I am designing a board which needs 5V and 3.3V rails, at about 1.5A. If I used two buck converters, I would use a lot of board area. So, is there a way to create a multi-output buck converter? I tried creating one by using two inductors (each of a different value), in a circuit simulator, but it doesn't work; the output ripple is through the roof and both rails sit at 5V or 3.3V (I kind of expected this.)
Electronic – Tapped buck converter
buckpowervoltage-regulator
Related Solutions
Core loss is caused by current induced into the magnetic core, which is effectively a secondary winding with a resistance across it.
Since the voltage going into this 'transformer' is a square wave, the current and voltage across the core resistance is also a square wave. If you remove the slope caused by current flowing into the buck converter's load then this low amplitude square wave is what's left.
Here is an LTspice simulation showing the effect. L2 and R2 represent the core 'secondary winding' and its resistance (shown as the equivalent 1:1 transformer - in reality it is a single turn with extremely low resistance):-
I(L1) is current going through the inductor. V(n002) is voltage across the core resistance.
Certainly the second option is the more cost effective. An LDO does not require as much external component as a buck-boost converter. This soultion saves some money by reducing the PCB size and the componenet costs.
On the other hand an LDO won't be as efficient as a buck-boost converter, but it all depends on the current demand of the bluetooth chip. I suppose we are talking about max 30 - 40 mA peak consumption here. That will be \$ \text{1.7 V} \times \text{40 mA} = \text{68 mW} \$ wasted energy. And it is the peak, most of the time the bluetooth chip will draw much less. So I would definietly go with the second option.
In general the switching converters efficiency is worse at light loads than at high (though you could find one with light-load-efficiency feature) so an LDO could be even better here.
Best Answer
You can sortof do it, but it's probably not worth the trouble. You would have to add a diode in front of the 5V output cap, and a pass element in front of the 3.3V output cap. The switcher regulates the 5V output, and a comparator turns on the pass element whenever the 3.3V output is low. In other words, the comparator via the pass element steals current that would otherwise go to the 5V supply, but the feedback system makes sure the 5V supply gets enough current to keep it at the right voltage. Like I said, probably not worth it.
To do this in one switcher somehow will require a inductor that can deliver 3A average. It won't be much worse to split that into two inductors that each only have to deliver 1.5V.
Another possibility is to have the switcher make 5V, then linearly regulate that to 3.3V. This wastes 2.6W in heat.
You could buck to make the 5V, then use a open loop fixed duty cycle buck converter with 3.3V/5.5V = 66% duty cycle to make the 3.3V from there.