How this instrument works
A bridge rectifier arranges four diodes so that both halves of an AC cycle push current through the load in the same direction — the negative swing gets flipped up onto the positive side instead of being discarded. Averaged over a full cycle, that folded sine wave works out to 2⁄π of its peak value, about 63.7 percent: integrate Vpeak·sin(θ) from 0 to π and divide by π, and the −cos(θ) antiderivative collapses to exactly 2Vpeak⁄π.
The two diode drops matter because a bridge, unlike a center-tapped full-wave rectifier, always routes current through two junctions in series — one carrying it into the load, one carrying it back out — no matter which half of the cycle is active. Silicon diodes cost about 0.7 V each this way; Schottky diodes, with their lower barrier, cost closer to 0.2–0.4 V, which is why they turn up in low-voltage supplies where every diode drop is a larger slice of the total.
The formula is a straight line, not a physical law, and it breaks down at its edges. Below roughly twice the diode drop, the AC peak never exceeds the combined threshold of the two junctions in the path, no current flows, and the real output is 0 V — not the negative figure the line predicts if you extrapolate it past where the diodes actually conduct. It also ignores load current: real diode drop creeps upward as current rises, so this is a starting estimate, not a bench reading.
- Enter the Peak AC input voltage (Vpeak) — the crest of the AC waveform feeding the bridge, not its RMS rating; multiply an RMS figure by √2 first if that's what your transformer datasheet gives.
- Enter the Diode forward voltage drop (Vdrop) — 0.7 V for standard silicon rectifiers, 0.2–0.4 V for Schottky diodes.
- Read off the Average DC output voltage (Vdc) — the mean level the bridge delivers before a reservoir capacitor smooths the ripple.
- Switch any field's unit between mV, V, or kV to match the values on your circuit's datasheet or meter.
Worked example — 170 V peak through silicon diodes
A 120 V RMS mains supply peaks at about 170 V (120 × √2 ≈ 169.7 V, the figure transformer datasheets round to). Feed that peak into a full-wave bridge built from four standard silicon diodes, each dropping 0.7 V, and the formula gives Vdc = 2 × 170 ⁄ π − 2 × 0.7 = 108.225361302 − 1.4 = 106.825361302 V.
That 106.825361302 V is what a multimeter reads across the bridge's output terminals before any regulator or filter capacitor smooths it — the number a technician checks first when a bench supply looks low. Swap in Schottky diodes at 0.3 V instead of silicon's 0.7 V on a larger 340 V peak (a 240 V RMS mains supply) and the same formula returns 215.850722605 V, exactly 0.8 V above what four silicon diodes would deliver at that same 340 V peak — the difference is always 2 × (0.7 − 0.3), independent of Vpeak.
Questions
Why does the formula subtract two diode drops instead of one?
Because current in a full bridge always passes through two diodes in series, no matter which half of the AC cycle it's in — one diode carries current into the load, another carries it back to the source. A half-wave rectifier only ever has one diode in the path, so its average is Vpeak ⁄ π − Vdrop; a center-tapped full-wave rectifier also has just one diode conducting at a time, giving 2Vpeak ⁄ π − Vdrop. The bridge trades that extra diode drop for not needing a center-tapped transformer.
Where does the 2 ⁄ π factor come from?
It's the average value of a full-wave rectified sine wave. Integrating Vpeak·sin(θ) over one half-cycle, 0 to π, and dividing by π gives (Vpeak⁄π)[−cos θ] from 0 to π, which works out to 2Vpeak⁄π ≈ 0.637 Vpeak. A bridge folds the negative half of the AC cycle up onto the positive side, so the whole waveform — not just alternate halves — carries this same mean value before the diode drops are subtracted.
What diode drop should I use for Vdrop?
0.7 V is the standard figure for a silicon rectifier diode, such as the 1N4001 family, at typical load current. Schottky diodes drop closer to 0.2–0.4 V, which is why they show up in low-voltage, high-efficiency supplies where every tenth of a volt matters. The 0.7 V figure is itself an approximation — real diode drop rises slowly with current — so treat the result as a design estimate, not a lab-grade reading.
Why does Vdc go negative if Vpeak is set to zero?
Because the formula is a straight-line approximation that assumes the diodes are always conducting; it has no built-in floor at zero. Below roughly twice Vdrop of peak input, the AC swing never exceeds the two diodes' combined threshold, no current flows at all, and the true output is 0 V, not the negative number the line predicts. Read any result below zero as 'the rectifier does not turn on at this input,' not as a literal reverse voltage.
How is this different from a half-wave rectifier calculation?
A half-wave rectifier lets only one polarity of the AC cycle through, so its average is Vpeak ⁄ π − Vdrop — half the multiplier and only one diode drop. A bridge rectifies both halves of the cycle, doubling the average value and the ripple frequency, at the cost of routing current through two diodes instead of one. For the same 170 V peak and 0.7 V drop, a half-wave design would average only about 53.4 V — roughly half the bridge's 106.8 V.
Does the load current change the answer?
Not in this formula, which assumes a fixed diode drop and no series resistance in the transformer or diodes. In practice both Vdrop and the effective Vpeak sag a little as load current rises, so a real supply loses a few percent that this idealized, no-load formula doesn't capture. Use it for sizing and sanity-checking a design, then confirm with a meter once the circuit is built.