How to troubleshoot a boost converter (ADP1613)?

Question

I have attempted to put together a schematic utilizing the ADP1613 boost converter, and I am having no luck at all. I don't have a good understanding of how this device really works, so I don't know how to troubleshoot it.

I lifted the schematic directly from the datasheet (page 16), replacing a couple of values, b/c I did not have the exact parts. The changes in values are very small and within the ranges specified in the datasheet:

Here's what the physical circuit looks like:

I am applying a 5V battery to Vin, and I am getting 40V at Vout, whereas what I configured the booster for is 5V output: 1.235V × (1 + 37.4/12.4) = 4.96V. This measurement was taken with no load attached to Vout.

I tried to examine the Switching Output pin with a 10X oscilloscope probe, but the chip heated up very quickly and I had to unplug it.

If I supply a load that draws 410mA, Vout drops to 8.5V.

What am I doing wrong here? What is the correct way to troubleshoot a device like this?

EDIT 1: Could the problem be that I am running the regulator w/o a load? If the chip was not capable of 0% duty cycle, wouldn't the voltage continue building to the maximum the inductor can handle, until something blows? Just guessing...

Answer 1

So, the input voltage is 5V and you're trying to get 5V output?

If that is the case, a boost isn't going to work out for you. A boost can only output a higher (or equal) voltage than the input. In fact it can't supply anything less than the input voltage (minus the diode drop), since current from the input will pass through L1 and D1 to Vout. This is true always. If the switch is switched at some duty cycle, then higher voltage will show up on the output. If Vin ranges both higher and lower than Vout, you will have to use a buck-boost or sepic converter.

You are correct that with no (or very light load) a very high voltage can show up on the output. Look at the circuit from pin 5 of the AD1613 to Vout. There is just a diode and a capacitor. A peak detector. So, without load, even a whisper thin PWM to the switch will put energy into the inductor that will transfer to Cout with nowhere to go until a leakage path is found. This is why it is common to put a zener diode, rated a little higher than Vout, across Cout.

Answer 2

From some of the posts and some further reading, I now understand that part of my problem was not applying any load to the boost regulator. Apparently this regulator is not capable of 0% duty cycle, so the voltage on the inductor just keeps building.

I have also realized that since my voltage can be slightly higher or lower than Vout, I need to use the Single-ended primary-inductor converter (SEPIC) configuration.

I have attempted to do this with another similar boost regulator and it worked much better, but still not functional unfortunately. I'll start a separate thread, b/c I am using a different part now (didn't have any more ADP1613s).

Troubleshooting a SEPIC regulator (MIC2296)

Answer 3

Summarizing my comments into an answer: A boost converter will not work for what you're after. A boost converter changes some of your current into higher voltage. You state that you already have 5V so you won't gain anything out of a power converter because the battery just doesn't have the wattage necessary for your application. You need more battery if you have it needing it to source a higher amount of power for hours at a time.

P=V*I
Buck and boost and any other kind of converters just alter the voltage at the expense of current or the converse. No electronic converter supplies more power.

The chemical reaction happening in the battery literally cannot source the current fast enough to supply your load properly. When draining a battery this fast, it's usually quite hard on batteries, which is why they generally have maximum ratings in the amount of power you should pull from them.