Green, and C1-Blue Simulated in LTspice.

Shown in FIG. 6 below is the input voltage indicate by the blue trace and the current in R1 is shown by the green trace.

FIG. 6: Traces Input Voltage-Blue and R1 Current-Green Simulated in LTspice.

There is some resemblance between the resistor calculated voltage and current and capacitor voltage and that of the values obtained by the LTspice simulation.