GATE 2024 EE Q.65: solution and answer

GATE 2024 · EEQ.65 · 2 marks · NAT

A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications: rated armature voltage 210 V; rated armature current 10 A; rated speed 1200 rpm; armature resistance 1 Ω; input to the converter bridge 240 V at 50 Hz. The armature of the DC motor is fed from the full-controlled bridge and the field current is kept constant. Assume that the motor is running at 600 rpm and the armature terminals of the motor are suitably reversed for regenerative braking. If the armature current of the motor is to be maintained at the rated value, the triggering angle of the converter bridge in degrees should be ____________ (rounded off to 2 decimal places).

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Answer and solution

ANSWER

The triggering (firing) angle is $α \approx 114.6 1^{\circ}$ .

Official answer: 113.00 to 116.00Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

Rated operating point: $E_{a} = V_{t} - I_{a} R_{a} = 210 - 10 \times 1 = 200 V$ at $1200 rpm$ .
The field current is constant, so $E_{a} \propto N$ . At $600 rpm$ : $E_{a} = 200 \times \frac{600}{1200} = 100 V$ .
In regenerative braking the machine acts as a generator (armature terminals reversed) and the bridge works as a line-commutated inverter. With the rated current $I_{a} = 10 A$ flowing from the machine into the supply, $E_{a} = ∣ V_{d c} ∣ + I_{a} R_{a}$ . Hence $∣ V_{d c} ∣ = 100 - 10 \times 1 = 90 V$ . The converter output (in the direction of the original motoring polarity) is $V_{d c} = - 90 V$ .
Peak supply voltage: $V_{m} = 2402 = 339.41 V$ .
For a single-phase fully controlled bridge with firing angle $α$ (continuous current): $V_{d c} = \frac{2 V _{m}}{π} cos α = \frac{2 \times 339.41}{π} cos α = 216.08 cos α V$
Set $216.08 cos α = - 90$ , so $cos α = - 0.4165$ .
$α = cos^{- 1} (- 0.4165) = 18 0^{\circ} - 65.3 9^{\circ} = 114.6 1^{\circ}$ . Since $α > 9 0^{\circ}$ , the bridge is indeed in inverting mode, consistent with regeneration.

KEY IDEA

In regenerative braking the DC machine becomes a generator and the fully controlled bridge runs as an inverter ( $α > 9 0^{\circ}$ , $V_{d c} < 0$ ). Find $E_{a}$ from speed scaling, subtract the armature drop to get $∣ V_{d c} ∣$ , and then use $V_{d c} = \frac{2 V _{m}}{π} cos α$ .

WATCH OUT FOR

Using the motoring relation $V_{d c} = E_{a} + I_{a} R_{a} = 110 V$ , which gives $α \approx 5 9^{\circ}$ (a rectifier angle). Under regeneration the drop must be subtracted from $E_{a}$ , and the bridge voltage is negative, so $α$ lies above $9 0^{\circ}$ . Another slip is to use $V_{d c} = \frac{V _{m}}{π} (1 + cos α)$ , which is the half-controlled formula.

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