Ripple (electrical)

Full-wave center-tapped rectifier with capacitor filter

Reducing ripple is only one of several principal considerations in power supply filter design.[nb 1] The filtering of ripple voltage is analogous to filtering other kinds of signals. However, in AC/DC power conversion as well as DC power generation, high voltages and currents or both may be output as ripple. Therefore, large discrete components like high ripple-current rated electrolytic capacitors, large iron-core chokes and wire-wound power resistors are best suited to reduce ripple to manageable proportions before passing the current to an IC component like a voltage regulator, or on to the load. The kind of filtering required depends on the amplitude of the various harmonics of the ripple and the demands of the load. For example, a moving coil (MC) input circuit of a phono preamplifier may require that ripple be reduced to no more than a few hundred nanovolts (10⁻⁹V). In contrast, a battery charger, being a wholly resistive circuit, does not require any ripple filtering. Since the desired output is direct current (essentially 0 Hz), ripple filters are usually configured as low pass filters characterized by shunt capacitors and series chokes. Series resistors may replace chokes for reducing the output DC voltage, and shunt resistors may be used for voltage regulation.

If the RC time constant is large in comparison to the period of the AC waveform, then a reasonably accurate approximation can be made by assuming that the capacitor voltage falls linearly. A further useful assumption can be made if the ripple is small compared to the DC voltage. In this case the phase angle through which the rectifier conducts will be small and it can be assumed that the capacitor is discharging all the way from one peak to the next with little loss of accuracy.[1]

Ripple voltage from a full-wave rectifier, before and after the application of a smoothing capacitor

With the above assumptions the peak-to-peak ripple voltage can be calculated as:

The definition of capacitance ${\displaystyle C}$and current ${\displaystyle I}$are[2]

${\displaystyle {\begin{aligned}&Q=CV_{\text{pp}}\\&Q=I_{\text{ave}}t_{\text{ave}},\end{aligned}}}$

where ${\displaystyle Q}$is the amount of charge. The current and time ${\displaystyle t}$is taken from start of capacitor discharge until the minimum voltage on a full wave rectified signal as shown on the figure to the right. The time ${\displaystyle t_{\text{ave}}}$would then be equal to half the period of the full wave input.

${\displaystyle t_{\text{ave}}={\frac {t_{\text{fullwave}}}{2}}={\frac {1}{2f}}}$

Combining the three equations above to determine ${\displaystyle V_{\text{pp}}}$ gives,

${\displaystyle V_{\text{pp}}={\frac {Q}{C}}={\frac {I_{\text{ave}}t_{\text{ave}}}{C}}}$

Thus, for a full-wave rectifier:[3]

where

• ${\displaystyle V_{\mathrm {pp} }}$ is the peak-to-peak ripple voltage
• ${\displaystyle I}$ is the current in the circuit
• ${\displaystyle f}$ is the source (line) frequency of the AC power
• ${\displaystyle C}$ is the capacitance

For the RMS value of the ripple voltage, the calculation is more involved as the shape of the ripple waveform has a bearing on the result. Assuming a sawtooth waveform is a similar assumption to the ones above. The RMS value of a sawtooth wave is ${\displaystyle {\frac {V_{\mathrm {p} }}{\sqrt {3}}}}$ where ${\displaystyle V_{\mathrm {p} }}$ is peak voltage. With the further approximation that ${\displaystyle V_{\mathrm {p} }}$ is ${\displaystyle {\frac {V_{\mathrm {pp} }}{2}}}$, it yields the result:[4]

${\displaystyle V_{\mathrm {rms} }={\frac {V_{\mathrm {pp} }}{2{\sqrt {3}}}}={\frac {I}{4{\sqrt {3}}fC}}={\frac {V_{\mathrm {DC} }}{4{\sqrt {3}}fCR}}}$ where ${\displaystyle V_{\mathrm {DC} }=IR}$

where

• ${\displaystyle \gamma }$ is the ripple factor
• ${\displaystyle R}$ is the resistance of the load
• For the approximated formula, it is assumed that X_C ≪ R; this is a little larger than the actual value because a sawtooth wave comprises odd harmonics that aren't present in the rectified voltage.

Another approach to reducing ripple is to use a series choke. A choke has a filtering action and consequently produces a smoother waveform with fewer high-order harmonics. Against this, the DC output is close to the average input voltage as opposed to the voltage with the reservoir capacitor which is close to the peak input voltage. Starting with the Fourier term for the second harmonic, and ignoring higher-order harmonics,

${\displaystyle V_{\mathrm {2f} }(t)={\frac {4V_{\mathrm {AC_{p}} }}{3\pi }}\cos(2\omega t)}$

the ripple factor is given by:[5]

${\displaystyle {\begin{aligned}V_{\mathrm {rms} }={}&{\sqrt {{\frac {1}{T}}\int _{0}^{T}\left({\frac {4V_{\mathrm {AC_{p}} }}{3\pi }}\cos(2\omega t)\right)^{2}dt}}\cdot Z_{\mathrm {RL} }\\&{\text{where }}Z_{\mathrm {RL} }{\text{ is the impedance of the RL filter formed by the choke and load}}\\[8pt]={}&{\frac {4V_{\mathrm {AC_{p}} }}{3\pi }}{\sqrt {{\frac {1}{\pi }}\left[{\frac {t}{2}}+{\frac {\sin 2\omega t}{4\omega }}\right]_{0}^{\pi }}}\cdot {\frac {R}{\sqrt {R^{2}+X_{\mathrm {L} }^{2}}}}={\frac {4V_{\mathrm {AC_{p}} }}{3\pi }}{\sqrt {\frac {1}{2}}}\cdot {\frac {R}{\sqrt {R^{2}+X_{\mathrm {L} }^{2}}}}={\frac {4V_{\mathrm {AC_{p}} }}{3{\sqrt {2}}\pi }}\cdot {\frac {R}{X_{\mathrm {L} }}}.\end{aligned}}}$

For $R\ll X_{L},$

${\displaystyle {\begin{aligned}V_{\mathrm {DC} }={}&{\sqrt {\left({\frac {2V_{\mathrm {AC_{p}} }}{\pi }}\right)^{2}-V_{\mathrm {rms} }^{2}}}={\frac {2V_{\mathrm {AC_{p}} }}{\pi }}\quad {\text{because }}V_{\mathrm {rms} }^{2}=K{\frac {R^{2}}{X_{\mathrm {L} }^{2}}}{\text{ is negligible for }}R\ll X_{L}.\\[8pt]\gamma ={}&{\frac {V_{\mathrm {rms} }}{V_{\mathrm {DC} }}}=\left.{\frac {4V_{\mathrm {AC_{p}} }}{3{\sqrt {2}}\pi }}\cdot {\frac {R}{2\omega L}}\right/{\frac {2V_{\mathrm {AC_{p}} }}{\pi }}={\frac {R}{3{\sqrt {2}}\omega L}}.\\[8pt]&{\text{Substituting }}X_{\mathrm {L} }=2\omega L,{\text{where }}L{\text{ is the inductance of the choke}};{\text{ in the more familiar form,}}\\\approx {}&0.471{\frac {R}{X_{\mathrm {L} }}},{\text{ for }}R\ll X_{L}.\end{aligned}}}$

This is a little less than 0.483 because higher-order harmonics were omitted from consideration. (See Inductance.)

There is a minimum inductance (which is relative to the resistance of the load) required in order for a series choke to continuously conduct current. If the inductance falls below that value, current will be intermittent and output DC voltage will rise from the average input voltage to the peak input voltage; in effect, the inductor will behave like a capacitor. That minimum inductance, called the critical inductance is ${\displaystyle L={\frac {R}{2\pi (3f)}}}$ where R is the load resistance and f the line frequency. This gives values of L = R/1131 (often stated as R/1130) for 60 Hz mains rectification, and L = R/942 for 50 Hz mains rectification. Additionally, interrupting current to an inductor will cause its magnetic flux to collapse exponentially; as current falls, a voltage spike composed of very high harmonics results which can damage other components of the power supply or circuit. This phenomenon is called flyback voltage.

The complex impedance of a series choke is effectively part of the load impedance, so that lightly loaded circuits have increased ripple (just the opposite of a capacitor input filter). For that reason, a choke input filter is almost always part of an LC filter section, whose ripple reduction is independent of load current. The ripple factor is:

${\displaystyle \gamma ={\frac {V_{\mathrm {rms} }}{V_{\mathrm {DC} }}}={\frac {\sqrt {2}}{3}}\cdot {\frac {1}{4\omega ^{2}CL}}\approx 0.471{\frac {X_{\mathrm {C} }}{X_{\mathrm {L} }}}}$

where

• ${\displaystyle \omega =2\pi f}$

In high voltage/low current circuits, a resistor may replace the series choke in an LC filter section (creating an RC filter section). This has the effect of reducing the DC output as well as ripple. The ripple factor is

${\displaystyle \gamma ={\frac {V_{\mathrm {rms} }}{V_{\mathrm {DC} }}}={\frac {\left(1+{\frac {R}{R_{\mathrm {L} }}}\right)}{3{\sqrt {2}}\omega CR}}={\frac {1}{3{\sqrt {2}}\omega CR}}\approx 0.471{\frac {X_{\mathrm {C} }}{R}}}$ if R_L >> R, which makes an RC filter section practically independent of load

where

• ${\displaystyle \omega =2\pi f}$
• ${\displaystyle R}$ is the resistance of the filter resistor

Similarly because of the independence of LC filter sections with respect to load, a reservoir capacitor is also commonly followed by one resulting in a low-pass Π-filter.[6] A Π-filter results in a much lower ripple factor than a capacitor or choke input filter alone. It may be followed by additional LC or RC filter sections to further reduce ripple to a level tolerable by the load. However, use of chokes is deprecated in contemporary designs for economic reasons.

A more common solution where good ripple rejection is required is to use a reservoir capacitor to reduce the ripple to something manageable and then pass the current through a voltage regulator circuit. The regulator circuit, as well as providing a stable output voltage, will incidentally filter out nearly all of the ripple as long as the minimum level of the ripple waveform does not go below the voltage being regulated to.[7] Switched-mode power supplies usually include a voltage regulator as part of the circuit.

Voltage regulation is based on a different principle than filtering: it relies on the peak inverse voltage of a diode or series of diodes to set a maximum output voltage; it may also use one or more voltage amplification devices like transistors to boost voltage during sags. Because of the non-linear characteristics of these devices, the output of a regulator is free of ripple. A simple voltage regulator may be made with a series resistor to drop voltage followed by a shunt zener diode whose Peak Inverse Voltage (PIV) sets the maximum output voltage; if voltage rises, the diode shunts away current to maintain regulation.

Ripple is undesirable in many electronic applications for a variety of reasons:

• ripple represents wasted power that cannot be utilized by a circuit that requires direct current
• ripple will cause heating in DC circuit components due to current passing through parasitic elements like ESR of capacitors
• in power supplies, ripple voltage requires peak voltage of components to be higher; ripple current requires parasitic elements of components to be lower and dissipation capacity to be higher (components will be bigger, and quality will have to be higher)
• transformers that supply ripple current to capacitive input circuits will need to have VA ratings that exceed their load (watt) ratings
• The ripple frequency and its harmonics are within the audio band and will therefore be audible on equipment such as radio receivers, equipment for playing recordings and professional studio equipment.
• The ripple frequency is within television video bandwidth. Analogue TV receivers will exhibit a pattern of moving wavy lines if too much ripple is present.[8]
• The presence of ripple can reduce the resolution of electronic test and measurement instruments. On an oscilloscope it will manifest itself as a visible pattern on screen.
• Within digital circuits, it reduces the threshold, as does any form of supply rail noise, at which logic circuits give incorrect outputs and data is corrupted.