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Filter Capacitors and Ripple

A bridge or center-tap rectifier converts alternating current into a pulsating direct current — a series of humps that repeatedly rise to the peak voltage and fall back toward zero. If you tried to power a ham radio transceiver directly from this pulsating output, the result would be a loud 120 Hz hum in the audio, ripple modulation on the transmitted carrier, and likely microcontroller crashes as the supply voltage swings up and down. Before DC is useful, it must be smoothed.

The filter capacitor is the component that does this smoothing. It acts as an energy reservoir, charging up to the peak voltage when the diodes conduct and then releasing that stored energy to the load during the brief periods when the diodes are off. The larger the capacitor, the more energy it stores, and the less the voltage falls between pulses. Understanding how to calculate the required capacitor value, how to choose the right component, and how to design the supporting circuit is essential knowledge for anyone who wants to build or repair ham radio power equipment.

What you will learn: How filter capacitors smooth pulsating DC, the ripple voltage formula and its derivation, how to calculate the capacitor size for any load, how to choose electrolytic capacitors, the purpose and design of bleeder resistors, and why ripple matters in ham radio equipment.
Rectifier circuit with filter capacitor across load showing charge and discharge curves with ripple voltage labeled

Bridge rectifier with filter capacitor C across load R_L. During the diode conduction phase the capacitor charges rapidly to the peak voltage. During the off phase the capacitor discharges slowly through the load. The small voltage drop during discharge is the ripple voltage V_r.

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How the Filter Capacitor Works

Picture a water tower supplying a town. The supply pump (the rectifier) runs in bursts, pushing water into the tank each time it operates. Between bursts, households draw water from the tank. If the tank is large, the water level barely drops between pump cycles and the pressure at the taps stays nearly constant. If the tank is small, the level drops noticeably during each pause and pressure fluctuates. The filter capacitor is exactly this water tower, and the ripple voltage is the fluctuation in water level.

In electrical terms, the capacitor charges during the short interval each cycle when the rectifier diodes are conducting and the rectified voltage rises above the current capacitor voltage. As soon as the rectified waveform begins to fall below the capacitor voltage, the diodes stop conducting — the diodes can only pass current one way. The capacitor is now the sole power source for the load. It discharges slowly, supplying current, and its voltage gradually falls. The cycle repeats 120 times per second in a full-wave rectifier.

The Charge Phase

During the charge phase, the transformer secondary and diodes form a low-impedance path that charges the capacitor rapidly to the peak voltage of the rectified waveform. The charging current can be very large — sometimes 10 to 20 times the average load current — because the capacitor is being recharged in a short burst. This high peak current is why diodes in filter capacitor circuits must be rated for peak repetitive surge current, not just average DC current.

The Discharge Phase

During the discharge phase, the capacitor supplies current to the load resistor (and anything else connected across it). If the load draws a constant current I_L, the voltage falls approximately linearly with time: ΔV = I_L × Δt / C. The larger the capacitor C, the smaller the voltage drop ΔV for the same current and discharge time. The smaller the load current, the smaller the drop. A 10,000 µF capacitor supplying 1 A will discharge far less than a 1,000 µF capacitor supplying 5 A.

Close-up ripple waveform showing Vpeak, Vripple, sawtooth charge and discharge pattern with ripple formula labeled

Close-up of the filter capacitor output waveform. The capacitor charges rapidly to V_peak at each rectifier pulse and then discharges slowly through the load until the next pulse. The peak-to-peak ripple voltage V_r is determined by load current, ripple frequency, and capacitance.

View Larger

Ripple Voltage Formula

The ripple voltage formula follows directly from the basic relationship between capacitor charge and voltage. Recall that for a capacitor: Q = C × V, which means a charge Q flowing in or out of a capacitor of value C will change its voltage by V = Q / C. If the load draws a constant current I_L over a time interval Δt, the charge removed from the capacitor is Q = I_L × Δt. The resulting voltage drop (the ripple) is:

V_ripple = I_L × Δt / C

For a full-wave rectifier at 60 Hz, the time between peaks is Δt = 1/120 s ≈ 0.00833 s
Substituting Δt = 1/f (where f is the ripple frequency):

V_ripple ≈ I_L / (f × C)

Where:
I_L = load current in amperes
f = ripple frequency in Hz (120 Hz for full-wave at 60 Hz utility power, 60 Hz for half-wave)
C = capacitance in farads

This approximation is accurate when the ripple is small compared to the peak voltage (typically less than 10% of V_peak), which is the case in any well-designed supply. When ripple approaches 30% or more of V_peak, a more accurate calculation involves the exact charging waveform, but the approximation is good enough for all practical power supply design work.

Worked Example — Calculating Ripple

Given: I_L = 2 A, f = 120 Hz (full-wave), C = 10,000 µF = 0.01 F

V_ripple = I_L / (f × C)
V_ripple = 2 / (120 × 0.01)
V_ripple = 2 / 1.2
V_ripple = 1.67 V peak-to-peak

This is a reasonable ripple level. After a linear voltage regulator, this ripple would be reduced by 40–60 dB (a factor of 100 to 1000), so the regulated output would see only 1.7–17 mV of ripple.

Capacitor Sizing Formula

To design a filter capacitor for a specific ripple requirement, rearrange the ripple formula to solve for C:

C = I_L / (f × V_ripple)

Where V_ripple is the maximum allowable ripple voltage (peak-to-peak)

Worked Example — Sizing a Capacitor

Given: Need less than 0.5 V ripple at 3 A load, full-wave supply (f = 120 Hz).

C = I_L / (f × V_ripple)
C = 3 / (120 × 0.5)
C = 3 / 60
C = 0.05 F = 50,000 µF

This is a large capacitor. In practice, for a regulated supply the ripple requirement is much looser because the regulator removes most of it. The capacitor only needs to ensure the regulator's input voltage stays above its minimum (dropout) voltage throughout the ripple cycle. Practical regulated supplies typically use 1,000–4,000 µF per amp of load current.

The Ham Shack Rule of Thumb

For a linear-regulated supply (the type covered in the next lesson), the capacitor's primary job is not to produce a smooth output directly — it just needs to keep the unregulated voltage high enough for the regulator to do its job. The widely accepted rule of thumb is 1,000 to 4,000 µF per amp of load current. A 10 A regulated supply would use 10,000–40,000 µF. Commercial ham radio power supplies commonly use 20,000–50,000 µF for 20 A supplies.

For an unregulated supply (one with no voltage regulator, just a transformer, rectifier, and capacitor), you need to calculate the ripple directly and ensure it is low enough for the load. Many older tube-type amplifier power supplies and some battery chargers fall into this category.

Filter Capacitor Calculator

Filter Capacitor Size and Ripple Calculator

Calculate the required capacitor size for a given ripple voltage, or calculate the ripple produced by a given capacitor. Uses the formula C = I_L / (f × V_r).

Enter load current and either ripple voltage or capacitance, then click Calculate.

Filter Capacitor Output Voltage

A common point of confusion for beginners is why the output voltage of a rectifier-with-filter-cap is so much higher than the transformer's rated voltage. The reason is that the filter capacitor charges to the peak voltage of the rectified waveform, not the RMS voltage. The transformer's nameplate voltage is the RMS value, which for a sine wave is about 70.7% of the peak.

The actual DC output voltage with a large filter capacitor is approximately:

V_DC ≈ V_peak − V_ripple/2
V_DC ≈ (V_RMS × 1.414) − diode_drops − V_ripple/2

Example: 12 V RMS transformer, bridge rectifier, 10,000 µF cap, 1 A load at 120 Hz:
V_peak = 12 × 1.414 = 16.97 V
After bridge drops: 16.97 − 1.4 = 15.57 V
V_ripple = 1 / (120 × 0.01) = 0.83 V
V_DC ≈ 15.57 − 0.83/2 = 15.57 − 0.42 ≈ 15.2 V

Note: 15.2 V from a "12 V" transformer is normal and expected.

This peak-charging behavior has an important design implication: when you design a regulated supply, the unregulated voltage before the regulator is always significantly higher than the regulated output voltage. The regulator must be chosen and heat-sinked accordingly, because the excess voltage is dissipated as heat. This is covered in detail in the next lesson on linear voltage regulators.

Selecting Electrolytic Capacitors

Filter capacitors for power supply applications are almost universally aluminum electrolytic capacitors, because they offer the highest capacitance per unit volume and per dollar. However, they have important limitations that must be respected. Choosing the wrong capacitor can result in failure, fire, or in extreme cases, rupture.

Voltage Rating

The voltage rating of the capacitor must exceed the peak voltage that will appear across it, with a safety margin. The peak voltage is (V_RMS × 1.414) minus diode drops. For a 15 V RMS secondary with a bridge rectifier, the peak is about 19.8 V. Use a capacitor rated for at least 25 V — ideally 35 V or 50 V. Never operate an electrolytic capacitor close to its voltage rating; a 10–50% margin is standard practice.

Capacitance Value

Calculate using the formula above or use the ham shack rule of thumb. Electrolytic capacitors have a wide tolerance — typically ±20% — so the actual capacitance may vary significantly from the marked value. This is why calculated values are always a minimum, and rounding up to the next available standard value is correct practice.

ESR — Equivalent Series Resistance

Real capacitors are not ideal components. They have an internal resistance called Equivalent Series Resistance (ESR). In filter applications, ESR causes two problems. First, it adds to the ripple voltage — the ripple across the total impedance (capacitive reactance plus ESR) is larger than across an ideal capacitor. Second, the ripple current flowing through the ESR dissipates heat inside the capacitor. If the ripple current exceeds the capacitor's rating, the capacitor overheats and eventually fails from electrolyte boiling.

Standard aluminum electrolytic capacitors have ESR values from 0.05 to 1 Ω. Low-ESR capacitors (also called "low-impedance" or "switching supply" types) have ESR below 0.02 Ω and are preferred wherever significant ripple current flows. For a 15 A power supply with 2 V peak-to-peak ripple, the ripple current can be several amps — use low-ESR capacitors and check the datasheet ripple current rating.

Temperature Rating

Electrolytic capacitors are rated for 85°C or 105°C maximum operating temperature. The 105°C rating is strongly preferred for power supply applications, especially inside metal enclosures where temperatures can climb well above ambient. Capacitor life approximately doubles for every 10°C reduction in operating temperature, so a 105°C cap running at 70°C inside a supply will outlast an 85°C cap by a factor of four or more.

Physical Size

Large capacitance values at moderate voltages require physically large capacitors. A 10,000 µF 50 V electrolytic is typically 30 mm in diameter and 50 mm tall. A 47,000 µF 50 V capacitor is much larger. Leave adequate space in your enclosure and ensure the capacitor can be adequately cooled. Capacitors should not be mounted directly against hot components.

Polarity

Electrolytic capacitors are polarized — the positive terminal must connect to the higher potential and the negative terminal to the lower potential (ground). Reversing an electrolytic capacitor destroys it rapidly, often explosively, with hot electrolyte ejected from the vent strip at the top of the can. Always verify polarity before applying power. The negative lead is marked with a stripe bearing minus signs on the capacitor body, and the positive lead is typically longer before trimming.

Capacitors in Parallel

You can connect capacitors in parallel to obtain a total capacitance equal to the sum of individual values. This is common practice when a single large capacitor is not available, or when you want to reduce effective ESR. Parallel capacitors share ripple current in proportion to their capacitance, and their ESR values also combine in parallel. Three 3,300 µF capacitors in parallel give 9,900 µF with one-third the ESR of any individual unit — a practical advantage in high-current supplies.

⚖ Experiment: Observe Ripple with Different Capacitor Values

This experiment directly demonstrates how capacitor size controls ripple voltage. You will measure ripple with no capacitor, then with small and large capacitors, and observe the dramatic reduction in ripple as capacitance increases.

You will need:
  • Four 1N4007 diodes (bridge rectifier)
  • A 12 V AC wall adapter (AC output, not DC)
  • Three capacitors: 100 µF 25V, 1000 µF 25V, and 4700 µF 25V electrolytic (observe polarity)
  • A 100 Ω resistor, 2 W (as load)
  • A digital multimeter with AC and DC voltage ranges
  • A breadboard and jumper wires
  1. Build the bridge rectifier on the breadboard using four 1N4007 diodes as described in the previous lesson. Connect the 100 Ω resistor across the DC output as a load.
  2. Measure DC output voltage and AC ripple (set meter to AC volts) with no filter capacitor. Record both readings. You should see significant AC ripple — it will be a large fraction of the DC voltage.
  3. Add the 100 µF capacitor across the load (positive lead to positive output, negative to ground). Measure and record DC voltage and AC ripple again.
  4. Replace with the 1000 µF capacitor. Measure and record.
  5. Replace with the 4700 µF capacitor. Measure and record.
  6. Calculate predicted ripple for each value: V_r = I_L / (f × C), where I_L ≈ V_DC / 100 Ω, f = 120 Hz. Compare predicted vs measured values.
What you should see:

The AC ripple reading should decrease dramatically as capacitance increases — roughly in inverse proportion to the capacitance. The DC voltage will also rise as the capacitor charges to nearer the peak voltage. With the largest capacitor you should see very little AC ripple and a DC output significantly higher than the transformer's nameplate voltage. This confirms the ripple formula and demonstrates the peak-charging effect of filter capacitors.

Bleeder Resistors

A bleeder resistor is a high-value resistor connected permanently across the output of a power supply. Despite being continuously connected and wasting a small amount of power, it serves several important functions.

Purpose 1: Safety Discharge

When a power supply is switched off, large filter capacitors can hold a dangerous charge for minutes or even hours. A 47,000 µF capacitor charged to 35 V stores almost 29 joules of energy — enough to cause a serious shock. A bleeder resistor provides a controlled discharge path so the capacitor voltage drops to a safe level within seconds after power is removed.

Purpose 2: Regulator Stability

Some voltage regulators require a minimum load current to regulate properly. Without a minimum load, the output voltage can rise above specification. A bleeder resistor provides this minimum load even when the external circuit is disconnected.

Purpose 3: Voltage Overshoot Control

On startup, the sudden charging of filter capacitors can cause brief voltage overshoots that stress other components. A bleeder resistor helps dampen this transient.

Bleeder Resistor Calculation

I_bleeder = 1% to 10% of full-load current (1% is common for safety discharge; 10% for tight voltage regulation without a regulator)

R_bleeder = V_out / I_bleeder

P_bleeder = V_out² / R_bleeder (must size resistor wattage accordingly)

Example: V_out = 24 V, I_full_load = 2 A, I_bleeder = 1% = 20 mA
R_bleeder = 24 / 0.02 = 1200 Ω (use 1.2 kΩ standard value)
P_bleeder = 24² / 1200 = 576 / 1200 = 0.48 W — use a 1 W or 2 W resistor

Multiple-Section Filters (LC and RC)

A single capacitor provides significant filtering, but for extremely clean DC — as needed by sensitive receiver circuits or audio preamplifiers — additional filter sections can be added.

LC Filter (Choke-Input or Capacitor-Input with Choke)

Adding an inductor (choke) in series between the rectifier and the output capacitor creates an LC low-pass filter with much steeper rolloff than a capacitor alone. The inductor presents high reactance to the 120 Hz ripple (XL = 2πfL), blocking the AC component while passing the DC. The following capacitor then further smooths what little ripple gets through.

A 1–5 H choke with 10,000 µF of output capacitance provides filtering that is orders of magnitude better than the capacitor alone. This combination was standard in all vacuum tube power supplies and is still used in high-quality audio amplifier power supplies. For a ham shack HF rig operating near sensitive receive equipment, an LC filter can make the difference between clean audio and 120 Hz hum.

RC Filter

For low-current circuits (such as the bias supply for a preamplifier), a simple series resistor followed by a capacitor forms an RC filter. The resistor and capacitor form a voltage divider for AC but pass DC almost unchanged (the small DC voltage drop across the resistor is the only loss). The additional ripple rejection from an RC section is approximately:

Ripple attenuation ≈ Xc / R at the ripple frequency
For R = 100 Ω, C = 1000 µF, f = 120 Hz:
Xc = 1 / (2π × 120 × 0.001) = 1.33 Ω
Attenuation = Xc / R = 1.33 / 100 = 0.0133 = −37.5 dB

This is excellent ripple rejection with very simple components, suitable for low-current preamp bias supplies.

The disadvantage of an RC filter is the DC voltage drop across the resistor: at even 100 mA of load current through a 100 Ω resistor, the voltage drop is 10 V. RC sections are therefore limited to very light loads (below 10–20 mA).

Why Ripple Matters in Ham Radio

Ripple in the power supply causes real, audible, and measurable problems in ham radio equipment. Understanding these problems helps you diagnose supply issues in the field.

Audio Hum

If ripple appears on the DC supply of an audio amplifier stage — such as a microphone amplifier or the AF (audio frequency) output stage of a transceiver — the ripple is amplified along with the audio signal. The result is 60 Hz or 120 Hz hum in the received or transmitted audio. This is one of the most common power supply complaints in older and homebrew equipment. The cure is always more filtering (larger capacitor, LC section) or a voltage regulator.

Transmitter Hum Modulation

If ripple reaches the oscillator or modulator stages of a transmitter, it modulates the carrier. This produces sidebands at 60 Hz and 120 Hz offsets from the carrier — audible as a "buzz" on the transmitted signal, and reportable under FCC regulations as spurious emissions. Proper filtering and regulation eliminates this problem.

Microcontroller and DSP Instability

Modern transceivers contain digital signal processors and microcontrollers. These devices require a very stable supply voltage — often within ±5% or less. Large ripple on the supply line can cause digital logic errors, display glitches, frequency instability in synthesizers, and intermittent resets. Most modern rig power supply stages include low-dropout regulators and careful decoupling to prevent these issues.

The Correct Design Philosophy

The recommended approach in ham radio power supply design is: filter well before the regulator, then regulate. The filter capacitor's job is to maintain enough voltage above the regulator's minimum for reliable regulation throughout the ripple cycle. The regulator then removes the remaining ripple with 40–60 dB of attenuation. A well-designed regulated supply can have less than 1 mV of ripple at the output even with 2–3 V of ripple entering the regulator.

Frequently Asked Questions

Why does my voltmeter read a higher voltage than expected from the transformer?

Your multimeter is reading the DC output of the filter capacitor, which is close to the peak of the rectified waveform. Transformer ratings are given in RMS volts, and peak voltage is 1.414 times the RMS value. A "12 V" transformer produces a 16.97 V peak. After subtracting 1.4 V for the bridge diodes and accounting for light loading, you will measure approximately 14–16 V DC — which is perfectly normal and expected. The capacitor charges to the peak, not the average.

My 12 V transformer is producing 16 V DC after the bridge and capacitor — is it broken?

No, it is working exactly as designed. A 12 V RMS secondary produces a 16.97 V peak. Subtract 1.4 V for the bridge diodes to get approximately 15.6 V. Under light load with a large filter capacitor, the capacitor stays near the peak and you measure close to 15–16 V. At full rated load, transformer losses and the diode drops will bring it down somewhat. This is normal unregulated supply behavior. If you need exactly 12 V or 13.8 V, you need a voltage regulator after the filter capacitor.

Can I use capacitors in parallel to get a higher µF value?

Yes. Capacitors in parallel have their values added directly: three 3,300 µF capacitors in parallel give 9,900 µF. This is common practice in power supplies. The parallel combination also reduces ESR by the number of capacitors, which is a significant advantage in high-current supplies. Make sure all capacitors in parallel have the same or higher voltage rating than the supply voltage, and observe polarity on each one.

What happens if the capacitor voltage rating is too low?

The capacitor will be subjected to voltage above its rated maximum. Initially it may appear to work, but the oxide layer inside the capacitor degrades rapidly under over-voltage. The result is a gradual increase in leakage current and eventually catastrophic failure — often involving the capacitor venting hot electrolyte or even rupturing explosively. Always use a capacitor rated at least 20–50% above the peak voltage it will see. For a 20 V peak supply, use a 35 V or 50 V rated capacitor.

Test Your Knowledge

Answer the questions below to check your understanding. Every answer can be found in the lesson above.

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