Overheating a 7805 Regulator in a Small Power Supply
A trainee is building a fixed 5 V, 0.4 A supply for a sensor board. A 14 V AC transformer feeds a bridge rectifier and a 2200 µF reservoir capacitor, giving about 18 V DC at the intended load and nearly 19.5 V at light load. The regulator is a 7805 in a TO-220 package, mounted without a heatsink inside a closed ABS case where the ambient temperature reaches 45 °C. The datasheet gives junction-to-case thermal resistance 5 °C/W, junction-to-ambient without a heatsink 65 °C/W, and maximum junction temperature 125 °C, with thermal shutdown acting above that. The load is a sensor board that draws 0.4 A continuously. On the first test the output starts at 5 V, then falls to about 2 V after several minutes, and the regulator case is too hot to touch. The trainee proposes adding a small fan and increasing the output capacitor.
1. Calculate the regulator's dissipation and junction temperature without a heatsink. Explain the observed output collapse.
Read the worked answer
Dissipation P = (V_in − V_out) × I = (18 − 5) × 0.4 = 5.2 W.
Without a heatsink, the junction rise is P × RθJA = 5.2 × 65 = 338 °C. With 45 °C ambient, the junction would reach about 383 °C, far above the 125 °C maximum and above the thermal-shutdown point.
The sequence is therefore expected: the regulator heats until the internal thermal-shutdown circuit reduces or removes the output, the sensor board stops drawing full current, the device cools, the output returns, and the cycle repeats. The fall to 2 V is the regulator dropping out under thermal stress, not a capacitor problem.
The input voltage is also unnecessarily high. A 7805 needs only about 7-8 V at its input; the extra 10-11 V is being turned into heat. Adding a fan treats the symptom and adds a mechanical failure point.
2. Select a heatsink that keeps the junction at or below 100 °C at 45 °C ambient, allowing 1 °C/W for the mounting insulator. Check the design at the light-load input voltage of 19.5 V as well.
Read the worked answer
Allowed rise = 100 − 45 = 55 °C. At 0.4 A and 18 V input, total thermal resistance required = 55 / 5.2 = 10.6 °C/W.
Subtract RθJC (5 °C/W) and the insulator (1 °C/W): RθSA ≤ 10.6 − 6 = 4.6 °C/W.
Check the worst case at 19.5 V input: P = (19.5 − 5) × 0.4 = 5.8 W. With a 4.6 °C/W heatsink, rise = 5.8 × (5 + 1 + 4.6) = 61.5 °C, giving a junction of about 106.5 °C - marginally above the 100 °C target, so choose a 3.5-4 °C/W heatsink or reduce the input voltage.
Better still, reduce the input: use a 9 V transformer, add a preregulator, or replace the linear stage with a small buck converter. Keep the 0.33 µF input capacitor and 0.1 µF output capacitor close to the pins, and bolt the heatsink with thermal compound and an insulating kit.
3. The trainee suggests adding a fan. Evaluate that proposal and give a better set of modifications with reasons.
Read the worked answer
A fan does not address the root cause: 5.2-5.8 W is being dissipated because the input is far above the regulator's dropout requirement. A fan adds noise, dust ingress, a bearing that will fail, and a new dependency; if it stops, the failure returns silently.
The correct fixes, in order of preference:
- Reduce the input voltage - a 9 V transformer gives about 12 V DC, so dissipation falls to (12 − 5) × 0.4 = 2.8 W, which a modest heatsink can handle.
- Replace the linear regulator with a buck converter, which would dissipate well under 1 W.
- If the 7805 is retained, fit a heatsink of 3.5-4 °C/W or better, use thermal compound, and ventilate the case.
Also verify the reservoir capacitor and bridge rating, add reverse-polarity protection if the load is pluggable, and measure the case temperature after 30 minutes. The goal is a design that remains safe if the room gets hotter or the load increases slightly.