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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: enter image description here

Here's what the physical circuit looks like: enter image description here enter image description here

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...

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  • \$\begingroup\$ What diode is that? \$\endgroup\$ – Vladimir Cravero Apr 24 '14 at 11:54
  • \$\begingroup\$ @VladimirCravero: SCHOTTKY 40V 2A \$\endgroup\$ – Val Blant Apr 24 '14 at 12:03
  • \$\begingroup\$ Why are you trying to convert 5V to 5V? You'll have better results if you tie pin 3 low. :-) Seriously, most boost regulators have minimum load requirements, but 410 mA seems a bit much. It's behaving as though the feedback path isn't working. Are you sure all of the breadboard connections are good? \$\endgroup\$ – Dave Tweed Apr 24 '14 at 12:08
  • \$\begingroup\$ @DaveTweed: B/c 5V will not be 5V when I apply a large load. When the circuit is loaded, the battery voltage drops to around 3V, which shuts down a bunch of stuff on my board. I need to keep the voltage at 5V, at the expense of current. \$\endgroup\$ – Val Blant Apr 24 '14 at 12:11
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    \$\begingroup\$ The boost converter approach is entirely valid for burst current draw. However, I strongly recommend that for test purposes you put a dummy load in all the time. What have you done with pins 3 and 7? You MUST NOT leave inputs unconnected unless the datasheet says you can. Also, check the voltage at pin "FB". \$\endgroup\$ – pjc50 Apr 24 '14 at 15:19
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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.

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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)

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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.

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  • \$\begingroup\$ I understand that the boost converter cannot provide more power. Trading current for voltage is what I am after, and that is what boost converter is supposed to do. \$\endgroup\$ – Val Blant Apr 25 '14 at 1:31
  • \$\begingroup\$ I don't think I explained something well. The reason your voltage drops at all is because a battery is a non-ideal voltage source. If it was ideal, it would be able to source infinite current at 5V. Since it's non-ideal, the more current you pull from it, the more the battery voltage drop is. You're out of current, so your voltage is drooping. Where do you expect to get more current from to make more voltage? \$\endgroup\$ – horta Apr 25 '14 at 10:14
  • \$\begingroup\$ Well, I have a very rudimentary understanding of how boost regulators work, but I don;t think I need more current to make more voltage. The extra voltage is created when the inductor starts collapsing the magnetic field through the load in series with the batteries. When the inductor is building the magnetic field, the load is powered by the capacitor, which has been charged at a higher voltage by the inductor and batteries together in the previous step. Obviously that means even less current for the load, which is totally fine. \$\endgroup\$ – Val Blant Apr 26 '14 at 3:34

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