Download (Solution)Assignment ent 162 ch456

ENT 162 Analog Electronics
Assignment:
Topic: Power Amplifiers, Amplifier Frequency response, Thyristor, Operational
Amplifier
Due Date: 29 March 2010 (Monday)
Power Amplifiers
1. Explain Class B operation.
The class B operation is illustrated in Figure above, where the output waveform is
shown relative to the input in terms of time (t). The Q-Point Is at Cutoff .The class B
amplifier is biased at the cutoff point so that ICQ = 0and VCEQ = VCE(cutoff)' It is brought
out of cutoff and operates in its linear region when the input signal drives the
transistor into conduction. This is illustrated in Figure below with an emitterfollower circuit where, the output is not a replica of the input.
2. Describe crossover distortion and its cause.
Crossover Distortion. When the dc base voltage is zero, both transistors are off and
the input signal voltage must exceed VBE before a transistor conducts. Because of
this, there is a time interval between the positive and negative alternations of the
input when neither transistor is conducting. The resulting distortion in the output
waveform is called crossover distortion.
3. Calculate the voltage gain and power gain shown in Figure 1.
Figure 1
Solution:
re’= 0.45Ω
Av = Rc||(re’+RE1)
Ac=50/(0.45+4.7)=9.71
Rin=200.6 Ω-Ap = 189
4. Refer to the class AB Amplifier shown in Figure 2 operating with a single power
supply.
a. Solve the dc parameters VB(Q1), VB(Q2), ICQ, VCEQ (Q1), VCEQ (Q2).
b. Assuming the input voltage is 10 V pp, calculate the power delivered to the
load resistor.
Figure 2.
5. A certain class C amplifier transistor is on for 10 percent of the input cycle. If Vce(sat) =
0.18 V and Ic(sat) = 25 mA, calculate the average power dissipation for maximum
output.
Amplifier Frequency Response
6. Explain why the coupling capacitors do not have a significant effect on gain at
sufficiently high-signal frequencies.
7. Calculate the Miller input and Miller output capacitance shown in Figure 3 and
Figure 4.
Figure 3
Solution:
Figure 4
Solution:
8. A certain amplifier exhibits an output power of 5 W with an input power of 0.5 W.
Solve the power gain in dB.
Solution:
9. Determine the critical frequencies associated with the low-frequency response of
the BJT amplifier shown in Figure 5. Identify the dominant critical frequency. Sketch
the Bode plot.
Figure 5
Solution:
10. Determine the bandwidth of the amplifier shown in Figure 5.
Incomplete. Excluded.
Thyristor
11. Explain how an SCR is used to control current.
Solution:
An SCR circuit that permits current to be switched to a load by the momentary
closure of switch SW I and removed from the load by the momentary closure of
switch SW2. Assuming the SCR is initially off, momentary closure of SWI provides a
pulse of current into the gate, thus triggering the SCR on so that it conducts current
through RL. The SCR remains in conduction even after the momentary contact of SWI
is removed if the anode current is equal to or greater than the holding current, IH.
When SW2 is momentarily closed, current is shunted around the SCR, thus reducing
its anode current below the holding value, IH. This turns the SCR off and reduces the
load current to zero.
12. Explain the turn on and turn off operation of an SCS in terms of its transistor
equivalent.
Solution:
The basic operation of the SCS can be understood by referring to the transistor equivalent,
shown in Figure a.
To start, assume that both QI and Q2 are off, and therefore that the SCS is not conducting. A
positive pulse on the cathode gate drives Q2 into conduction and thus provides a path for Q.
base CUITent. When Ql turns on, its collector CUITent provides base current for Qb thus
sustaining the Oil state of the device. This regenerative action is the same as in the turn-on
process of the SCR and the 4-1ayer diode and is illustrated in Figure b. The SCS can also be
turned on with a negative pulse on the anode gate, as indicated in Figure a. This drives QI
into conduction which, in turn, provides base current for Q2' Once Q2 is on, it provides a
path for Q, base current, thus sustaining the on state. To turn the SCS off, a positive pulse is
applied to the anode gate. This reverse-biases the base-emitter junction of QI and turns it
off. Qb in turn, cuts off and the SCS ceases conduction, as shown in Figure b. The device can
also be turned off with a negative pulse on the cathode gate, as indicated in part (b). The
SCS typically has a faster turn-off time than the SCR.
13. In a certain UJT, rB1’ = 2.5 kOhm and rB2’ = 4 kOhm. Calculate the intrinsic standoff
ratio.
Solution:
Operational Amplifier
14. Derive the expression of inverting and noninverting amplifiers.
Solution: inverting amplifiers
Noninverting amplifiers
The product AolB is typically much greater than 1, so the equation simplifies to
Therefore ,
15. A noninverting amplifier has an Ri of 1.0 k Ohm and Rf of 100 k Ohm. Calculate Vf
and B if Vout = 5 V.
Solution:
Vf = 1000 / (1000 + 100000) 5 = 0.05 V
B = 1000 / (1000 + 100000) =9.9E-3