Bridge Rectifier Peak Output
Vdc = Vpeak − 2 · Vdiode
Quick answer
This calculator solves Vdc(peak) = Vpeak − 2 × Vdiode for DC Output (peak). With Transformer Peak 26.4 V, Diode Drop 56.4 V, the result is -86.4 V. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
In a full bridge rectifier with a reservoir capacitor, two diodes conduct in series on each half cycle, so the capacitor charges to the transformer peak voltage minus two diode drops. This peak value, not the averaged Vdc, is what a capacitor-input filter actually delivers under light load. It sets the no-load output and the voltage rating of downstream components.
Variables
- Vdc — Peak DC at the reservoir capacitor (V)
- Vpeak — Peak secondary voltage = √2 × Vrms (V)
- Vdiode — Forward drop per diode (~0.7–1 V)
Practical Notes
Use ~1 V per diode for ordinary silicon at current, or ~0.4 V for Schottky. Under load the average output drops below this peak by the ripple; size the transformer for the loaded, not the no-load, voltage.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| Vdc | DC Output (peak) | V | pV, nV, µV, mV, V, kV |
| Vpeak | Transformer Peak | V | pV, nV, µV, mV, V, kV |
| Vdiode | Diode Drop | V | pV, nV, µV, mV, V, kV |
Step by step
- 1. What you know: Transformer Peak 26.4 V, Diode Drop 56.4 V
- 2. The relation: Vdc(peak) = Vpeak − 2 × Vdiode
- 3. Substitute: Vpeak = 26.4 V, Vdiode = 56.4 V
- 4. Result: DC Output (peak) = -86.4 V
Worked examples
Minimum range
Given: Transformer Peak 0.726 V, Diode Drop 1.55 V
Result: DC Output (peak) = -2.374 V
Low range
Given: Transformer Peak 7.26 V, Diode Drop 15.5 V
Result: DC Output (peak) = -23.74 V
Typical
Given: Transformer Peak 26.4 V, Diode Drop 56.4 V
Result: DC Output (peak) = -86.4 V
High range
Given: Transformer Peak 0.506 kV, Diode Drop 1.08 kV
Result: DC Output (peak) = -1.654 kV
Maximum range
Given: Transformer Peak 5.06 kV, Diode Drop 10.8 kV
Result: DC Output (peak) = -16.54 kV
Solving for each variable
Solve for DC Output (peak)
Transformer Peak 0.726 V, Diode Drop 1.55 V → -2.374 V
Solve for Transformer Peak
DC Output (peak) 0.33 V, Diode Drop 1.55 V → 3.43 V
Solve for Diode Drop
DC Output (peak) 0.33 V, Transformer Peak 0.726 V → 0.198 V
How DC Output (peak) changes with Transformer Peak
| Transformer Peak (input) | DC Output (peak) (result) |
|---|---|
| 2.64 V | -110.2 V |
| 6.6 V | -106.2 V |
| 13.2 V | -99.6 V |
| 19.8 V | -93 V |
| 26.4 V | -86.4 V |
| 39.6 V | -73.2 V |
| 52.8 V | -60 V |
| 105.6 V | -7.2 V |
| 0.1848 kV | 72 V |
| 0.264 kV | 151.2 V |
The same result in other units
| Unit | Value |
|---|---|
| mV | -86400 |
| V | -86.4 |
| kV | -0.0864 |
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<p><a href="https://vbatpower.com/formulas/bridge-rectifier-output/">Bridge Rectifier Peak Output</a> Calculator by vbatpower.com</p> Frequently asked questions
How do you calculate Bridge Rectifier Peak Output?
Use Vdc(peak) = Vpeak − 2 × Vdiode. With Transformer Peak 26.4 V, Diode Drop 56.4 V, DC Output (peak) works out to -86.4 V.
What is the formula for Bridge Rectifier Peak Output?
The formula is Vdc(peak) = Vpeak − 2 × Vdiode. In a full bridge rectifier with a reservoir capacitor, two diodes conduct in series on each half cycle, so the capacitor charges to the transformer peak voltage minus two diode drops.
How do you find Transformer Peak when you know the other values?
Rearrange for Transformer Peak: with DC Output (peak) 0.33 V, Diode Drop 1.55 V, Transformer Peak is 3.43 V.
How do you find Diode Drop when you know the other values?
Rearrange for Diode Drop: with DC Output (peak) 0.33 V, Transformer Peak 0.726 V, Diode Drop is 0.198 V.
What units does Bridge Rectifier Peak Output use?
DC Output (peak) is shown in mV, V, kV. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
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