Power Electronics
Computer Lab: Uncontrolled AC-DC Converters (Diode Rectifiers)
Diode Rectifiers
Task 1: Single phase half wave diode Rectifier (File name: Diode_singlephase1) General
condition: VIN=100 V
 Resistive load
Consider the circuit diagram of a single phase half wave diode rectifier:
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vD (t)
+
iout(t)
D
vin(t)
R
vout(t)
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1-1 Check the output voltage and current waveforms with a 10 Ω resistive load. 1-2
Check voltage across the diode
1-3 Check the current drawing from the source. What is the problem with the current
waveform?
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1-4 Double click on resistor and increase it to 100 Ohms. What’s the difference?
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1-5 Put a DC voltage of 50 V in series with the resistor and run the simulation again. Compare the
results with the pervious simulation. What is the pick output voltage? When diode starts conducting?
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1-6 Remove the DC source and make the circuit as it was in 1-3. Now double click on diode and
increase the forward voltage to 0.8 V. Check the output voltage and input current. What’s the
difference with the case that diode has zero forward voltage?
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 R-L load
1-7 Double click on the resistor and change the branch type to a 5mH inductor in series with
a 10 Ω resistor and compare the output waveform results with the pure resistive load.
(Remove the DC source)
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1-8 Increase the inductance value to 50 mH and describe what happens.
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1-9 As shown in following figure, add a freewheeling diode and compare the output voltage
and current waveforms with the previous results. Observe the effect of freewheeling diode
on load current.
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1-10 What is the current waveform through the freewheeling diode? Sketch it?
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Task 2: Single Phase full wave diode Rectifier (File name: Diode_singlephase2)
General condition: VIN=100 V , R=100 Ohms
 Resistive load
Consider the circuit diagram of a single phase full wave diode rectifier:
2-1 Check the output voltage and current waveforms with a pure resistive load. What are the main
advantages comparing with a half-wave rectifier?
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2-2 Compare the input and output voltage by looking at Scope 1. What is the difference?
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2-3 Now double click on diodes and reduce the forward voltage to 0 V. Now compare the input and
output voltage again by looking at Scope 1.
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2-4 Set the diodes forward voltage to 0.8 V. Add a 1000 µF capacitor (passive filter) in parallel with
load and compare the output voltage.
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2-5 What is the disadvantage of having a capacitive filter regarding to input current. Compare with
2-1 case.
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2-6 Increase the capacitance value up to 2200 µF and compare the output voltage ripple.
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2-7 Decrease the resistance value down to 10 Ohms and compare the results.
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 Inductor at load side
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2-8 Now add an inductor (1 mH) at the load side as below. What are the differences regarding to input
current and output voltage ripple?
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2-9 What is the main practical issue of having an inductor for a circuit which runs at low frequency?
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Task3: Three phase full wave diode Rectifier (File name: Diode_threephase)
General condition: VIN=100 V
Consider a three phase full wave diode rectifier as shown below:
5
+
iout(t)
D3
D1
vb(t)
D5
va(t)
R vout(t)
vc(t)
D4
D2
D6
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3-1 Check the output voltage and current waveforms with a pure resistive load. What are the main
differences comparing with a full-wave diode rectifier with a same resistive load?
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3-2 Such a circuit configuration is known as 6-pulse diode rectifier. Why?
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3-3 Add a 1000 µF capacitor (passive filter) in parallel with load and compare the output voltage.
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