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MEO Class II: MEKG MEKM MEP MET SC&S SSEP 📊 Trends
Q7 (16 Marks) Electrical Circuits & Calculations 🔥 Repeated 8x in exams
MET • Written Exam

(a) By means of a schematic circuit diagram illustrate the peak rectifier. If the supply voltage is v(t) = Vm Sin wt, what is the voltage across the load resistor? (6)

(b) A battery-charging circuit is shown below in Fig. The forward resistance of the diode can be considered negligible and the reverse resistance infinite. The internal resistance of the battery is negligible. Calculate the necessary value of the variable resistance R so that the battery charging current is 1.0 Α. (10)

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Part (a)

Peak rectifier (peak detector):

  • A peak rectifier consists of a diode in series with a capacitor and a load resistor. The diode rectifies the a.c. input and charges the capacitor to the peak value of the input voltage.
  • Circuit: a.c. source -> diode -> node (capacitor to earth, load resistor to earth). The output is taken across the capacitor/load.
  • Operation: during the positive half cycle the diode conducts and charges the capacitor to the peak value Vm. When the input falls below the capacitor voltage, the diode becomes reverse biased and stops conducting; the capacitor discharges slowly through the load resistor. If the time constant (R x C) is large compared with the period, the capacitor holds the voltage near Vm, so the output is approximately the peak value.
  • If the supply voltage is v(t) = Vm sin(wt), the voltage across the load resistor is approximately the peak value Vm (for an ideal diode with negligible forward drop and a large time constant). The output is a d.c. voltage close to Vm, with a small ripple.
Part (b)

Battery-charging circuit:

  • The circuit is a half-wave (or full-wave) rectifier feeding a battery through a variable resistor R. The diode forward resistance is negligible and reverse resistance infinite; battery internal resistance negligible.
  • The charging current is to be 1.0 A. The battery has a fixed e.m.f. (say E_b). The rectified supply provides a peak voltage Vm. The charging current flows only when the instantaneous rectified voltage exceeds the battery e.m.f.
  • For a half-wave rectifier, the mean charging current is given by the average of (v - E_b)/R over the conducting period.
  • The necessary value of R is found from: R = (V_mean - E_b) / I_charge, where V_mean is the mean rectified voltage available. For example, if the supply peak is Vm and the battery e.m.f. is E_b, then R = (Vm - E_b)/1.0 ohm (for a simple d.c. equivalent), or using the mean value of the rectified waveform.
  • The variable resistor is adjusted so that the charging current is exactly 1.0 A. (The exact numerical value depends on the supply voltage and battery e.m.f. given in the figure; the method is to set R so that the mean charging current equals 1.0 A.)
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