GATE 2011 Electrical Engineering (EE) Power Electronics (2011)

SSolution

Current Source Inverter (CSI) requirements:

Key characteristics:

• DC current source input (constant current)
• Load is induction motor (inductive)
• Switches must handle bidirectional voltage
• Switches carry unidirectional current

Why series diode is needed:

1. Reverse voltage blocking:

• During commutation, load voltage can reverse
• MOSFET body diode would conduct in reverse
• Series diode blocks reverse voltage
• Prevents short circuit through body diode

2. Current commutation:

• Current source must have continuous path
• Series diode ensures current flows only forward through MOSFET
• Prevents circulating currents

3. Configuration:

Why not anti-parallel diode:

• Anti-parallel provides freewheeling path
• Used in VSI, not CSI
• Would short-circuit the current source

Correct answer: C

SSolution

SCR turn-off process:

1. Current goes to zero:

• Anode current reduced to zero
• But charge carriers still present in junctions
• SCR not yet ready to block forward voltage

2. Recombination time needed:

• Excess carriers must recombine
• Takes finite time
• If forward voltage applied too soon: SCR turns on again

3. Circuit turn-off time (\(t_c\)):

• Time for which reverse bias must be maintained
• Allows complete charge carrier recombination
• SCR regains forward blocking capability
• Typically 50-100 \(\mu\)s for regular SCRs

Definition:

Circuit turn-off time = Minimum duration for which reverse voltage must be applied by commutation circuit to ensure SCR turns off completely.

Comparison with other times:

Turn-off time (\(t_q\)): Time from current zero to when SCR can block forward voltage

Current fall time: Time for current to reach zero

Holding current time: Different concept - minimum current to maintain conduction

Correct answer: C

The SCR must be reverse biased by the commutation circuit for sufficient duration to ensure complete turn-off.

SSolution

Base circuit:

Take the top rail as the 0 V reference. The emitter is tied to \(-12\) V, so the base sits at \(-12 + V_{BE} = -11.3\) V. Walking from the rail through the 15 k\(\Omega\), the 1 k\(\Omega\) and the diode to the base:

The zener node then sits at \(-15\text{k} \times 0.6625\) mA \(= -9.94\) V, only 2.06 V above the \(-12\) V rail, so the 5 V zener never breaks down and carries no current.

Active or saturated?

In the active region \(I_C = \beta I_B = 30 \times 0.6625 = 19.9\) mA, which across the 2.2 k\(\Omega\) would need \(19.9 \text{ mA} \times 2.2 \text{ k}\Omega = 43.7\) V. Only 12 V is available, so the transistor is in saturation.

Collector current in saturation: \(V_{CE(sat)} = 0.2\) V puts the collector at \(-12 + 0.2 = -11.8\) V, and the 2.2 k\(\Omega\) runs from the 0 V rail to that node:

Verify saturation: \(I_C/\beta = 0.179\) mA \(< I_B = 0.6625\) mA, so the base drive is more than enough.

Correct answer: D

SSolution

Voltage commutated chopper operation:

Main thyristor (M):

• Carries load current when ON
• Also carries capacitor charging current during turn-on

Auxiliary thyristor (A):

• Used for commutation
• Discharges capacitor to reverse-bias M
• Carries only commutation current

Maximum current through M:

Peak current = Load current + Capacitor discharge current

Capacitor voltage charges to source voltage: \(V_C = V_s = 200\) V

When A fires (for commutation):

Maximum current through M:

Maximum current through A:

Auxiliary thyristor only carries commutation current (capacitor discharge):

(During commutation, load current transfers through A)

Correct answer: A

SSolution

Circuit components:

• Zener diode: 10V rating (upper path)
• Regular diode: 0.7V forward drop
• Battery: 5V

Positive input:

\(D\) becomes forward biased as soon as the output reaches the battery voltage plus one diode drop:

Above that, \(D\) holds the output at 5.7 V and the surplus is dropped across the 1 k\(\Omega\). The zener is reverse biased with only 5.7 V across it, so it never reaches its 10 V breakdown and takes no part on this side.

Negative input:

For negative output the zener is forward biased and behaves as an ordinary diode:

In between: both devices are off, no current flows in the 1 k\(\Omega\), and \(V_o = V_i\).

Transfer characteristic:

• For \(V_i < -0.7\)V: \(V_o = -0.7\)V (clamped)
• For \(-0.7 < V_i < 5.7\)V: \(V_o = V_i\) (linear)
• For \(V_i > 5.7\)V: \(V_o = 5.7\)V (clamped)

Correct answer: C

The output is clipped at +5.7V (upper) and -0.7V (lower).