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I'm trying to use the LT3048-5 to get a steady 5V output from a battery that outputs 4.2->3.0V

I connected together the following setup from the datasheet:

enter image description here

But what I'm getting when I input 4.2V is ~3.7V (Normally 0.5V less than the input for typical input voltages). The LDOOUT value and the BSTOUT have the same value with this input voltage, however, if I turn up the voltage to ~6V I actually get a regulated 5V output (~6V BSTOUT).

I'm confused, shouldn't I be seeing ~6.1V at the BSTOUT pin boosted from a proper input voltage and then a steady 5V at the LDOOUT pin?

And there's a part of the datasheet which seems strange to me (http://cds.linear.com/docs/en/datasheet/3048fa.pdf): under Electrical Characteristics, BSTOUT-VIN Regulation Voltage: there is a condition that LDOUT < VIN That seems contradictory to the idea of a boosting regulator to me

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  • \$\begingroup\$ Are you sure about the inductor and the boost capacitor? \$\endgroup\$
    – Ignacio Vazquez-Abrams
    May 11, 2016 at 23:31
  • \$\begingroup\$ You haven't said anything about the inductor you're using - is it a recommended part from the datasheet? Linking to its data may be useful \$\endgroup\$
    – user16324
    May 11, 2016 at 23:41
  • \$\begingroup\$ Inductor chosen: digikey.ca/product-search/en?keywords=490-5116-1-nd The recommended part wasn't convenient to get so I tried to get something with similar characteristics, but I'm admittedly not sure what's important when selecting an inductor beyond the inductance and the current rating. \$\endgroup\$
    – Colin
    May 11, 2016 at 23:49
  • \$\begingroup\$ I read the spec sheet. This part requires a 100uA minimum load and a max load of 40mA. This IC is a voltage reference, not a power supply per se. I also saw the LDOUT < Vin limit, but the reasons were not clear. \$\endgroup\$
    – user105652
    May 12, 2016 at 2:34
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    \$\begingroup\$ The spec for BSTOUT-Vin regulation voltage applies when LDOOUT < Vin because "The error amplifier, A1, regulates BSTOUT (and LDOIN) to 1.1V above LDOOUT or VIN, whichever is higher" (p. 10; because a boost can only produce output voltage greater than its input voltage) \$\endgroup\$
    – The Photon
    May 12, 2016 at 4:57

1 Answer 1

Reset to default
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This is going to hurt, but I am guessing you forgot to pull the EN pin high. (or you did pull it high but there is a broken connection somewhere)

Here's the chip block diagram from the datasheet: enter image description here

With the EN pin low, the boost converter is disabled but the LDO portion not affected (at least, if the block diagram is honest). Then DC conduction through the inductor and the switching diode (between the SW and BSTOUT pins) will produce a voltage on BSTOUT about 0.5 to 0.7 V below the input voltage (like you saw). This would normally put the LDO into dropout, but when you get this voltage sufficiently above 5 V, the LDO will work normally. Again, this is exactly the behavior you observed.

If you just left the EN pin floating, there is nothing in the datasheet that says which way the pin will default. So you will need to actively pull it high to enable the chip.

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  • \$\begingroup\$ Hmm, that's a really good explanation of the behavior if the EN pin was low. However, I have the EN pin and Vin shorted, and so a Vin of anywhere above 1.25V will pull EN high. I checked that the two pins are shorted; maybe internally there was an issue with the EN connection \$\endgroup\$
    – Colin
    May 12, 2016 at 21:50
  • \$\begingroup\$ Dang. Well, I'll leave the answer here for future readers. \$\endgroup\$
    – The Photon
    May 12, 2016 at 22:22

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