All:

I have successfully prototyped a new product based on the ADP2303 (and based very closely on the recommended 4 layer layout). It's output is set at 5.0V, and uses a 4.7 uH inductor (as spec'd by the Table 11 in the datasheet). Thus far, while I've tested with varying output currents, I was most concerned with the maximum current of 3.0 A and didn't really do much testing at really light loads.

My questions concern the proper use/interpretation of the formula for inductance (pg 18 of the datasheet):

Since my goal is to have a limited number each of fixed-output supplies (5.0V, 3.3V, and 2.5V, for example), I started running thru some Excel spreadsheets based on the above formula to get a "feel" for what inductor values/ranges would be needed for each board. I found, somewhat to my surprise, that the inductor size increases dramatically as the output current drops. As an example, for Vin = 20.0, Vout = 5.0, the inductor values I calculated are 6.3 uH @ 3 A out, 12.6 uH @ 1.5 A, 37.8 uH @ 0.5A, and 189 uH 2 0.1 A.

I've rechecked the math and my spreadsheet and am convinced that this is exactly what the formula predicts. (The "left half" of the formula dominates the increase, due to Iload being in the denominator.) But does this mean that if I choose an inductor size, such as the 4.7 uH inductor for Vout = 5.0 and Vin approx 12 V, that if my product is run at a 200 mA load that the inductor is going to actually be much too small? (The formula indicates that the inductor should be 72 uH for these specs.) Will this cause premature failure of my product (probably either the converter or the inductor)? Conversely, is there any issue (other than the obvious parts cost "waste") if I choose a large inductor "to be safe"?

If this scenario (increasing inductance for decreasing load) is correct, how is it possible to design and implement a fixed output buck converter based on the ADP2303? Have I missed something really basic? What are the best practices for the design of a power supply that would be stable and reliable over a wide current range (from very small load to full load)?

Thanks very much for whatever assistance is offered.

Dave

When all you need is lower output current then a larger inductance with lower current rating for the inductor makes sense. This reduces the ripple current in the inductor and allows you to use smaller output capacitors for the same ripple voltage seen on the output resulting in a smaller more cost effective design.

If you size your supply for higher output currents using the design recommendations then it will still work fine when loaded lightly. However your efficiency will go down compared to a supply sized for the current that is actually drawn.

Klaus