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1. Draw the output voltages of each rectifier for the indicated input voltages, as shown in Figure 1.The 1n4001 and 1n4003 are specific rectifier diodes?

Figure 1

Solution

The peak output voltage for circuit (a) is

() = () 0.7 = 5 0.7 = 4.30 where () = peak output voltage

() =peak input voltage

The peak output voltage for circuit (b) is

() = () 0.7 = 100 0.7 = 99.3 The output voltage waveforms are shown in Figure 2. Note that the barrier potential could have

been neglected in circuit (b) with very little error (0.7 percent); but, if it is neglected in circuit (a), a

significant error results (14 percent).

Figure 2 output voltages for the circuits in figure 2. Obviously, they are not shown on the same scale.

2. Determine the peak value of the output voltage for figure 3 if the turns ratio is 0.5.

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Figure 3

Solution

() = () = 156 The peak secondary voltage is (()) ,

() = () = 0.5(156 ) = 78 The rectified peak output voltage is

() = () 0.7 = 78 0.7 = 77.3 Where () is the input to the rectifier.3. (a) Determine the peak value of the output voltage for figure 4 if n=2 and () = 312 ?

(b) What is the PIV across the diode?

(c) Describe the output voltage if the diode is turned around.

Figure 4

Solution

() = () = 312a) The peak secondary voltage is

() = () = 2(312 ) = 624

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The rectified peak output voltage is

() = () 0.7 = 624 0.7 = 623.3 Where () is the input to the rectifier.

b) PIV =()

= 624 V

c) The output voltage if the diode is turned around will give negative half-cycles rather thanpositive half cycles

4. (a) show the voltage waveforms across each half of the secondary winding and across RL when a100V peak sine wave is applied to the primary winding in figure 5

Figure 5

Solution

(a) the transformer turns ratio n= 0.5. The total peak secondary voltage is

() = () = 0.5(100 ) = 50 There is a 25 V peak across each half of the secondary with respect to ground. The output load

voltage has a peak value of 25 V, less the 0.7 V drop across the diode. The waveforms are shown in

Figure 6

Figure 6

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(b) Each diode must have a minimum PIV rating of

PIV = 2() + 0.7 = 2(24.3 V) + 0.7 V = 49.3 V5. Determine the peak output voltage for the bridge rectifier in figure 7. Assuming the practical

model, what PIV rating is required for the diodes? The transformer is specified to have a 12 V

rms secondary voltage for the standard 110 V across the primary.

Solution

Figure 7

The peak output voltage (taking into account the two diode drops) is

() = 1.414 = = 1.414(12 ) 17 () = () 1.4 = 17 1.4 = 15.6

The PIV rating for each diode is

PIV = () + 0.7 = 15.6 V + 0.7 V = 16.3 V6. Describe the output voltage waveform for the diode limiter in Figure 8

Figure 8

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Solution

The circuit is a positive limiter. Use the voltage-divider formula to determine the bias voltage.

= + = 220

100 + 22012 = 8.25

The output waveform is shown in Figure 9. The positive part of the output voltage waveform is

limited to + 0.7 .

Figure 9.

7. What would you expect to see displayed on an oscilloscope connected across RL in the limitershown in Figure 10.

Figure 10

Solution

The diode is forward biased and conducts when the input voltage goes below -0.7 V. so, for the

negative limiter, determine the peak output voltage across RL by the following equations:

() = + () = 1.0 1.1 10 = 9.09

The scope will display an output waveform as shown in figure 11.

Figure 11 output voltage waveform for figure 10

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8. What is the output voltage that you would expect to observe across RL in the clamping circuit ofFigure 12? Assume that RC is large enough to prevent significant capacitor discharge.

Figure 12

Solution

Ideally, a negative dc value equal to the input peak less the diode drop is inserted by the clamping

circuit.

() 0.7 = (24 0.7) = 23.3 Actually, the capacitor will discharge slightly between peaks, and, as a result, the output voltage will

have an average value of slightly less than that calculated above.

The output waveform goes to approximately +0.7 V, as shown in figure 11.

Figure 13. Output waveform acrossRL

for figure 12