A 7805 Supply That Drops Out at 0.4 A
A technician builds a 5 V, 0.4 A bench supply using a 230 V/9-0-9 V, 500 mA transformer, a bridge rectifier, a 1000 µF 25 V reservoir capacitor and a 7805 regulator bolted to a 6 °C/W heatsink. At 0.4 A load the output is 4.6 V with about 1.8 V ripple. The regulator case reaches 78 °C in a 35 °C room. The DC voltage across the reservoir capacitor measures 8.1 V average and 6.3 V minimum on a true-RMS meter with DC coupling. The transformer secondary measures 8.6 V rms at load. The 7805 datasheet gives a 2 V dropout voltage and a 5 °C/W junction-to-case thermal resistance. The load is a digital circuit that draws a steady 0.4 A. The technician proposes adding a 10,000 µF capacitor and a larger heatsink.
1. Calculate the ideal peak voltage, the ripple voltage and the minimum regulator input. Compare this with the 7805 dropout requirement.
Read the worked answer
Ideal peak: 9 × √2 = 12.73 V; after the bridge, 12.73 − 2 × 0.7 = 11.3 V. Ripple for a full-wave capacitor filter is approximately Vr = I/(2fC) = 0.4/(2 × 50 × 1000 × 10⁻⁶) = 4.0 V peak-to-peak. Ideal minimum is 11.3 − 4.0 = 7.3 V.
The measured minimum is 6.3 V because the transformer is rated only 500 mA and the secondary reads only 8.6 V rms at load; its regulation and the rectifier loss reduce the peak. The 7805 needs about 2 V dropout, so its minimum input is about 7 V. At 6.3 V it is in dropout, and the output falls to 4.6 V. Increasing the capacitor alone raises the minimum but cannot fix the transformer's rms current overload and sag.
2. Assess the proposed 10,000 µF capacitor and larger heatsink. What supply changes would you make?
Read the worked answer
The 500 mA transformer is already overloaded. A capacitor-input rectifier draws an rms secondary current much higher than the DC output, typically 1.6–1.8 times, so 0.4 A DC means roughly 0.65–0.72 A rms. A larger capacitor makes the current pulses narrower and the rms current worse.
A suitable linear fix is a 9 V or 12 V transformer rated at least 1.5 A, a 2200–4700 µF reservoir capacitor, and a heatsink sized for the increased dissipation. A 12 V transformer gives about 15.6 V peak and about 10 W dissipation at 0.4 A, which is wasteful. A better fix is a 24 V or 12 V input buck converter to 5 V, or a low-dropout regulator with adequate headroom. Whichever is chosen, verify that the minimum input stays above dropout at low mains and full load.
3. Estimate the regulator dissipation and explain whether the measured 78 °C case temperature is consistent with the heatsink. What would you check?
Read the worked answer
Dissipation is approximately (Vin(avg) − 5) × I. With 8.1 V average and 0.4 A, P = 1.24 W. A 6 °C/W heatsink plus about 5 °C/W junction-to-case and 1 °C/W interface gives θJA ≈ 12 °C/W, so the case should rise only about 7 °C above ambient, to about 42 °C.
A measured 78 °C is inconsistent with 1.24 W and suggests the heatsink is not properly bonded, the actual current is higher, the input is higher, or the regulator is oscillating. Check the mounting torque and insulating washer, measure the input current, and inspect the output capacitor and layout for stability. If the worst-case dissipation rises to 2.5–4 W with a larger transformer, recalculate the heatsink and consider a switching regulator.