Worked examples for power diodes and single-phase diode rectifiers PDF

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Sem A: 2017-2018 ECS643U

Worked Examples

Power Electronics

Subject: Worked examples for power diodes and single-phase (diode) rectifiers

Lecturer: Kamyar Mehran

c Queen Mary, University of London, 2017-2018 

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ECS643U (2017-2018)

Question 1 The reverse recovery time of a diode is trr = 5 µs, and the rate of fall of the diode current is di /dt = 80 A/µs. If the softness factor is SF = 0.5, determine (a) the storage charge QRR , and (b) the peak reverse current IRR .

Solution: The storage charge: QRR =

1 di 2 t = 0.5 × (80 × 106 ) × (5 × 10−6 )2 = 1000µC 2 dt rr

(1)

and the peak reverse current IRR =

r

2QRR

di p = 2 × 1000 × 10−6 × (80 × 106 ) = 400A dt

(2)

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Question 2 The storage charge and the peak reverse current of a diode are QRR = 10000 µC and I RR = 4000. If the softness factor is SF = 0.5, determine (a) the reverse recovery time of the diode trr , and (b) the rate of fall of the diode current di /dt .

Solution: We can find di /dt from the peak reverse current IRR =

=⇒ 4000 =

r

r

2QRR

di dt

2 × 10000 × 10−6 ×

(3) di dt

di = 8 × 108 A/s dt then the reverse recovery time of the diode trr can be found from =⇒

QRR =

1 di 2 t 2 dt rr

=⇒ 10000 × 10−6 =

1

× (8 × 108 ) × trr2 2 =⇒ trr = 5µs

(4) (5)

(6) (7) (8)

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ECS643U (2017-2018)

Question 3 The forward voltage drop of a power diode is VD = 1.2V at ID = 300A. Assuming that n = 2 and VT = 25.7mV , find the reverse saturation current IS .

Solution: Applying the ’Shockley diode equation’, we can find the leakage (or saturation) current IS from −3

ID = IS (e VD /nVT − 1) = IS [e 1.2/(2×25.7×10 which gives IS = 2.17746 × 10−8 A.

)

− 1]

(9)

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Question 4

(a)

What are the types of power diode? (b) What is the reverse recovery current in power diodes? (c) What is the reverse recovery time in power diodes? (d) What are the effect of reverse recovery in power diodes? (e) What type of diode is mainly used for high-frequency and fast-switching applications? In this type of power diode, what are the main features of Silicon-Carbide (SiC)-based Schottky power diodes in comparison to Silicon-based Schottky power diodes?

Solution: (a) We have Schottky diodes, Fast-recovery diodes and Line-frequency diodes (b) An important characteristic of an non-ideal diode is reverse-recovery current. When a diode turns off, the current in it decreases and momentarily becomes negative before becoming zero. The diode continues to conduct due to minority carries that remain stored in the PN junction. (c) In reverse recovery current, the minority carries require a certain time to recombine with opposite charges and to be neutralized. We call this time reverse recovery time rrt . (d) The effects are switching losses increase especially in high frequency applications, voltage rating increase, over voltage (spikes) in inductive loads. (e) Schottky diodes are used mainly in high-frequency and fast-switching applications since they have low forward voltage. The SiC-based Schottky diodes have the following features:

• Lowest switching losses due to low reverse recovery charge; • Fully surge-current stable, high reliability and ruggedness; • Lower system costs due to reduced cooling requirements; • Higher frequency designs and increased power density solutions. • low device capacitance that enhances overall system efficiency, specially at higher switching frequencies.

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ECS643U (2017-2018)

Question 5 If we assume in the Figure 1 that the resistance R is negligible, the source voltage Vs = 220 V (constant time), and the load inductance L = 220 µH . (a) What is the role of free-wheeling diode Dm in this circuit? (b) draw the waveform of the load current if the switch is closed for a time 1t = 100 µs, and is then opened. (c) Determine the final energy stored in the load inductor.

Figure 1

Solution: (a) If the switch S1 is closed for time t1 , a current is established through the load; then if the switch is opened, a path must be provided for the current in the inductive load. Otherwise the inductive energy induces a very high voltage and this energy is dissipated as heat across the switch as sparks. This is done by connecting a free-wheeling diode Dm . (b) The circuit has a zero initial current. When the switch is closed at t = 0, the load current rises linearly and is expressed as i(t) =

Vs t L

(10)

and at t = t1 , I0 = Vs t1 /L = 200 × 100/200 = 100 A. (c) When switch S1 is opened at a time t = t1 , the load current starts to flow through diode Dm. Since there is no dissipative (resistive) element in the circuit, the load current is constant at I0 = 100A and the energy stored in the inductor is 0.5 LI20 = 1.1J. The

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current waveforms:

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ECS643U (2017-2018)

Question 6 The rectifier in Figure 2 has a purely resistive load of R, determine (a) The load voltage vL in the form of Fourier series, (b) the load voltage and load current waveforms, (c) the average value of the load voltage, (d) the average value of the load current, (e) rectification ratio, (f) ripple factor (RF), (g) Transformer utilization factor (TUF).

Figure 2

Solution:

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ECS643U (2017-2018)

Question 7 The circuit shown is Figure 3b, is a single-phase full-wave bridge rectifier with resistive, inductive load. Determine (a) the input current waveform if the load is designed with a low inductance, (b) the instantaneous input current waveform if the load is highly inductive, its fundamental component, and the input voltage. (c) if the load is highly inductive, how we can find the displacement factor (DF), (d) if the load is highly inductive, how we can find the input power factor (PF), and the total harmonic distortion (THD), (e) If the rectifier supplies a dc motor as shown in Figure 3b, determine the total harmonic distortion (THD), and the input power factor (PF) of the rectifier.

(a) full-wave bridge rectifier with RL load

(b) full-wave bridge rectifier with dc motor load

Figure 3: Single-phase full-bridge rectifier

Solution:

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End of questions...