Test: Half-Wave & Full-wave Rectifier - UPSC MCQ

 For a half wave rectifier, the I_DC=I_AVG=I_m/π
I= V_msinωt/(R_F+R_L)=I_msinωt
I_m =V_m/ R_F+R_L So, I_DC=I_m/π=V_m/(R_F+R_L).

The given op-amp is in inverting mode and this makes the output voltage to have a phase shift of 180°. The output voltage is now negative. So, the diode 1 is reverse biased and diode 2 is forward biased. Then output is clearly zero.

When the input of the inverted mode op-amp is positive, the output is negative.
The diode is reverse biased. The input appears at the output.

The standard form of a sine wave is V_msinωt. BY comparing the given information with this equation, V_m =50.
Power=V_m²/R_L=50*50/1000=2.5V.

 Comparing with the standard equation, V_m=200V.
Average value is given by, V_avg=V_m/π.
So, 200/π=63.694.

Efficiency of a rectifier is the effectiveness to convert AC to DC. For half wave it’s 40.6%. It’s given by, V_out/V_in*100.

PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, If the PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=V_m-V_d=5-0.7=4.3V.

Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.287.

The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 1.21.

Efficiency of a rectifier is the effectiveness to convert AC to DC. It’s obtained by taking ratio of DC power output to maximum AC power delivered to load. It’s usually expressed in percentage. For centre tapped full wave rectifier, it’s 81.2%.

The ripple voltage is (V_ϒ)_RMS=ϒV_DC /100.
V_DC=0.636*V_RMS* √2=0.636*220* √2=198V and ripple factor ϒ for full wave rectifier is 0.482.
Hence, (V_ϒ)_RMS=0.482*198 /100=0.954V.

We know that, P_DC=V_DC²/R_L. So, V_DC=(P_DC*RL)^1/2=10000^1/2=100V.
Here, ϒ=0.02
ϒ=V_AC/V_DC=V_AC/100.So, V_AC=0.02*100=2V.

The AC power input P_IN=I_RMS²(R_F+R_r).
I_RMS=I_m/√2=V_m/(R_f+R_L)√2=30/(1500+10)*1.414=13.5mA
So, P_IN=(13.5*10-3)2*(1500+10)=275.2mW.

The ripple factor ϒ= [(I_RMS/I_AVG)² – 1]^1/2. I_RMS =I_m /√2=V_m/(R_f+R_L)√2=200/1.01=198.
(Secondary line to line voltage is 800/2=400. Due to centre tap V_m=400/2=200)
I_RMS=198/√2=140mA, I_AVG=2*198/π=126mA. ϒ=[(140/126)²-1]^1/2=0.48. So, ϒ=48%.

In the output of the centre tapped rectifier, one of the half cycle is repeated. The frequency will be twice as that of input frequency. So, it’s 100Hz.

Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.693.

When the input is applied, a full wave rectifier should have a current flow. The flow should be in the same direction for both positive and negative half cycles. Only the third circuit satisfies the above condition.

 PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, if PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=2V_m-V_d = 10-0.7 = 9.3V.

The equation of sine wave is in the form V_msinωt. So, by comparing we get ω=100. Frequency, f =ω/2=50Hz. The output of centre tapped full wave rectifier has double the frequency of inpu. Hence, f_out = 100Hz.

Concept:

Half wave rectifier

A half-wave rectifier converts an AC signal to DC by passing either the negative or positive half-cycle of the waveform and blocking the other. Half-wave rectifiers can be easily constructed using only one diode, but are less efficient than full-wave rectifiers.
The maximum efficiency of a half-wave rectifier is 40.6%. The proportion of dc output power to input power is used to define a rectifier's efficiency.
A half-wave rectifier's efficiency will be:

V_DC = DC voltages
V_AC = AC voltages

Calculation