Question

(5) Capacitor and Transient Analysis 7 marks Fig. P-5A shows a switching circuit with switches Si and S2 that are opened and closed periodically. Fig. P-5B shows the steady state waveform of the output voltage V.(t). Compute the steady state values of V1 and V2. R: Vo(t) Vo(t) Vot 5V V2 --- -- 1kΩ 1uF C Vi Ve 5V Fig. P-5A Fig. P-5B (not to scale) (5a) 3 marks Let the initial voltage of the capacitor be V.(O)= V1. The switch Siis closed and S2 is opened. Compute V.(t) after 1 milli-second. (That is, compute V2 = V.(t=1ms).) (5b) 2 marks Att=lms, the switch Si is opened and S2 is closed. Compute Vo(t) after another 1 ms. (That is, compute V.(2ms). (50) 2 marks In the steady state, V.(2ms) should return to be V. Use results obtained in (5a) and (5b) and find the numerical values of V, and V2.

Answer 1

Question 5(a)

V.0) = V1

R = 1KN = 1 x 10312

C = 1uF = 1 x 10-6F

So the time constant is

T= RC = 1 x 103 x 1 x 10-6

T= 1 x 10-3s = 1ms

The voltage across a capacitor given the initial and final voltage across it is given by

vc(t) = V, -(V; - Videa

Where

is final voltage to which capacitor is getting charged

is initial voltage across the capacitor.

During the time from 0 to 1 ms

vc(1ms) = V2

V.0) = V1 = V

The final voltage to which the capacitor is trying to charge is

V.1 + V 2 = 10V

V = 10V

So

1ms vc1ms) = 10 - (10 - V. De Ims

V2 = 10 - (10 - V.)e-1

10 V = 10-- +-

10e - 10+ V, =

17.1828+V2 V2 2.71828

Question 5(b)

During the time from 1ms to 2 ms

vc (2ms) = V. (2ms)

Vi = V2

The final voltage to which the capacitor is trying to charge is

V.2 = 5V

Vi = 5V

So

V. (2ms) = 5- (5 – V,)eims

V. (2ms) = 5-(5-V,)e-1

V. (2ms) = 5 - +

V. (2ms) = 5e -5 + V

V. (2ms) = 3.161 + 0.368V,

Question 5(c)

V. (2ms) = V1

So

V = 3.161 +0.368 17.1828+V2 2.71828

V = 3.161 +2.326 + 0.135V,

V. = 6.343 V

Substituting this in the equation for V2

17.1828+V2 V2 2.71828

V2 = 8.655 V

This was simulated in NI MULTISIM as follows. [This is not asked]. I just did to verify the answers that we got

. . . . . . . . . . . . . . ' 1 . . . . . . . . . . . . . . . . . . ko 1: 1 . . . . . . . . . . . . . . . ' . PR . 02. .VS1. . .. . R$2. . . . . ..5V. C l . . . . . V3 . . . . . . . . . . . 1uF........ -5V 5V ......., IC=6.343V .. . VS2...... / 1 ms 2ms: ........... .. .5V . . . . . . . . . . . . . . . . . . . . . . . . .

The transient response obtained is

Grapher View File Edit View Graph Trace Cursor Legend Tools Help BMX & WENGA..100 g ou 1 Transient Transient Transient TransientTransient TransientTransient Transient A * - Design1 Transient VO(V) Cursor V (2) V (PRI) 1.0000m 8.6565 2.0000m 6.3433 1.0000m -2.3132 -2.3132k 1.0000k dy dy/dx 1/dx - -- 1m 2m 3m 4m 5m 6m 7m Time (s) V(2) (PR1)

On the red cursor, you can see the value at 1ms. The output voltage at 1 ms is 8.656V.  On the blue cursor, you can see the value at 2 ms. The output voltage at 2 ms is 6.3433V. This matches our calculations.