A power diode is in the forward conduction mode and the forward current is now decreased. The reverse recovery time of the diode is trr and the rate of fall of the diode current is di/dt. What is the stored charge?
The reverse recovery characteristic of a power diode is shown below. reverse recovery time trr composed of ta and tb
ta is the time between zero crossings and peak reverse current IRR and tb is measured from reverse peak IRR value to 0.25 IRR.
If the characteristics are assumed to be triangular (i.e. abrupt recovery) then from the figure charge stored is
QRR = area of the triangle
A G.T.O with anode fingers has
The short circuiting fingers, also called anode-shorts, leads to short-circuit of the emitter p+(Anode A) with base n of Q1 transistor as shown in Fig. For anode or emitter current Ia, effective emitter current IE1 is reduced because of anode short. This further decreases collector current IC1. Therefore, effective current gain α1 of Q1, now given by IC1/Ia gets reduced. So by anode-short structure, α1 is reduced but α2 remains unchanged as desired.
Anode-short, however, reduces reverse voltage blocking capability. With reverse biased GTO, junction that blocks reverse voltage is J3 only. Junction J1 blocks no voltage because of n+fingers in between p+ anode. As J3 junction has large doped layers p+, n+ on its two sides, J3 has lower breakdown voltage, of the order of about 20 to 30 V.
Four power semiconductor device are shown in the figure along with relevant terminals the device that can carry dc current continuously in the Direction shown when gated appropriately is (are)
In power transistor, during turn on process voltage and current change linearly from 400 V and 0 V and 0 A to 300 A respectively in 3 μS. the energy loss during turn on process in given by _____Joule (up to 2 decimal places)
Change in V and L are shown below
Energy loss during turn on process =
= 20 × 103 × 3 × 10-6
= 0.06 J
A power transistor has its switching waveforms as shown in fig. If the average power loss in the transistor is limited to 470 W, find the switching frequency at which this transistor can be operated – (in kHz)
Energy loss during turn – on
= 0.2 watt – sec.
Energy loss during turn – off
= 0.27 watt – sec
Total energy loss in one cycle = 0.2 + 0.27
= 0.47 watt – sec.
Average power loss in transistor
= switching frequency × energy loss in one cycle
⇒ Switching frequency =470/0.47
= 1000 Hz.
For a power transistor typical switching waveform shown in Fig. the various parameter of the transistor circuit are as under: Vcc = 220V, VCES = 2V, Ics = 80 A, td = 0.4 μs tr = 1 μs f = 5 kHz and ICEO = 2mA. Determine Average Power loss during turn-on time.
Average power loss during delay time with o < t < td
Average power loss during rise time is
Pd = 5 × 103 × 2 × 10−3 × 220 × 4 × 10−6 = 88mW
Pr = 5 × 103 × 80 × 1 × 10−6 = 14.933W
Total power loss during turn on time = 14.933 + 0.00088 = 14.9339 W
In a power electronics circuit, MOSFET is using as the switch, MOSFET with an ON state resistance 50 mΩ has with the following switching characteristics. The power losses associated with MOSFET are
P = Irms2 × 50 × 10-3 = 24 × 50 × 10-3 = 1.2 W
Find the time required (in hour) to deliver a charge of 400 Ah through a single phase half-wave diode Rectifier with an output current 200 A rms and with sinusoidal input voltage. Assume diode conduction over a half-cycle
Correct Answer :- 3.14
Explanation : For half-wave diode rectifier
R.M.S value of the output voltage
Vor = [1/2π ∫(0 to π) Vm2 sin2ωt d(ωt)]1/2
= Vm/(2π)1/2[∫(0 to π)(1-cos2ωt)/2 d(ωt)]1/2
Ior = Vm/2R
Average value of load current I0 = Vm/πR
Ior = Vm/2E
Io × time = 400 Ah
∴ Time required = (400*π)/400
= 3.14 hrs
A gate turn off (GTO) thyristor has capacity to
For the circuit shown in Fig R = 10 Ω L = 1 mH C = 5 μF and VS = 230 V. the circuit is initially relaxed determine conduction time of diode _________ μs
The roots of the characteristic equation are
Ringing Frequency in rod/sec = ωr =
Conduction time of diode = π/ωr
Conduction time =