Problem 4 A basic power supply circuit is shown in Figure P1-1. The f = 50 Hz AC voltage source is given by vi(t) = 100 sin(2x ft). The transformer windings are in the ratio 4: 1 (voltage reduction). The diodes are all 1N4007. The capacitor shown has value C = 100 μF. The load resistor is RL = 2.2 k. vi(t) A C B ww RL Figure P1-1: Basic power supply circuit where vi(t) = 100 sin(100mt), C=100 μF and RL = 2.2k. (a) [3 marks] Describe concisely the operation of this power supply covering the roles of trans- former, rectifier (diodes) and capacitor. Solution: This is discussed in L31-L33. (b) [2 marks] Sketch the voltage waveform at point A (with respect to ground), as a function of time, over a few periods. Solution: This is discussed in L31-L33. You can use PSPICE to verify your sketch. (c) [3 marks] Sketch the voltage waveform at point B (with respect to ground), as a function of time, showing the peak voltage V, and the shape of the voltage ripple (the magnitude of the ripple voltage is addressed in part (e) below). Also, indicate on the voltage waveform at point B the time periods when the capacitor is charging or discharging (or when the diodes are "on" or "off"). Solution: This is discussed in L31-L33. You can use PSPICE to verify your sketch. (d) [2 marks] An RC circuit has a natural exponential voltage characteristic v(t)=Vpe, where 7 = RC, for t> 0, depicted as the solid line in Figure P1-2. Show that the tangent line at t = 0, depicted by the dashed line in Figure P1-2, intercepts the horizontal (time) axis at = RC. Solution: dv(t) dt =-(Vp/T)e/0=-(Vp/7), so the voltage drops by -V, by t = r seconds. (e) [5 marks] Show that the peak-to-peak ripple voltage for the basic power supply circuit in Figure P1-1, V(peak-to-peak), experienced by the load resistor RL, can be approximated by V,(peak-to-peak) 2fRLC' Fv(t) Vp Vpe-t/T Figure P1-2: Exponential waveform and tangent line where Vp is the peak voltage and f is the frequency in Hz of the AC voltage source. Calculate the value for the peak-to-peak ripple voltage, V, (peak-to-peak), for the basic power supply circuit in Figure P1-1 where f = 50 Hz, C = 100 μF and RL = 2.2k. Solution: Se L32 and L33 for proof. Vp = 23.6V, f = 50 Hz, then 23.6 ≈ 1.0727 V 2 x 50 x 2200 x 0.0001