Richard, yes. I will do that. My goal was 5mV peak to peak out. I worked out I need ESR <= 5.4mOhm. That should be possible. But it does surprise me that the datasheet doesn't mention what kind of actual capacitance is needed? Surely this must be relevant. 1uF vs 1000uF is a big difference at 15V 7A. Right?
Have I missed something?
Thanks.M
The capacitance value comes out of your performance specs and the frequency and inductor choices.
Lets say you design for 100mA p-p ripple in the inductor and you want 5mV p-p from the capacitor. The ripple will be set by two capacitor parameters - the capacitance and ESR. Lets just assume half the ripple is from ESR and half from capacitance.
At 200KHz and 2.5mV capacitive ripple, you have a voltage slew rate (dv/dt) of 0.0025V * 200KHz = 500V/sec.
C = I/(dV/dT) = 0.1/500 = 200uF
For 2.5mV of ESR ripple, you need an ESR of less than
R
ESR = .0025V/0.1A = 0.25 milliohms
So at minimum, you want a single capacitor of combination that has a capacitance of 200uF or greater, and 25 milliohms or less.
Now the next issue is the voltage overshoot or undershoot.
I will just look at overshoot.
At 1A current through the inductor, then energy stored in the inductor is E = 1/2 L x I
2 = 1/2L joules.
If you suddenly drop the load down to zero, the only place left for this energy to go is the output capacitor.
For a capacitor, E = 1/2 C x V
2The biggest voltage overshoot will occur when the capacitor voltage starts at 0v
In this case the inductor energy all goes into the capacitor so
E = 1/2 L = 1/2 C x V
2So the capacitor output voltage V will be (L/C)
1/2If the inductor is 10uH and the capacitor is 200uF, then the voltage overshoot is (10
-5/2x10
-4)
0.5 = 0.7V
In other words with a 10uH inductor and a 200uF capacitor, if you had set the power supply to near 0V out at a current of 1A, and suddenly you drop to 0A, the voltage on the capacitor will jump up to 0.7V, and the it will only drop if there is a load to discharge it back to the regulated voltage near 0V.
I won't do the calculations here, but at 12V out, the same parts would have an overshoot of about 4mV if my quick calculation is right.
To reduce this overshoot, you can reduce the inductor, or increase the capacitor. If you increase the output capacitor, then a lot of energy is stored in that capacitor - if you short the output, you get many amps flowing momentarily instead of the 1A current limit.
Undershoot is caused by the fact that the bigger the inductor is, the lower the current ripple as you know, but the longer time it takes for the current to increase after a load current increase. This causes the output voltage to undershoot, which is again reduced with a big output capacitor.
There are other subtleties, but I hope I have shown how the choice of C is based on your design specs and your other regulator design choices.
The reason why they didn't spend much time on C, is that if you are designing a fixed voltage regulator, just whack in a big enough output capacitor.
If you are designing a variable input and output voltage regulator, optimizing performance can be complex. In comparison, a linear regulator has no problem (other then heat) at handling widely varying input and output voltages, and you can often use a much small output capacitor, which means the output can change much faster, and there is no large capacitor to discharge into you load under fault conditions.
Richard