Test: 1P Diode Rectifiers HW
20 Questions MCQ Test Power Electronics | Test: 1P Diode Rectifiers HW
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In the process of diode based rectification, the alternating input voltage is converted into
Rectification is AC to DC. In DIODE biased rectification, control is not possible.
Current is always in one direction only, but voltage can be bi-directional in case of an L load.
For a certain diode based rectifier, the output voltage (average value) is given by the equation
1/2π [ ∫Vm sin ωt d(ωt) ] Where the integral runs from 0 to π
1/2π [ ∫Vm sin ωt d(ωt) ] Where the integral runs from 0 to π
Integration is 0 to π from base period of 1/2π so it is a half wave R load.
For a single phase half wave rectifier, with R load, the diode is reversed biased from ωt =
Diode will be reversed biased in the negative half cycles.
For the circuit shown below,
The secondary transformer voltage Vs is given by the expression
Vs = Vm sin ωt
Find the PIV of the diode.
PIV = √2 Vs = Vm.
For the circuit shown below,
The peak value of the load current occurs at ωt = ?
Due to the L nature, load current is maximum when the diode will be com-mutated i.e at π.
Find the rms value of the output voltage for the circuit shown below.
Voltage across the secondary is given by Vm sinωt.
The above is a HW diode rectifier, the RMS o/p voltage equation is given by
Vor = √ [ (1/2π) ∫π Vm2sin2ωt. d(ωt) ] Solving above equation we get, Vor = Vm/2.
In a 1-Phase HW diode rectifier with R load, the average value of load current is given byTake Input (Vs) = Vm sinωt
Vo = √ [(1/2π) ∫π Vm sinωt. d(ωt)] Vo = Vm/π
I = Vo/R = Vm/πR.
In the circuit shown below,
The switch (shown in green) is closed at ωt = 0. The load current or capacitor current has the maximum value at ωt =
The instant switch is closed the load current will be zero due to the nature of the capacitor.
Find the average value of output current for a 1-phase HW diode rectifier with R load, having RMS output current = 100A.
I(rms) = Vm/2R
Therfore, Vm = 200R
I(avg) = Vm/πR = 200R/πR.
A 1-phase 230V, 1KW heater is connected across a 1-phase HW rectifier (diode based). The power delivered to the heater is
R = (230 x 230)/1000
V(rms) = (√2 x 230)/2
P = V(rms)2/R = 500W.
A 1-phase half wave diode rectifier with R load, has input voltage of 240 V. The input power factor is
Input p.f = V(rms)/Vs
Vrms is the RMS value of output voltage. Vrms = (√2 x 230)/2
Vs = 230
pf = 0.707.
A 1-phase half wave diode rectifier with R = 1 KΩ, has input voltage of 240 V. The diode peak current is
Diode peak current = peak current through the load = Vo/R = Vm/2R.
For the below given circuit, after the switch is closed the voltage across the load (shown open) remains constant.
Assuming that all initial conditions are zero. The element across the load would be a/an
As the voltage remains constant as soon as the switch is closed, the element is most likely to be a capacitor.
For the below given circuit,
After the supply voltage (Vs) is given the
The diode will be forward biased only when Vs will be greater than Vdc.
For the below given circuit,
With Vs = Vm sin ωt (secondary side). The expression for the average voltage is
Due to the FD, we get 1st quadrant operation.
Where, output voltage (Avg) = 1/2π [ ∫Vm sin ωt d(ωt) ], integration runs from 0 to 180 degrees.
The output voltage is the voltage across the resister and Vdc. Even if the current falls to zero, the output voltage will be equal to Vdc.
For the below given circuit,
Vs = 325 sin ωt (secondary side)
The ripple voltage is
Ripple voltage = √(Vrms2 + Vavg2)
Vrms = Vm/2
Vavg = Vm/π
For a single phase half wave rectifier, the rectifier efficiency is always constant & it is
Rectifier efficiency = Pdc/Pac
Pdc = (Vm x Im)/π2
Pac = 4/(Vm x Im).
For the below given circuit,
The power delivered to the load in Watts is
P = power delivered to the resister + power delivered to the emf source.
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