Actually, you can get all kinds of crazy-sounding numbers for capacitors' voltage slew rates. But if the equations you use take into account ESR and maybe ESL, the crazy-sounding numbers will be pretty close to correct. However, it's often only for a VERY short time.
Even just the reservoir caps in a PSU behave in ways that were, to me, somewhat unexpected. You can see for yourself by either running LT-Spice simulations or running the PSU spreadsheet that I uploaded as an attachment to the post at
http://www.diyaudio.com/forums/power-supplies/216409-power-supply-resevoir-size-167.html#post3287619 .
Attached is a small sample of the numbers that the spreadsheet plots after performing the numerical solution of the differential equations for a transformer-rectifier-capacitor-load power supply circuit (including transformer winding leakage inductance and resistance). The attachment shows the numbers during the initial part of the first startup pulse of a particular PSU.
It would be very easy to get almost any VOLTAGE slew rate you wanted, out of almost any cap (although they might explode if you over-do it). With small-value caps, it's quite easy.
Since i = C dv/dt, and dv/dt is the voltage slew rate, just solve for that:
dv/dt = i / C
Obviously, for SMALLER values of C, the same current, i, would cause a larger voltage slew rate.
If you put 10 amps into a 0.1 uF cap, it would give 10 A / 0.0000001 F = 100 million volts per second, which would be 100 Volts per microsecond.
But if you're talkiing about a larger capacitance, the ONLY way to get a high voltage slew rate to occur would be to put in or take out a VERY LARGE current.
BUT, I don't know why anyone would care about a cap's VOLTAGE slew rate. We use caps to provide CURRENT TRANSIENTS, by lowering the voltage across the cap by a very small amount (for example, when an output transistor is commanded to lower its channel resistance in order to let more current flow, which will lower the rail voltage slightly, making the decoupling and/or reservoir caps provide exactly the right surge of current). INITIALLY, for a short time, the cap will obey the impedance equation:
∆v / ∆i = ESR
or
∆i = ∆v / ESR
(but other circuit series resistance would need to be added to ESR, if the cap was in a circuit)
That equation means that for a lower ESR, the cap will have an initial transient current with a greater amplitude, but will reach that higher amplitude in the same amount of time. Or, you could say that it would reach the same current as before in less time. So it would be faster, in terms of current tranients.
Just playing around:
Actually, since initially, for a linear (ramp) change in voltage, versus time, the impedance relationship holds:
∆v / ∆i = ESR
But we also know that
∆i = C ∆v / ∆t
which can be rearranged to
∆v / ∆i = ∆t / C (impedance in seconds per Farad?)
we could also say
∆v / ∆i = ∆t / C = ESR (i.e. larger-value caps have lower ESR)
or
∆t = ESR ∙ C (an RC time constant?)
Also, an approximate ESR equation is
ESR = 0.02 / (C ∙ VR)
where VR is an electrolytic cap's voltage rating.
That gives
∆t = ESR ∙ C
and
∆t = 0.02 / VR
which seems to imply that electrolytic caps with higher voltage ratings should have lower ESR, for the same capacitance value.
Anyway, for a linear current ramp versus time of ∆i Amps in ∆t seconds, into a capacitor with ESR = Resr, the change in the cap's voltage would be:
∆v = ∆i ( ( ∆t / ( 2 C ) ) + Resr )
If it was a constant current instead of a ramp, for ∆t seconds, that would change to:
∆v = i ( ( ∆t / C ) + Resr )
or
∆v = i ∙ ( ∆t / C ) + Resr ∙ i
which also implies that caps with higher ESR would have a larger ripple voltage for a given PSU load current.
For the same load current, the ripple voltage amplitude with a cap with a higher ESR would be larger by the average load current times the difference in ESR valus.
