Checking a Student's 6 V Supply Before It Fails
A student builds a low-voltage DC supply for a project. A step-down transformer with a 12 V rms secondary feeds a single silicon diode (forward drop 0.7 V) in series with the load, and a 1000 microfarad electrolytic capacitor marked '16 V' is connected across the load. The load draws a steady 50 mA and the mains frequency is 50 Hz. On testing, the DC output measures close to 16 V rather than the 6 V the student expected, and the capacitor is noticeably warm. The student's logbook states that 'the capacitor charges to the rms value of the secondary, and 1000 microfarad is so large that the ripple can be ignored'. You are asked to check the design quantitatively, decide whether the capacitor is being used within its rating, and specify what must be changed to obtain a regulated 6 V output.
1. Calculate the peak voltage available after the diode, the DC output with its ripple, and judge whether the capacitor's 16 V rating is acceptable.
Read the worked answer
Secondary peak = 12 x sqrt(2) = 16.97 V. After the diode drop, the capacitor can charge only to 16.97 - 0.7 = 16.3 V. The logbook is wrong on this point: a capacitor in this position charges to the peak value (minus the diode drop), not the rms value. That is why the output is near 16 V and not near 12 V, let alone 6 V.
Ripple for a half-wave rectifier with a capacitor filter:
V_r(peak-to-peak) = I/(f C) = 0.050/(50 x 1000 x 10^-6) = 0.050/0.050 = 1.0 V.
The DC level is roughly the peak minus half the ripple, about 15.8 V. So the ripple is 1 V on 15.8 V, i.e. about 6 % - not negligible.
Capacitor rating. The capacitor is held at about 16.3 V while marked 16 V; it is at or beyond its rating on every cycle, and it is warm because of the ripple current through its equivalent series resistance. It will vent or dry out. The correct choice is a 25 V (or higher) unit of the same capacitance. Electrolytics should be derated; running at 100 % of the marked voltage is never acceptable.
2. Compare this half-wave circuit with a full-wave bridge rectifier for the same load and capacitor: what changes in the DC level and in the ripple?
Read the worked answer
Full-wave bridge. The capacitor recharges twice per mains cycle, so the ripple frequency is 2f = 100 Hz and
V_r = I/(2f C) = 0.050/(100 x 1000 x 10^-6) = 0.50 V - half the half-wave ripple, at double the frequency, which is also easier to smooth.
But two diodes conduct in series in each path, so the peak is 16.97 - 2(0.7) = 15.6 V, about 0.7 V lower than in the half-wave case, and the DC level is about 15.3 V.
So the bridge gives cleaner DC at a slightly lower level and, importantly, a much better transformer utilisation (both half-cycles are used). Either way the raw DC is around 15 V, nowhere near 6 V, and no amount of extra capacitance will bring it down - adding capacitance reduces ripple but always leaves the output near the peak. The student's expectation of 6 V was a design error, not a component error.
3. Design the regulating stage needed to deliver 6 V to the 50 mA load from about 15.6 V, choosing the series resistor and checking the zener dissipation. State the limitations of this design.
Read the worked answer
A shunt zener regulator is the simplest option. Choose a 6.2 V zener and allow a minimum zener current of 10 mA so that regulation survives some load variation.
Total current through the series resistor: I = 50 mA + 10 mA = 60 mA. Voltage across it: 15.6 - 6.2 = 9.4 V.
R = 9.4/0.060 = 157 ohm, so use a preferred value of 150 ohm.
Check. With 150 ohm: I = 9.4/150 = 62.7 mA. At full load the zener carries 12.7 mA; at no load it carries all 62.7 mA. Worst-case zener dissipation = 6.2 x 0.0627 = 0.39 W, so a 1 W zener is required, not the usual 400 mW part. Resistor dissipation = (9.4)^2/150 = 0.59 W, so a 1 W resistor.
Limitations. Efficiency is poor: the useful output is 6.2 x 0.050 = 0.31 W out of an input of about 15.6 x 0.0627 = 0.98 W, roughly 32 %. The output is only as good as the zener, the series resistor wastes most of the power as heat, and the whole regulator must be redesigned if the load current changes by more than a few tens of milliamps. A three-terminal regulator IC (for example a 7806 with its own heatsink) would give far better line and load regulation and is the appropriate choice for a real project supply. The capacitor must still be changed to 25 V, and the 16 V unit must not be reused.