publicweb.unimap.edu.my ~paul pdfs ent162 ent162 assignment 2 ans

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Assignment 2 1. An SCR half-wave rectifier has a forward breakdown voltage of 150 V when a gate current of 1 mA flows in the gate circuit. If a sinusoidal voltage of 400 V peak is applied, find: 1. Firing angle 2. Average output voltage 3. Average current for a load resistance of 200 Ω 4. Power output Assume that the gate current is 1mA throughout and the forward breakdown voltage is more than 400 V when I g = 1mA Solution. V m = 400 V, v=150 V, R L = 200 Ω 1. Now Or i.e. firing angle, 2. Average output voltage is 3. Average current, 4. Output power = 2. An a.c. voltage v=240 sin 314 t is applied to an SCR half-wave rectifier. If the SCR has a forward breakdown voltage of 180 V, find the time during which SCR remains off.

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Page 1: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

Assignment 2

1. An SCR half-wave rectifier has a forward breakdown voltage of 150 V when a gate current of 1 mA flows in the gate circuit. If a sinusoidal voltage of 400 V peak is applied, find:

1. Firing angle

2. Average output voltage

3. Average current for a load resistance of 200 Ω

4. Power output

Assume that the gate current is 1mA throughout and the forward breakdown voltage is more than 400 V when Ig = 1mA

Solution.

Vm = 400 V, v=150 V, RL = 200 Ω

1. Now

Or

i.e. firing angle,

2. Average output voltage is

3. Average current,

4. Output power =

2. An a.c. voltage v=240 sin 314 t is applied to an SCR half-wave rectifier. If the SCR has a forward breakdown voltage of 180 V, find the time during which SCR remains off.

Page 2: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

Figure 1

Solution.

The SCR will remain off till the voltage across it reaches 180 V. this is shown in Fig. 1. Clearly, SCR will remain off for t second.

Now

Here

Or

Or

3. The intrinsic stand-off ratio for a UJT is determined to be 0.6. if theinter-base resistance is 10 kΩ, what are the values of RB1 and RB2 ?

Solution.

Now

Or

Also

Or

Page 3: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

And

4. Calculate 1. Input impedance and 2. The voltage gain of the OP-AMP amplifier circuit of Fig.2.

Figure 2

Solution.

The input impedance of the OP-AMP amplifier is very high and when negative feedback is used, the impedance is increased even further. Hence, input impedance of a non-inverting OP-AMP amplifier can be thought of as infinite.

5. For the inverting amplifier of Fig. 3 R1 = 1 K and Rf=1M. assuming an ideal OP-amp amplifier, determine the following circuit values:

1. Voltage gain

2. Input resistance

3. Output resistance.

Page 4: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

Figure 3

Solution.

It should be noted that we will be calculating values of the circuit and not for the OP-AMP proper.

1.

2. Because of virtual ground at A, Rin = R1 = 1K.

3. Output resistance of the circuit equals the output resistance of the OP-AMP i.e., zero ohm.

6. Find the output voltage of an OP-AMP inverting adder for the following sets of input voltages and resistors. In all cases, Rf=1M.

Solution.

v1 = -3 V, v2 = +3 V. v3 = +2 V; R1 = 250 K, R2 = 500 K, R3 = 1 M

7. In the subtractor circuit of Figure 4, R1 = 5 K, Rf = 10 K, v1 = 4 V and v2 = 5 V. find the value of output voltage.

Figure 4

Solution.

Page 5: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

8. A 5 mV, 1kHz sinusoidal signal is applied to the input of an OP-AMP integrator of Fig.5 for which R=100 K and . find the output voltage.

Figure 5

Solution.

The equation for the sinusoidal voltage is

Obviously, it has been assumed that at t=0,

9. The input to the differentiator circuit of Fig.6 is a sinusoidal voltage of peak value 5 mV and frequency 1 kHz. Find out the output if R = 100 K and .

Page 6: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

Figure 6

Solution.

The equation of the input voltage is

As seen, output is a cosinusoidal voltage of frequency 1 kHz and peak value .

10.A certain differential amplifier has a differential voltage gain of 2000 and a common mode gain of 0.2. determine CMRR and express it in dB.

Solution.

11.A differential amplifier has an output of 1 V with a differential input of 10 mV and an output of 5 mV with a common-mode input of 10 mV. Find the CMRR in dB.

Differential gain, mV

= 100

Common-mode gain,

Page 7: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

12.The differential amplifier shown in Figure 7 has a differential voltage gain of 2500 and a CMRR of 30,000. A single-ended input signal of 500 µ2 r.m.s. is applied. At the same time, 1V, 50 Hz interference signal appears on both inputs as a result of radiated pick-up from the a.c. power system.

Determine the common-mode gain.

Find the CMRR in dB

Determine the r.m.s. output signal.

Determine the r.m.s interference voltage on the output.

Figure 7

Solution.

In Figure 7, the differential input voltage is the difference between the voltages on input 1 and that on input 2 is grounded, its voltage is zero.

Differential input voltage= 500 - 0 = 500

The output signal in this case is taken at output 1.

Page 8: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

The common-mode input is 1 V r.m.s. and the common-mode gain is .

Noise on the output =

13.Determine the bandwidth of each of the amplifiers in Figure 8. Both OP-AMP have an open-loop voltage gain of 100 dB and a unity-gain bandwidth of 3 MHz.

Figure 8 Figure 9

Solution

For the noninverting amplifier shown in Figure 8, the closed-loop voltage gain (ACL) is

For the inverting amplifier shown in Fig. 9

Page 9: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS

Bandwidth, BW=

14.A three stage OP-AMP circuit is required to provide voltage gains of +10,-18 and -27. Design the OP-AMP circuit. Use a 270 feedback resistor for all three

circuits. What output voltage will result for an input of 150

Figure 10

Solution.

Designing the above OP-AMP circuit means to find the values of R1, R2. And R3. The first stage gain is +10 so that this stage operates as noninverting amplifier.

Now

The second-stage gain is -18 so that this stage operates as an inverting amplifier.

The third-stage gain is -27 so that this stage operates as an inverting amplifier.

Overall voltage-gain, A = A1A2A3 = (10) x (-18) x(-27) = 4860

Output voltage,

Page 10: Publicweb.unimap.edu.My ~Paul Pdfs ENT162 ENT162 Assignment 2 ANS