December 2018 Q.1. Solution - Grad Plus

December 2018 Q.1. Solution

Q.1. a) With the help of two transistor analogy of SCR, briefly explain why gate loses his control once SCR is tumed ON? [5M]

Ans.

Fig (a) shows the thyristor as p-n-p and p-n-p transistor coupled in regenerative fashion and fig.(b) shows the equivalent to two-transistor analogy for thyristors The latching is explained by this analogy.

Powe electronics s 19 Page 3 - Grad Plus

From fig.(b) , it is seen that, the collector current IC1 of transistor T1 is the base current of transistir T2 and collector current IC2 of transistor T2 is the base current of transistor T1.

The collector current Ic of transistor is given as

Ic=α. IE + ICBO

where , α = common base current gain

ICBO– leakge current from collector to base with emitter open.

  • For transistor T1 , the collector current IC1 is, IC1 = α1 IA + ICBO1

where, α1= current gain

IA = emitter current which is anode current.

ICBO1 = leakage current for T1

  • Similarly for transistor T2, the collector current IC2 is,

IC22. IK + ICBO2

ICBO2=leakage current for T2

where, α2= current gain

  • Applying KCL to transistor T1

Incoming current = outgoing current

IA = IC1 + IC2

IA = α1 IA + ICBO1 + α2. IK + ICBO2 ——-(1)

Applying KCL to entire circuit we get,

Incoming current = outgoing current

IA + IG = IK

Substitute the value of IK in equation 1 , we get

IA= α1 IA + ICBO1 + α2 (IA+IG) + ICBO2

IA= α1 IA + ICBO1 + α2 IA+ α2 IG + ICBO2

IA= (α1 + α2) IA + α2 IG + ICBO2 + ICBO1

{1- (α1 + α2)] IA = α2 IG + ICBO2 + ICBO1

I_{A}=\frac{\alpha {2}I{G}+I_{CBO1}+I_{CBO2}}{{1-(\alpha {1}+\alpha {2})}}

From above equation we can say that anode current depends on gate current IG, leakage currents ICBO1 and ICBO2 and current gains α1 and α2 .

  • If (α1 + α2) tends to unity then the denominator {1- (α1 + α2)] approaches to zero , resulting in large value of anode current IA and the thyristor will turn on.
  • As the gain (α1 + α2) approaches unity, the circuit starts to regenerate and each transistor drives its mate into saturation. In this state, all junctions assumed to be forward biased and total potential drop across the device approximates that of a single p-n junction . Anode current is limited only by the external circuit resistance.
  • The current gain α varies with emitter current. During turn-on process, the rapid increase in current gain α with emitter current is highly essential and this is observed in silicon transistors. This effect leads to positive feedback effect or regenerative action and latching of thyristors.
Powe electronics s 19 Page 2 1 - Grad Plus


Q.1.b) Mention any two applications of de to de converter. Draw the diagram of a Buck do to de converter and draw the inductor voltage, inductor current and derive the voltage ratio. [5M]

Ans.

The two applications of dc to dc converter nare –

i) Used for traction motor control in electric automobiles trolley cars etc.

ii) Used in regenerative braking of dc motor.

  • The circuit diagram of buck dc-to-dc converter is –
Untitled Diagram Page 10 4 - Grad Plus
  • The waveform of inductor Voltage (VL) and inductor (IL) is –
Untitled Diagram Page 9 4 - Grad Plus
  • When switch is closed,

  • VL= Vs– Vo

-When switch is open ,

VL=- Vo

  • Duty cycle D=\frac{t_{on}}{T} = \frac{t_{on}}{t_{on}+t_{off}}

ton = D.T

Dt_{on} + Dt_{off} = t_{on}

Dtoff = ton– Dton

Dtoff = (1-D) ton

t_{off} = (1-D) \frac{t_{on}}{D} = (1-D) T

Rate of change of Current IL is

V_{L}=L\frac{dI_{L}}{dt} \bigtriangleup I_{L}=\int \frac{V_{L}}{L}dt

For on state

\bigtriangleup I_{L_{on}}=\int_{0}^{t_{on}} \frac{V_{s}-V_{o}}{L} dt = \frac{V_{s}-V_{o}}{L}(t)^{t_{on}} \bigtriangleup I_{L_{on}}= \frac{V_{s}-V_{o}}{L} t_{on} = \frac{V_{s}-V_{o}}{L}D.T

For off state

\bigtriangleup I_{L_{off}}= \int_{t_{on}}^{t_{on}+t_{off}} -\frac{V_{o}}{L} dt= -\frac{V_{o}}{L} (t)^{t_{on}+t_{off}}<em>{t</em>{on}} = -\frac{V_{o}}{L} \left ( t_{on}+t_{off}-t_{on} \right ) \bigtriangleup I_{L_{off}}= -\frac{V_{o}}{L} \; t_{off} = -\frac{V_{o}}{L} (1-D)T

Assuming the converter operates in steady state

∆ILon + ∆IL off = 0

\frac{(V_{s}-V_{o})}{L} DT -\frac{V_{o}}{L} (1-D)T=0 \frac{(V_{s}-V_{o})}{L} DT =\frac{V_{o}}{L} (1-D)T

VsD – VoD = Vo – VoD

VsD = Vo

D=\frac{V_{o}}{V_{s}}

This is the voltage ratio.


Q.1.c) What are the advantages of PWM rectifier as compared to controlled rectifier using SCR? What are its applications? Illustrate the diagram of a single phase PWM rectifier. [5M]

Ans.

The advantage of PWM rectifiers are :- It has bidirectional power flow, nearly sinusoidal input current, lopw harmonic distortion of line current, reduced capacitor size due to continuous current.

  • The controlled thyristors (SCR) rectifiers overload the supply network with higher harmonics and reactive power consumption.Using PWM rectifiers, there will be no consumption of reactive power and so the harmonics are reduced.
  • The application of PWM rectifier are – used for the charge3r circuit of an electric vehicle, used for harmonic reduction in smart Grid, used in traction application etc.
  • The diagram of single phase PWM rectifier is shown below.
Untitled Diagram Page 11 2 - Grad Plus

Q.1.d) illustrate the diagram of a single-phase half bridge inverter and draw the output voltage waveform for square wave mode of operation. Such an inverter is connected to a resistive load of 24 with de input voltage of 24V each, Determine: (i) RMS output voltage (1) Output power and (1) Peak blocking voltage of each switch. [5M]

Ans.

The diagram of single phase half bridge inverter is shown below.

Untitled Diagram Page 5 4 - Grad Plus
  • The output voltage waveform for square wave mode of operation is shown below.
Untitled Diagram Page 10 3 - Grad Plus
  • Given :- R= 2.4 Ω, Vs = 24 V

To Find :- i) Vorms ii) Po iii) PIV

Solution – i) For square wave output, rms value is equal to amplitude of square wave.

RMS output voltage (VoRMS) =

\frac{V_{s}}{2}=\frac{24}{2}= 12 \; V

ii) Output power (Po) = \frac{V_{oRMS}^{2}}{R}=\frac{(12)^{2}}{2.4}=60 \; W

iii) Peak blocking voltage of each switch (PIV) = Vs

PIV = 24 V


Q.1.e) Compare Silicon Carbide and Gallium Nitride devices. [5M]

Ans.

Sr. No. Silicon CarbideGallium Nitride
1.Lower electron mobility i.e. 900 cm2/ Vsec. Higher electron mobility than silicon carbide i.e. 2000 cm2 / Vsec.
2.Higher thermal conductivity than Gallium Nitride i.e. 5 watts / cm2 K. Lower thermal conductivity i.e. 1.3 watts/cm2 K.
3.The bandgap is 3.2 eV.The bandgap is 3.4 eV.
4. The critical field is lower i.e. 3 x 106 V/cm The critical field is higher i.e. 3.5 x 106 V/cm which allows devices to operate at higher voltage and lower leakage currents.
5. It has lower electron saturation velocity i.e. 22 x 106 cm/sec but it is also used for higher frequency operation. It has higher electron saturation velocity i.e. 25 x 106 cm/sec used for higher frequency operation.

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