Output capacitor for a boost converter in DCM

Question

I'm designing a boost DC/DC converter for a backlight (it needs to be constant current but it's not relevant to the question).

Starting from a 5V preregulated rail I need to go to about 25V 30mA; nothing too exciting.

Usually I work with the TI handbooks, in this case SLVA372; even if it's not explicit it's pretty clear they target for a CCM converter. However the converter I need to use (the LT1618) recommends (as in, we give starting compensation values for this only and no other support other than "measure it and fix it yourself", as usual for Linear) a maximum of 10µH inductor. Control is mostly a constant current mode with some tricks in the EA for setting either the voltage or maximum current.

Even for a 40% current ripple the inductor is more than 50µH so I guess it's safer to stuck to a 10µH and let it go: in fact it doesn't go exactly in DCM (average 190mA and 300mApp ripple).

That said, having never explicitly designed a boost which such a working point, other than the usual efficiency issue (I don't care I have a huge stable supply), I think that most of the remaining design should be the same:

• Duty cycle is the same since it only depends on input and output voltages;
• The current peaks and limits are calculated in the same way (with the new current ripple of course);
• Input capacitor is the same (well, it's mostly dictated by the switch itself);
• I think the output capacitor should be bigger.

In CCM we have phase 1 when the inductor charges and the capacitor supplies the load and phase 2 when the inductor supplies the load and recharges the capacitor; however in DCM the inductor has no more remaining energy stored and the output capacitor still needs to feed the load.

So (still I guess!) the output capacitor shouldn't depend only on Iout×D but on Iout×(D+Dd) (where Dd is the part of the switching cycle when the inductor is discharged but and the switching cycle isn't started yet)

Am I missing something or am I in the right direction?

Answer 1

I'd rather use a chip with a MOSFET switch instead of a BJT (also much cheaper).

5V to 25V 30mA, so switch duty cycle is 80% on, 20% off. With 1.2MHz frequency, that's 6.6µs on, 160ns off.

Input current should be 20mA*25/5=100mA plus losses.

With 10µH inductor, we get a current ripple of di = vin/L Ton = 333mA, so it will run in discontinuous mode.

It is neither difficult nor expensive to find a 47µH inductor for this current. In this case, ripple current would be about 80mA, which is much better.

Now, to answer the question itself, running in discontinuous mode will result in higher inductor peak current. This means more I2R losses. The diode conduction time will be lower, as it will turn off before the switch turns back on. When it the diode is off, only the cap powers the load, but that won't change much in your design, because the cap is would already be powering the load 80% of the time in continuous mode when the switch is on and the diode is also off. It's not a big change to go to 90% of the time, so the capacitor value shouldn't change.

The input cap sees a ripple current equal to the inductor ripple current. So it is higher in discontinuous mode.

The output cap is either charged by the inductor current, or discharged by the load. So it sees a ripple current equal to the Inductor peak current. This also means a higher ripple current in discontinuous mode. This can be a problem for electrolytic caps.