Question

3. ข"Pshr delivered a full sine wave lwith positive and negative peaks) to the circuits of Fig. 2 & 3 marks a) Compare the Du, and res waveforims in the scope images. Half the peaks in Vasp disappeared they do not appear In the positlve or negative pcaks from s? waveform. Were the observed ad peaks the b Briefly explain why only half the AC peaks were observed across the load. c) What is the maximum power that could be delivered to a load by adjusting R in the circuits of Fig. 2 & 3, as a percentage of the total AC power available from the small transformer? Include a brief explanation. Hint: Think PWM. PART C Basic AC Operation (Phase Control Circuit) 2- 500 kQ Rs C1060 y 12.0 v C1 0.47 μ F lamp type 7327 Figure 3 PARTB Basic AC Operation (Self-Triggering Circuit) 500 kQ A. C106D supy 12.0 Vrm lamp type 7327 Figure 2

Answer 1

(a) As the supply voltage is a full sine wave it would look like as shown in figure below

The two circuits are firing circuits or gate triggering circuits for the Thyristor. Once the thyristor is on it will conduct but it is a unidirectional switch so it will conduct only in one direction. Assuming L1 to be positive polarity terminal, only positive pluses would appear across the load.

So, the load voltage waveform will only contain positive peaks and will look like as shown in figure below

(b) As Thyristor or SCR is a unidirectional switch so it will conduct only in one direction. therefore, only half the AC peaks will be observed across the load.

(c) For maximum transfer of power  $R_1=0$ , but even with this condition only half or 50% of the AC power can be transfered to the load.

This is because thyristor will conduct for only half the time as it is a unidirectional switch.