1) Determine the peak output voltage for the bridge rectifier in Figure 1 shown.
ID: 1715767 • Letter: 1
Question
1) Determine the peak output voltage for the bridge rectifier in Figure 1 shown. Assuming the practical model, what PIV rating is required for the diodes? The transformer is specified to have a 12 Vrms secondary voltage for the standard 120 V across the primary.
Figure 1. Bridge Full Wave Rectifier
2) In the full-wave center-tapped rectifier circuit, the transformer has a turns ratio of 1:2. The transformer primary winding is connected across an AC source of 230V (rms), 50 Hz. The load resistor is 50. For this circuit, determine the DC output voltage, peak-to-peak ripple in the output voltage, and output ripple frequency. Show all work.
3) In an NPN transistor, the emitter and collector are both N-type materials.
True
False
4) The dc beta is defined as the dc collector current divided by the dc emitter current.
True
False
5) When the ground side of each voltage source is connected to the emitter of a bipolar junction transistor, it is called a common collector.
True
False
6) The middle region of a transistor is the ________.
7) A transistor has a collector current of 12 mA and a base current of 40 µA. What is the current gain of the transistor? Show all work.
8) A transistor has a current gain of 260. If the base current is 90 µA, what is the collector current?
Explanation / Answer
1) Peak output voltage is Vm= Vsec(rms)*1.414 = 16.968 V; PIV is Voltage across each diode when it is not conducting is Vm = 16.968 V;
2) Given to consider center-tapped rectifier and whose turns ratio if 1:2 with 230 V rms primary voltage. Secondary voltage of each across on half of secondary is given by Vrms = 230 V. Each secondary Voltage of center tap coil will be 230 V (rms). Vm(peak) = 230*1.414 = 325.22 V.
a) DC output voltage = 2*Vm(peak)/pi = 2*325.22/3.14159 = 207.04 V, Vo(average output voltage) = 207.04 V.
Vrms(output voltage) = Vm(peak)/1.414 = 325.22/1.414 = 230.
b) Peak-to-peak ripple voltage Vr = Vo(average output)/Vrms(output) = squrroot(Vrms(output)2 - Vo(average output)2)
(squrroor(2302- 207.042)) = 100.1720 V.
c) Output ripple frequency is given by fr=2*f = 2*50 = 100 Hz. because in one cycle of input there are two cycles of outptut.
3) True
4) False , because Ic / Ie = alpha forward current gain.
5) False, it is common emitter
6) base region
7) Gain of transistor is given by beta = Ic / Ib = 12*10-3/40*10-6 = 300
8) Current gain beta = Ic/Ib = 260, Ic = 260*90*10-6 = 23.4 mA.
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