Unit 02 DIODE APPLICATION, Electronic Engineering Qs & Solutions AKTU

Prepare yourself for your Electronic engineering exams with crucial questions and answers from Unit 1: Semiconductor Diodes for AKTU students. For evaluate success, improve your knowledge of diode characteristics, rectifiers, and other topics.

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Important Questions For Electronic Engineering:
*Unit-01     *Unit-02    
*Unit-03    *Unit-04 
*Unit-05    *Short-Q/Ans
*Question-Paper with solution 21-22 

Q1. For the circuit shown in Fig., determine I₁ I₂, I₃, I₄, V_O. 

Ans. 

Q2. Sketch V_O, V_DC for the network of Fig., and determine the peak inverse voltage of each diode. 

Ans. 

Peak inverse voltage of each diode = V_m = 100 V

Q3. Determine V_O and the required PIV rating of each diode for the configuration of Fig.

Ans. 

Q4. Explain the function of the circuit of Fig. and draw the output waveform. 

Ans. 1. During positive half cycle 

Q5. Draw the output waveform for the following circuits for the input waveforms, given in F. (a) and (b): 

Ans. For Fig. (a),

During negative cycle of input :

1. Diode will be ON, capacitor starts charging with polarity as shown in Fig. 

2. The diode will act as short circuit, therefore

                                       V_O =0 and V_C= +10 V 

During positive half cycle : 

Output waveform of Fig. (b): 

2. During entire negative half cycle, diode will be reverse biased. Therefore 

Q6. What is voltage multiplier using p-n junction diode ? Explain the operation of voltage doublers. 

Ans.

• 1. The term “voltage multiplier” refers to a circuit that uses a rectifier circuit to produce a higher DC output voltage than AC input voltage. 
• 2. The output of voltage doubler is twice the peak input voltage. 
• 3. There are two types of voltage doubler circuits: 

i. Half wave voltage doubler: The circuit diagram is shown in Fig.

• 1. During positive half-cycle of the input voltage, diode D₁ is forward biased (ON) and diode D₂ is reverse biased (OFF). Capacitor C₁ charges to the peak value of secondary voltage V_m with polarity.
• 2. During negative half-cycle, diode D₂ is ON while diode D₁ is OFF. Capacitor C₂ is charged upto a voltage i.e., the sum of peak-supply voltage and the voltage across C₁.
• 3. During positive half-cycle: V_C1 = V_m   

ii. Full wave voltage doubler : The circuit diagram is shown in Fig.

                                            V_C1 = V_m

2. During negative half cycle, D₁ is OFF and D₂ is ON. Capacitor C₂ is charged to peak value V_m. Thus output voltage with no load connected : 

                                              V_O = V_C1 + V_C2                                                V_O = 2V_m

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