The bridge rectifier calculator above turns a transformer secondary voltage into the DC voltage you will actually measure at the output, once the diode drops and the ripple have been taken out. It handles the three standard topologies, works in either direction between ripple and capacitance, and reports the minimum peak inverse voltage the diodes have to survive. Those are the four numbers that decide whether a linear supply works or lets out smoke.
Arb Digital builds free tools that answer the whole question rather than the easy half of it. Most rectifier pages stop at multiplying the RMS voltage by the square root of two, which is the unloaded answer and almost never the one you want. This page starts there and then subtracts the forward drops, computes the ripple from the load current and the capacitance, and gives the average DC figure a regulator downstream would actually see.
What This Bridge Rectifier Calculator Does
A rectifier converts alternating current to direct current by allowing conduction in one direction only. A single diode gives half-wave rectification, passing every other half-cycle and wasting the rest. Two diodes fed from a centre-tapped secondary give full-wave rectification, using both halves. Four diodes in a bridge give full-wave rectification from an ordinary two-wire secondary, at the cost of a second forward drop in the current path.
On its own a rectifier produces a lumpy output that touches zero twice per cycle. A reservoir capacitor across the output fills in the gaps: it charges to near the peak on each conduction pulse and then supplies the load on its own until the next peak arrives. The voltage sag during that interval is the ripple, and the calculator's job is to relate it to the capacitance, the load current and the ripple frequency.
Topology changes the arithmetic in two ways. It sets how many diode drops sit in the path, one or two, and it sets the ripple frequency. Full-wave circuits refill the capacitor twice per mains cycle, so the ripple frequency is double the line frequency and half the capacitance is needed for the same ripple. A half-wave circuit refills once per cycle and needs twice the capacitor for the same result, which is one of several reasons it is rarely used above a few milliamps.
How to Use It
- Enter the secondary RMS voltage, not the mains voltage. This is the figure printed on the transformer's output winding. For a centre-tapped part, use the half-winding voltage.
- Set a realistic forward drop. Silicon rectifiers are conventionally taken at 0.7 V, but the real figure rises with current. Look it up on the datasheet curve at your working current if the margin is tight.
- Enter the current the load actually draws, not the transformer rating. Ripple depends on how fast the capacitor is being emptied, so the load current is the driving term.
- Choose the capacitor direction. Design mode gives the capacitance for a ripple you specify; check mode gives the ripple for a capacitor you already have.
- Size the diodes from the PIV figure and add margin. The reported value is the minimum the parts must withstand, before any allowance for mains surges or transformer transients.
The Formula: How Rectifier Output Is Calculated
Start from the peak. Section 15.2 of OpenStax University Physics Volume 2, on simple AC circuits, defines the RMS voltage as the peak voltage divided by the square root of two, so the peak is the RMS value multiplied by 1.41421. A 12 V RMS secondary therefore peaks at 16.97 V.
Subtract the conducting diodes. Section 9.7 of OpenStax University Physics Volume 3, on semiconductor devices, describes the p-n junction diode as the element that permits current in one direction only and notes that diodes are called rectifiers because they turn alternating current into direct current. A bridge has two of them in series with the load at every instant, so at 0.7 V each the peak output is 16.97 − 1.4 = 15.57 V.
Then take the ripple. The capacitor supplies the load between refills, and for the modest ripple used in practice the discharge is close enough to linear that the standard approximation applies: the peak-to-peak ripple equals the load current divided by the product of the ripple frequency and the capacitance. With 500 mA, a 120 Hz ripple frequency for a full-wave circuit on a 60 Hz line, and a target of 1 V peak to peak, the required capacitance is 0.5 ÷ (120 × 1) = 4,167 µF. The average DC output is then the peak minus half the ripple, 15.57 − 0.5 = 15.07 V. The exponential discharge behind that approximation is set out in section 10.5 of the same volume, on RC circuits.
Why the Output Is Not 1.414 Times the RMS Voltage
Measure a rectifier output with nothing connected and you will see almost exactly the peak: the capacitor charges to the top of each cycle and, with no load, has nothing to discharge into. Connect the load and the reading drops, sometimes alarmingly. Two separate effects are at work and both are in the calculator.
The first is the diode drops, which are fixed and predictable. Two silicon junctions cost about 1.4 V regardless of the secondary voltage, which is trivial on a 40 V rail and severe on a 6 V one. Schottky diodes roughly halve that penalty, which is why low-voltage supplies often use them despite the higher reverse leakage.
The second is the ripple, and it grows with load current and shrinks with capacitance. The useful DC level is the peak minus about half the ripple, but what matters for a regulator downstream is the bottom of the ripple, not the average. A regulator that needs two volts of headroom must still have it at the lowest point in the ripple trough, at the highest load, at the lowest mains voltage the supply must tolerate. Designing to the average figure is a common way to build a supply that works on the bench and fails in service.
There is a third effect this calculator does not model: transformer regulation. A small transformer's secondary voltage falls under load, often by ten to twenty per cent between no load and full load, because of winding resistance. The no-load voltage printed on a cheap transformer is frequently higher than its rated figure, so the honest approach is to measure your own secondary at the current you intend to draw.
Sizing the Reservoir Capacitor
Capacitance and ripple trade off directly. Halving the ripple doubles the capacitance, and there is no clever way around that in a capacitor-input filter. The practical stopping point is usually set by physical size, by cost, or by the inrush current a very large capacitor demands at switch-on.
Voltage rating matters as much as capacitance. The capacitor sees the peak output voltage, not the average, and it sees the unloaded peak whenever the load is disconnected or switched off. A rail that measures 15 V under load may sit at 17 V with the load removed, so a 16 V capacitor is not adequate. The usual rule is to pick a rating at least a third above the highest voltage the capacitor can ever see.
Electrolytic capacitors also carry a ripple-current rating, and in a capacitor-input filter the ripple current is substantially higher than the DC load current, because the capacitor is refilled in short high peaks rather than continuously. Exceeding that rating heats the capacitor internally and shortens its life sharply. Our capacitance converter handles the unit changes between farads, microfarads and nanofarads, and the capacitor energy calculator shows how much energy a given reservoir actually stores.
Peak Inverse Voltage and Inrush Current
Every diode spends half the cycle reverse biased, and it has to hold off whatever voltage appears across it. In a bridge with a capacitor-input filter, the non-conducting diodes see roughly the peak secondary voltage. In a half-wave circuit with a reservoir capacitor, the single diode sees the peak of the secondary plus the capacitor voltage on the other side, which is close to twice the peak. A centre-tapped full-wave circuit is the same, because each diode sees its own half-winding plus the other half in series.
The calculator reports these minima. Real designs use a comfortable multiple, because mains supplies carry transients that a bare minimum rating will not survive. Common 1N400x rectifiers are cheap enough that going several steps up the voltage range costs almost nothing.
The other stress is inrush. At the instant power is applied, an empty reservoir capacitor is effectively a short circuit, and the only things limiting the charging current are the transformer's winding resistance and the diodes' bulk resistance. Peak inrush of tens of amps is normal in a supply with a large reservoir, which is why diodes are chosen with a surge rating well above the average current, and why larger supplies use a thermistor or a resistor to soften the first few cycles. Sizing the series parts is ordinary Ohm's law work, covered by our Ohm's law calculator and the voltage drop calculator.
Transformer Rating and the Power Factor Trap
A capacitor-input rectifier does not draw a sine wave from the transformer. It draws nothing at all for most of the cycle and then a tall narrow spike near each peak, while the capacitor is being refilled. The RMS value of that spiky current is much larger than its average, typically by a factor between 1.5 and 2, which means the transformer heats up more than the DC output power suggests.
The practical consequence is that a transformer's volt-amp rating should be roughly 1.6 to 1.8 times the DC watts you intend to draw. A supply delivering 15 V at 0.5 A is only 7.5 watts of DC, but asking a 7.5 VA transformer to provide it will overheat the winding. The same effect is why the fuse in the primary should be a slow-blow type: the inrush at switch-on would clear a fast fuse of the correct running rating.
If the ripple is still too large after the reservoir capacitor has been made as big as is sensible, the answer is a regulator rather than more capacitance. A linear regulator removes the remaining ripple at the cost of dissipating the difference as heat, and a switching converter does it far more efficiently. Our DC-DC converter calculator covers the switching case, and the voltage divider calculator handles the feedback network many regulators need.
Arb Digital builds free tools like this one because useful pages earn attention. If you want tools, calculators or content built for your own audience, we can help.
Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Using the full end-to-end voltage of a centre-tapped winding — each diode only sees half of it, so entering the whole figure doubles the calculated output.
- Forgetting that a bridge costs two diode drops — about 1.4 V of silicon, which is a small tax on a 40 V rail and a serious one on a 5 V supply.
- Rating the reservoir capacitor for the loaded voltage — it sees the unloaded peak whenever the load is switched off, which can be a couple of volts higher.
- Designing a regulator's headroom from the average DC figure — what matters is the bottom of the ripple trough at maximum load and minimum mains voltage.
- Choosing a transformer by DC watts — a capacitor-input filter draws short high current peaks, so the volt-amp rating needs to be well above the DC power delivered.
Related Free Tools From Arb Digital
Continue the design with the Ohm's law calculator for load and bleeder resistors, the voltage divider calculator for regulator feedback networks, and the voltage drop calculator for wiring losses. Capacitor questions are covered by the capacitor energy calculator, the capacitor charge time calculator and the capacitance converter. For switched-mode alternatives see the DC-DC converter calculator, and for reading the parts in front of you, the resistor colour code calculator. The full free online tools hub lists everything Arb Digital publishes.
Frequently Asked Questions
Multiply the secondary RMS voltage by 1.414 to get the peak, subtract two diode forward drops for the bridge, then subtract about half the peak-to-peak ripple. A 12 V RMS secondary with 0.7 V diodes and 1 V of ripple gives roughly 15.1 V of usable DC.
Current flows through two of the four diodes in series at every instant, one on the way out to the load and one on the way back. Half-wave and centre-tapped full-wave circuits only ever have one diode in the path, so they lose a single drop.
Divide the load current by the product of the ripple frequency and the ripple voltage you will accept. A full-wave circuit on a 60 Hz line has a 120 Hz ripple frequency, so 0.5 A with 1 V of allowed ripple needs about 4,200 microfarads. Halving the ripple doubles the capacitance.
In a bridge with a capacitor filter the non-conducting diodes see roughly the peak secondary voltage. In half-wave and centre-tapped circuits they see about twice the peak. Those are minimums, and real designs use a substantial margin to survive mains transients.
Usually transformer regulation. A small transformer's secondary voltage sags under load because of winding resistance, often by ten to twenty per cent. Measure the secondary at your actual working current rather than trusting the printed rating.
Roughly 1.6 to 1.8 times the DC watts you intend to draw. A capacitor-input filter refills the reservoir in short tall current pulses, so the RMS current in the winding is much higher than the DC load current and the transformer heats accordingly.
Only for very light loads. It uses one diode instead of four, but it refills the capacitor once per cycle instead of twice, so it needs double the capacitance for the same ripple and it puts a DC component through the transformer winding.
This tool is provided for educational and estimating use, and it models an idealised rectifier without transformer regulation, winding resistance or diode recovery effects. Mains-connected equipment carries a risk of fatal electric shock and fire. Any circuit connected to the supply mains should be designed, built and inspected by a suitably qualified person, in accordance with the wiring rules that apply where you are.
