7805 Regulator Dropout, Heating and a Series Resistor
A trainee builds a 5 V, 0.8 A supply for a microcontroller board. The parts are a 230 V/9 V transformer rated 1 A, a bridge rectifier, a 2200 µF filter capacitor, and a 7805 regulator bolted to a clip-on heatsink with θSA = 50 °C/W. The 7805 has θJC = 5 °C/W and the thermal pad gives θCS = 1 °C/W. At no load the output is 5.02 V. At 0.8 A the output falls to 4.55 V and the heatsink is too hot to touch after five minutes. The input DC voltage is measured as 6.6 V average with 1.8 V peak-to-peak ripple. Ambient temperature is 32 °C. The 7805 data sheet requires Vin ≥ 7.5 V for regulation and gives TJ(max) = 125 °C. The trainee proposes adding a 1 Ω, 2 W resistor in series with the regulator input to reduce its heating.
1. Calculate the 7805 power dissipation and junction temperature with the existing heatsink. Is it in dropout? Support your answer with the measured input.
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Use P = (Vin - Vout) × I. At the measured operating point, P = (6.6 - 4.55) × 0.8 = 1.64 W. The total thermal resistance is θJC + θCS + θSA = 5 + 1 + 50 = 56 °C/W. Tj = Ta + P × θtotal = 32 + 1.64 × 56 = 32 + 91.8 = 123.8 °C. That is just below the 125 °C maximum, with almost no margin; the heatsink is too small. The regulator is in dropout: the data sheet requires Vin ≥ 7.5 V, but the average is 6.6 V and the ripple trough is about 6.6 - 0.9 = 5.7 V. The output of 4.55 V instead of 5 V confirms that regulation has been lost. The heating is not the only fault; the input is also inadequate.
2. Evaluate the 1 Ω series resistor proposal. Calculate its effect at 0.8 A.
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The resistor would drop V = IR = 0.8 × 1 = 0.8 V, so the regulator input would fall from 6.6 V to about 5.8 V. That is further below the 7.5 V minimum, so dropout would become worse and the output would fall below 4.55 V. The resistor dissipates I²R = 0.8² × 1 = 0.64 W, so a 2 W rating is electrically okay, but it wastes power and does not solve the problem. It may appear to reduce regulator dissipation only because the output is lower and the input is lower, not because the design is correct. Reject this proposal. The correct approach is to raise and stabilise the input voltage and improve the heatsink, or use a switching regulator.
3. Recommend changes to deliver 5 V at 0.8 A with TJ below 100 °C. Include a filter capacitor and heatsink calculation.
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First raise the input above the dropout with margin. Use a transformer with better regulation or a larger filter capacitor. For about 1 Vpp ripple at 0.8 A, C = I/(2fVpp) = 0.8/(2×50×1) = 8000 µF, so use 10,000 µF rather than 2200 µF. Check the transformer rms current and VA rating; a 9 V, 2 A transformer is more realistic than the 1 A unit. Suppose the improved average input is about 9.5 V. Then P = (9.5 - 5) × 0.8 = 3.6 W. For TJ < 100 °C: 100 = 32 + 3.6 × (5 + 1 + θSA), so 68/3.6 = 18.9 = 6 + θSA, giving θSA < 12.9 °C/W. Use a 10 °C/W heatsink with thermal compound. Recheck TJ and Vin at low mains. If the heat budget is still tight, use a 5 V switching regulator, which would dissipate far less.