Solved Problems on Power Electronics
Q1. In Fig.1 the switch is closed. If the triac has fired, what is the current through 50Ω resistor when (i) triac is ideal (ii) triac has a drop of 1V ?
Fig.1
Solution:
(i) Since the triac is ideal and it is fired into conduction, the voltage across triac is 0V. Therefore, the entire supply voltage of 50V appears across 50Ω resistor.
Current in 50Ω = 50V/50Ω = 1A
(ii) When triac is fired into conduction, voltage across 50Ω resistor = 50V – 1V = 49V
So current in 50Ω = 49v/50Ω = 0.98A
Q2. In Fig.2 the switch is closed. A diac with breakover voltage VBO = 30V is connected in the circuit. If the triac has a trigger voltage of 1V and a trigger current of 10mA, what is the capacitor voltage that triggers the triac ?
Fig.2
Solution:
When switch is closed, the capacitor starts charging and voltage at point A increases.
When voltage VA at point A becomes equal to VBO of diac plus gate triggering voltage VGT of the triac, the triac is fired into conduction.
so, VA = VBO +VGT = 30 V + 1 V =31 V
This is the minimum capacitor voltage that trigger the triac.
Q3. The intrinsic stand-0ff ratio for a UJT us determined to be 0.6. If the inter-base resistance is 10 KΩ, what are the values of RB1 and RB2 ?
Solution :
RBB = 10 KΩ, η = 0.6
Now, RBB = RB1 +RB2
Or 10 = RB1 +RB2
Also η = RB1/(RB1+RB2)
Or 0.6 = RB1/10
So RB1 = 0.6 × 10 = 6 KΩ
and RB2 = 10 – 6 = 4 KΩ
Q4. A unijunction transistor has 10 V between the bases. If the intrinsic stand off ratio is 0.65, find the value of stand off voltage. What will be the peak-point voltage if the forward voltage drop in the pn junction is 0.7 V ?
Solution:
VBB = 10 V, η = 0.65, VD = 0.7 V
Stand off voltage = η VBB = 0.65 × 10 =6.5 V
Peak-point voltage VP = η VBB + VD = 6.5 +0.7 = 7.2 V
Q5. Determine the maximum and minimum peak-point voltage for a UJT with VBB = 25 V. Given that UJT has a range of η =0.74 to 0.86.
Solution:
VBB = 25 V, ηmax = 0.86, ηmin = 0.74
So VP(max) = ηmax VBB + VD
= (0.86 ×25) + 0.7 = 22.2 V
VP(min) = ηmin VBB + VD
= (0.74 × 25 ) + 0.7 = 19.2 V
Q6. Fig.3 shows the UJT circuit. The parameters of UJT are η = 0.65 and RBB = 7 KΩ. Find (i) RB1 and RB2 (ii) VB2B1 and VP.
Fig.3
Solution:
η = 0.65 , VS = 12 V, RBB = 7 KΩ
(i) η = RB1/RBB
So RB1 = η × RBB = 0.65 × 7 = 4.55 KΩ
Also RB2 = RBB – RB1 = 7 -4.55 = 2.45 KΩ
(ii) Fig.4 shows the UJT circuit model. Because UJT is OFF, IE =0. We can find VB2B1 by the voltage-divider rule.
Fig.4
VB2B1 = [VS/(R1 + R2 + RBB)] × RBB
= [12 V/ ( 100 + 400 + 7000) Ω] × 7000 Ω
= 11.2 V
The voltage VP required turn on the UJT is
VP = η VB1B2 +VD = (0.65 × 11.2 V) +0.7 V =7.98 V
Q7. Fig.5 shows the relaxation oscillator. The parameters of the UJT are RBB = 5 KΩ and η = 0.6.
(i) Determine RB1 and RB2 at IE = 0.
(ii) Calculate the voltage VP necessary to turn on the UJT.
(iii) Determine the frequency of oscillations.
Fig.5
Solution:
VS = 12 V, RBB = 5 KΩ, η = 0.6
(i) η = RB1/RBB
So RB1 = η × RBB = 0.6 × 5KΩ = 3 KΩ
Also RB2 = RBB – RB1 = 5 KΩ – 3 KΩ = 2 KΩ
(ii) The voltage VP required to turn on the UJT can be found from the UJT circuit model shown in Fig.6. Referring to Fig.8, we have,
Fig.6
VP = VD + Voltage drop across (RB1 + R2)
= VD + [( VS/(RBB+R2)) × (RB1 +R2)]
= 0.7 V + [ (12 V/ (5KΩ +0.1 KΩ)) × (3 +0.1 )KΩ]
= 0.7 V + 7.294 V
= 8 V
(iii) Time period, t = R1 C loge(1/1-η)
Here, R1 = 50 KΩ = 50 × 10³Ω, C = 0.1 μf , η = 0.6
So t = 50 × 1o³ ×0.1 × 10-6× loge (1/1-0.6)
= 4.58 × 10-³s = 4.58 ms
hence Frequency, f = 1/t
= 1/ 4.58ms = 218 Hz
Q8. The circuit shown in Fig.7 uses variable resistor RE to change the frequency of pulses delivered at Vout. The variable resistor is initially set at 5 kΩ and is then adjusted to 10 kΩ. Determine the frequency of the voltage spikes produced for (i) 5 kΩ setting and (ii) 10 kΩ setting.
Fig.7
Solution:
(i) Time period, t = RE C loge (1/1-η)
Here, RE = 5 KΩ = 5 × 10³ Ω, C= 0.2 μF, η = 0.54
So t = 5 × 10³ × 0.2 × 10-6 × loge (1/1-0.54)
= 0.78 × 10-3 s = 0.78 ms
Hence, Frequency, f = 1/t = 1/ 0.78 ms = 1282 Hz
(ii) t = 10 × 10³ × 0.2 × 10-6 × loge (1/1-0.54)
= 1.55 × 10-3 s = 1.55 ms
Hence, Frequency, f = 1/t = 1/ 1.55 ms = 645 Hz
