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Free Bridge Rectifier & Filter Capacitor Calculator Electronics & Embedded
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Free Bridge Rectifier & Filter Capacitor Calculator

Calculate peak DC voltage, ripple voltage, required smoothing capacitor (µF), diode PIV, and capacitor RMS ripple current for linear power supplies.

Transformer & Load Specifications

V_rms
A
µF

📊 Rectified DC & Ripple Performance

Average DC Output Voltage (V_dc)
22.28 V
Peak Voltage: 24.06 V | Valley: 20.51 V
Peak-to-Peak Ripple (ΔV)
3.55 V_pp
4.6% Ripple Factor (RMS)
Capacitor RMS Ripple Current
3.85 A_rms
Capacitor heating current
Minimum Diode PIV
25.5 V
Peak Inverse Voltage rating
Capacitor Voltage Rating
35 VDC
≥ 1.3× peak DC voltage
💡 The 2,000 µF per Ampere Rule of Thumb
In analog linear power supplies, a classic design rule is to allocate 2,000 µF to 2,500 µF of capacitance for every 1 Ampere of load current. This keeps peak-to-peak ripple below 10% of the rail, leaving plenty of headroom for post-regulators (LM317, 7812).

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How Full-Wave Bridge Rectifiers Convert AC to DC

A full-wave diode bridge (Graetz bridge) rectifies both halves of an AC sine wave into pulsating unipolar DC at double the line frequency (120 Hz for North America, 100 Hz for Europe). A large electrolytic reservoir capacitor smooths these pulses into steady DC.

1. Peak Output Voltage Formula

Because two diodes conduct simultaneously in series during each half-cycle:

V_peak = (√2 × V_RMS) - (2 × V_diode) ≈ (1.414 × V_RMS) - 1.4V

2. Peak-to-Peak Ripple Voltage

Between consecutive charging peaks ($T_{ripple} = 1 / (2 imes f_{line})$), the filter capacitor discharges into the load:

V_ripple(p-p) = I_load / [2 × f_line × C]

The average DC voltage available to your downstream regulator is:
V_DC,avg = V_peak - [V_ripple(p-p) / 2]

3. Diode Peak Inverse Voltage (PIV)

During the reverse-biased cycle, each diode experiences the peak transformer voltage:
PIV ≥ √2 × V_RMS. Always select diodes rated for at least $1.5 imes$ to $2 imes$ this minimum to absorb AC line voltage surges.

Frequently Asked Questions

Why does the transformer secondary RMS rating read lower than the DC output?

A 12V AC transformer outputs 12V RMS. The peak of that wave is 12 * 1.414 = 16.97V. Subtracting two diode drops (~1.4V) leaves ~15.5V peak DC. When lightly loaded, the filter capacitor charges to this peak, reading higher than the AC RMS rating.

What causes electrolytic capacitors to blow up in power supplies?

The two main culprits are: (1) exceeding the capacitor's DC voltage rating, and (2) exceeding its maximum RMS ripple current rating. During each 120 Hz recharge pulse, high peak current flows into the capacitor, heating internal electrolytes until safety vents rupture.

How much voltage headroom does a linear regulator (like 7812) need?

Standard 78xx regulators require a 2.0V to 2.5V dropout voltage. For a clean 12V output, the valley of your ripple voltage (V_peak - V_ripple) must NEVER drop below 14.5V under worst-case full load and low AC wall voltage.

What is the inrush current when the power switch is first turned on?

At the instant of turn-on, an uncharged 10,000 uF capacitor looks like a dead short circuit. A massive inrush surge (often 20A to 50A) flows through the diodes and transformer. NTC thermistors or soft-start relays are commonly used to protect the bridge rectifier.