A half-wave rectifier circuit with a 1 kΩ load operates from a 120 V (rms value), 60 Hz household supply through a 10-to-1 step-
1 answer:
Answer:
(a) 16.27 Vpk
(b) 48.7%
Explanation:
The transformer is assumed to be an ideal 10:1 voltage divider with no internal impedance. The diode is assumed to be modeled in the forward direction by a perfect 0.7 V voltage drop with no internal impedance. That means the frequency of the supply voltage is irrelevant.
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<h3>(a)</h3>The peak voltage will be 0.7 V less than the transformer secondary peak voltage:
((120 V)√2)/10 -0.7 V ≈ 16.27 V
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<h3>(b)</h3>The fraction of the amplitude for which the diode is non-conducting is ...
0.7/(12√2) ≈ 0.041248
The period of conduction is symmetrical about the peak of the waveform, so it is convenient to use the arccos function to find the (half) conduction angle:
arccos(0.041248) ≈ 87.64°
As a fraction of half the cycle, this is ...
conduction fraction ≈ 87.64°/180° ≈ 48.7%
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Complete Question:
Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.
Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.
Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!
Answer & Explanation:
[Find the attachments]
Step 1 :
Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.
Emitter and base, collector and base are reverse biased then BJT in cut off region.
Three sketches one below the other is shown in Figure 1.
[find the figure in attachment]
Step 2:
Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.
Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.
[For Steps 3 4 5 6 and 7 find attachments]
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