A transformer, a diode bridge, a filter capacitor and a linear regulator are the four stages of the classic mains power supply. This one turns 12 V AC into a steady 5 V for logic circuits and microcontrollers.
T1 steps the mains down to 12 V AC and isolates the rest of the circuit from it. F1 on the primary blows if the transformer or anything after it fails short. A 12 V RMS winding peaks at about 12 V × √2 ≈ 17 V.
BR1 holds four diodes in the diamond of a full-wave bridge. When the top of the secondary is positive, current flows through the upper-right diode to the + corner, through the load and back to the − corner through the lower-left diode. On the other half-cycle it flows through the lower-right and upper-left diodes. Either way the + rail only ever sees positive pulses, two per mains cycle, and two diode drops are lost on the way: about 15.6 V at the peak.
C1 charges to the peak of each pulse and feeds the load in between, so the voltage sags only a little before the next pulse tops it up. That sag is the ripple, about I / (2fC): 2.3 V at 0.5 A on 50 Hz mains, or 1.9 V on 60 Hz.
The 7805 needs its input a couple of volts above its output; check the dropout in your regulator's datasheet. Here the input never falls below about 13 V, so the output holds at 5 V. The cost is heat: the regulator burns off the difference, about 4.7 W at 0.5 A, so it needs a heatsink. C2 at the output improves the regulator's response to sudden changes in load.
Vpeak ≈ VAC × √2 − 2 × VF = 12 V × 1.414 − 1.4 V ≈ 15.6 VΔV ≈ I / (2·f·C) = 0.5 A / (2 × 50 Hz × 2200 µF) ≈ 2.3 VP ≈ (Vin,avg − Vout) × I ≈ (14.4 V − 5 V) × 0.5 A ≈ 4.7 WVin,avg is roughly the peak voltage minus half the ripple.
C ≈ I / (2·f·ΔV)This is the smallest capacitance that meets the target, so round up. For 1 V of ripple at 0.5 A on 50 Hz: 0.5 A / (100 Hz × 1 V) = 5000 µF, so a 5600 µF part; 4700 µF would give about 1.06 V. Electrolytics are often −20 % from nominal, which calls for 6800 µF.
| Ref | Part | Value | What it does |
|---|---|---|---|
| BR1 | Bridge rectifier | DB107 | DB107 bridge rectifier (1 A, 1000 V). Four 1N4007 diodes wired as a bridge do the same job. |
| C1 | Electrolytic capacitor | 2200 µF | 2200 µF electrolytic filter capacitor, rated 25 V or more. |
| C2 | Capacitor | 100 nF | 100 nF ceramic capacitor at the regulator output. |
| F1 | Fuse | 250 mA | Slow-blow fuse on the primary. Size it for the transformer's primary current and switch-on surge. |
| T1 | Transformer | 12 V AC | Transformer with a 12 V AC secondary, rated about 1 A for 0.5 A of DC output. |
| U1 | IC / module | 7805 | 7805 5 V, 1 A linear regulator (LM7805, L7805), on a heatsink. |
| V1 | AC source | Mains | Mains input. |
Recovered from the drawing itself: every pin on each net below is joined by a wire, a junction or a shared supply or ground symbol.
| Net | Connects |
|---|---|
| N1 | T1.p1, F1.1 |
| N2 | T1.p2, V1.2 |
| N3 | T1.s1, BR1.ac1 |
| N4 | T1.s2, BR1.ac2 |
| N5 | F1.2, V1.1 |
| N6 | BR1.plus, C1.+, U1.IN |
| GND | BR1.minus, C1.-, U1.GND, C2.2 |
| +5V | U1.OUT, C2.1 |
A single diode uses only one half of each AC cycle. The bridge uses both halves, which halves the time between pulses, halves the ripple for the same capacitor and needs no centre-tapped transformer.
C ≈ I / (2 · f · ΔV) for a full-wave rectifier, where ΔV is the ripple you can accept. Keep the bottom of the ripple above the regulator's minimum input voltage.
About 12 V × √2 − 1.4 V ≈ 15.6 V at the peak with a filter capacitor and a light load. It falls as the load and the ripple grow.
Not from a 7805, which is rated about 1 A and is usually limited by heat before that. For more current or less heat, use a switching regulator.
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