LM7812 supply sags only under load: regulator or headroom?
A trainee at a PSEB NSQF electronics workshop assembles a fixed 12 V, 0.75 A regulated supply: a 230 V/14 V, 1 A transformer, a four-diode bridge (1N4007), a single 1000 µF/25 V filter capacitor, and an LM7812 with a small clip-on heatsink. On the bench the unloaded output measures 11.95 V, which looks perfect. When a 16 Ω, 10 W dummy load (about 0.75 A) is connected, the output falls to 8.6 V, the heatsink stays only warm, and a DMM on AC shows 0.9 V of ripple at the output. The trainee's immediate conclusion is that the regulator IC is counterfeit and wants to replace it. The instructor asks for the DC voltage at the regulator input pin under both conditions before any part is changed; the trainee measures 15.2 V unloaded and 9.8 V loaded. Mains is 231 V.
1. Using the transformer, bridge and filter values, show by calculation whether the LM7812 is faulty. State what the two input-voltage readings prove.
Read the worked answer
Peak secondary voltage = 14 V × 1.414 ≈ 19.8 V. A bridge drops about 1.4 V, so the filter peak is ≈18.4 V.
At 0.75 A the ripple is I/(2fC) = 0.75/(2 × 50 × 1000 × 10⁻⁶) = 7.5 V peak-to-peak. The capacitor valley is therefore about 18.4 − 7.5 = 10.9 V, and the average DC is roughly 18.4 − 3.75 ≈ 14.6 V.
The LM7812 needs about 14.5 V at its input (12 V output + ≈2 V dropout). The measured 9.8 V loaded input is well below that, so the IC has simply run out of headroom and is passing the sagged input through. The regulator is not faulty. The unloaded 15.2 V fits the same circuit; replacing the IC will reproduce exactly the same fault.
2. Recommend a filter capacitance that keeps the regulator input above dropout at 0.75 A, and state two practical side effects of that change.
Read the worked answer
Allow Vin(min) ≈ 15 V, so permitted ripple = 18.4 − 15 = 3.4 V peak-to-peak.
C = I/(2f × Vripple) = 0.75/(100 × 3.4) ≈ 2200 µF. The next standard value is 3300 µF; 4700 µF would give ripple ≈1.6 V and a valley near 16.8 V, which also covers low mains.
Side effects: (1) switch-on inrush and repetitive peak diode current rise sharply, stressing the 1N4007 bridge and possibly the fuse and transformer; (2) physical size and cost increase, and the capacitor must be rated at least 25 V (35 V is safer), with a low-ESR type preferred. Transformer heating also rises because the charging current is drawn in short, high peaks.
3. The shop has no large capacitor in stock. Give an alternative that removes the root cause, and explain why it is better than simply fitting a bigger capacitor.
Read the worked answer
The root cause is insufficient secondary voltage, so raise it: use a 15 V or 16 V, 1 A transformer. A 16 V secondary gives a peak after the bridge of ≈21.2 V, so even with the existing 1000 µF (7.5 V ripple) the valley is ≈13.7 V - still marginal, so pair it with 2200 µF, giving a valley of ≈17.8 V.
This is better because it restores genuine headroom rather than relying on one large component, and it keeps regulator dissipation reasonable: with about 16 V average input, dissipation is (16 − 12) × 0.75 ≈ 3 W, manageable on a small heatsink. Check the worst case at no load and high mains (231 V + 10%), where the input may approach 24 V and dissipation reaches ≈9 W - size the heatsink for that. Also allow for transformer regulation and 10% mains sag when choosing the secondary.