Question

Begin by connecting the waveform generator to the series combination of a 0.2uF capacitor (use a decade capacitor box) and a 1 ks resistor (use a decade resistor box). Generate a 500 Hz, 8 Vp- p square wave with a 4 Volt offset and observe this waveform on the oscilloscope's Channel 1. Monitor the voltage drop across the capacitor on Channel 2. Figure 1 shows the schematic for the RC circuit 1k M R1 0.20uF AD2: Wave C1 channel 2 Generator Figure 1: RC Circuit Note: An offset square wave input is used so that the student can see results for a natural response (Vin=0 V "short circuit") and a step response (Vin= 8 V). Notice that as the capacitor begins a step response, the voltage drop across it asymptotically approaches 8 Volts. As the capacitor undergoes a natural response, the voltage across it asymptotically approaches 0 Volts.

You can use any software of your choice

Answer 1

1. A clock voltage of 8 Vpp is taken and a sc voktage of 4V is taken for providing dc offset.

Vin VC V R1 w V 1kΩ 2 V1 + 2 8V 500Hz 3 C1 0.2uF V2 4V 0

Waveforms-

Initial capacitor voltage = 4V

Final capacitor voltage= 12V

So time constant will be calculated for time required to reach 63% of 4V+( 12-4)V = 4+0.63*8 = 4+5.04V = 9.04V

TOjent 1 C1 C2 13 —Vin: V(1) —VC: V(2) 12 11 10 9 Voltage (V) 8 7 6 5 4 3 0 1.5m 2m 500u 1m Time (s) Cursor 2. 2.0000 ms, 4.0000 V Cursor 1 199.27 us, 9.0418 V AX 1.8007 ms 1/A 555.

From cursor 1 data time constant = 199 micro sec.

Theoritical time constant = RC = 1k*0.2u

= 0.2msec

= 200micro sec.

2.

General equation for charging-

$V_c=V_o(1-e^{(-t/\tau)})$

Where tau is time constant.

Vo = 8V

Let's check capacitor voltage at t= 200 micro sec.

Vc = 8(1-e^-1)

Vc = 5.05V

Consider dcoffset Vc = 4+5.05 =9.05V

Which matches with graph.

3.

For R =3.9kohm

Time constant will increase so capacitor will charge upto less Voltage than 12V.

Waveform for two cycle-

Transient 1 13 -Vin: V(1) -VC: V(2) 12 11 10 9 Voltage (V) 00 7 6 5 4. 3 O 500u 1m 1.5m 3m 3.5m 4m 2m 2.5m Time (s)

For C =0.1uF

Time constant further decreases by ten times, capacitor voltage reaches to more than previous case-

Transient 1 13 Vin: V(1) -VC: V(2) 12 11 10 9 Voltage (V) 8 7 6 5 4 3 o 500u 1m 1.5m 3m 3.5m 4m 2m 2.5m Time (s)

RL circuit-

PR1 Vin V R1 А 1kΩ 2 + V1 8V 500Hz 3 L1 0.1H V2 4V 0 1

Time constant will be time required for inductor current to reach 9.04A

nsient 1 C1 13 C2 13m -Vin: V(1) ..... PR1:1(R1) 12 12m 11 11m 10 10m 9 9m Voltage (V) 8 8m Current (A) 7 7m 6. 6m 5 5m 4. 4m 3 3m 2m 0 500u 1.5m 1m Time (s) Cursor 1 103.56 us, 9.1633 mA Cursor 2 1.9275 ms, 4.0000 V AX 1.8239 ms 1/C 54

Time constant= 100 micro sec approx

Theoretically= L/R = 0.1/1k = 100 micro sec.

For R=3.9k

Transient 1 13 3.25m -Vin: V(1) ..... PR1: 1(R1) 12 3m 11 2.75m 10 2.5m 9 2.25m Voltage (V) 8 2m Current (A) 7 1.75m 6 1.5m 5 1.25m 4. 1m 3 750u 2m O 500u 1.5m 1m Time (s)

For L=4.7mH(negligible time will be required for steady state reaching)

Transient 1 13 3.25m -Vin: V(1) ..... PR1: 1(R1) 12 3m 11 2.75m 10 2.5m 9 2.25m Voltage (V) 8 2m Current (A) 7 1.75m 6 1.5m 5 1.25m 4 1m 3 0 750u 2m 500u 1.5m 1m Time (s)