Ripple current in capacitors

[davmax]

There can be some misleading information on web sites. Take the www.lowESR.com web site it states the folowing:


The output capacitor (or capacitor bank) will have a ripple voltage (seen as noise on the supply voltage) proportional to the ESR of the capacitor (or capacitor bank).

V = I X R ... Ripple voltage = Current X ESR

If the current increases from 4A to 20A, within a circuit, the ripple voltage will increase by a factor of FIVE. This increased noise cannot be tolerated, so the ESR of the capacitors must be reduced.

This is very misleading in that it states that increasing ripple current through ESR will increase ripple voltage.
In any circuit it is the voltage source that determines the current through a resistor. It is not the reverse that the current determines the voltage. The only exception is the case of a current source developing and voltage across a resistor.

In a practical power supply the ripple voltage is developed in relation to the values of filter capacitor, the presence or not of a filter choke, the load and the frequency of ripple. The ripple appears as an AC voltage at the load and across the filter capacitor. The capacitor will bypass the AC ripple in proportion to the ESR, a high ESR will pass less current than a low ESR. The ESR is simple a resistance connected across and AC voltage source(neglecting the the capacitor reactance than is typically a very low value) . It cannot increase the ripple voltage, but a low ESR will decrease the ripple by virtue of the ripple voltage source resistance.

It was surprisng to find such a statement on a NIC site.

[linuxguru]

Re: Ripple current in capacitors

For the range of values of capacitances normally found on the secondary side of typical PSUs and VRMs, the reactive impedance is so small at the switching frequency (say 100 kHz) that the capacitors are effectively AC shorts. In these cases, the dominant component of the ripple voltage is the ESR * ripple current through the output cap. The ripple current is a function of a lot of things, including the secondary inductance, transformer turns ratio, primary voltage swings, PWM duty factor, etc.

A typical value may be 25 milliohm ESR and a (loaded) ripple current of 2A, giving rise to a 50 mV ripple voltage on (say) the +12v rail. In this case, if the output cap is replaced with another with the same capacitance, but a lower ESR of (say) 10 milliohms, the ripple voltage would drop to 20 mV.

So the original article is correct for the case of capacitors being driven by AC current sources (including most PWM forward, flyback and current-mode converters).

[davmax]

Re: Ripple current in capacitors

Thanks Linuxguru for your input. I actual picked up that I made a mistake in the last statement of my post , namely that an increase in ESR will not increase ripple. It does increase ripple to the load (CPU), as should be very evident, but does not increase the source ripple voltage.

The output of a VRM for example contains two components DC voltage and ripple an AC voltage that is passed by the caps. That is why I am dealing with AC.

The help make all this easier for everybody I have put together an equivalent circuit diagram and explanation. It helps me sort out my thoughts too.

To make it easier to understand ripple in a power supply a diagram is helpful to explain.



The diagram below has a ripple voltage source Er and it has a source impedance Zr represented by Rs (series resistance) and reactance of the output choke XL. A filter capacitor is represented by Xc and ESR (ESL is omitted) and the load is RL. RL can be represented by a CPU.



DIAGRAM is a seperate attachment due a 19K restriction on attached documents.







The voltage source Er is determined by the DC load current. With no load the filter capacitor charges to the ripple peak value and there is no ripple. Therefore there is maximum AC ripple at maximum DC load. The source ripple Er will also increase if the filter capacitor loses capacitance. (More ripple to a CPU)



Only the AC ripple will now be considered. At any load the ripple voltage appearing across the load and filter capacitor Er LOAD will be determined by Er and the ratio of Zr and the resistance represented by the parallel resistances of ESR and RL (the value of Xc is taken as too small to affect ripple through ESR). The ripple currents IrLOAD and Irc will be proportional the resistive values of ESR and RL.



A few of scenarios

• If the load resistance decreases the value of Er will increase and the proportion of IrLOAD will also increase. (More ripple to a CPU)
• If the value of ESR increases Er will not increase but ErLOAD will increase because of decreased voltage drop across Zr. This will cause a larger ripple flow through RL. (More ripple to a CPU).
• If the value of ESR decreases then ErLOAD and I rLOAD will decrease. (Less ripple to a CPU)
• If the filter capacitor is comprised of a number of components the ripple current through each will be proportional to their respective ESR value.

[linuxguru]

Re: Ripple current in capacitors

Thanks - that was consistent and clear. One could simplify a bit by omitting Rs and treating Zr as a pure inductance XL, but your more general explanation is both intuitive and accurate.

[davmax]

Re: Ripple current in capacitors

Thank you very much Linuxguru your comments are appreciated.

[999999999]

Re: Ripple current in capacitors

The situation cannot be oversimplified until it is considered only a voltage source and resistance because it is a circuit with feedback, that feedback being there deliberately to maintain a constant voltage by _changing_ the voltage as needed from the measured value to the target value.

You do not have lower heat to any significant extent in a higher ESR cap. Practically speaking it is always the opposite unless the part was just impossibly unworkable in the circuit.

[999999999]

Re: Ripple current in capacitors

I again urge you to not try to rewrite what cap manufacturers state to try to align it to your prior ideas - ideas that need experimental backing, the good ole scientific method appplied, before being put out for public consumption. You've made a prediction but before going overboard with trying to claim cap manufacturers have it wrong, it is time to test that hypothesis.

Also, the current does determine the voltage, because it's following an inductor in typical cases. Think about what this higher ESR part is doing in filtering, the resultant waveform. Larger ripple.

There is a simple test that will clear this right up. Take a board with high ESR cap (I mean a new cap placed at that point in the circuit), and large load (to maximize potential differeces enough to measure), and measure the cap temp. Now replace that with a significantly lower ESR part of same size and uF value. Maybe the temp difference won't be enough to discriminate, but if it is large enough, the lower ESR part will be cooler.