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I have a simple LT3663 circuit that I'm using to regulate 15v down to 13.8v.

enter image description here

("24v" is actually 15v in, and "15v" is 13.8v out.) This works almost perfectly. It current limits properly, etc. However, for some reason, the LT3663 only outputs 10.4v, at any current draw below the limit. It regulates perfectly at this voltage. Messing around with R18 and R20 values didn't change anything at all.

Layout:

enter image description here

It's...a bit ugly. However, I think it's pretty close to the reference in terms of placement and grounding and routing.

Why is this?

EDIT:

Something very odd is happening, indeed. I just pulled R18 off the board completely (grounding FB), and the vreg still maintains 10.2v.

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  • \$\begingroup\$ How does your board layout look? Did you follow the recommended pcb layout? And are you using the right type of diode on SW? And the inductor is chosen for its specs or just any 15 µh? \$\endgroup\$
    – Passerby
    Mar 9, 2016 at 22:42
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    \$\begingroup\$ @Passerby Oh no not the kittens! I'm sorry, puppies. I'll try and clean it up. Board layout is reasonably close to stock, I'll post a pic. Inductor has a saturation of 1.1a (only running about 200ma), resistance of 200 milliohm. NRS5020T150MMGJ \$\endgroup\$
    – 0xDBFB7
    Mar 9, 2016 at 22:51
  • \$\begingroup\$ -1 for the messy schematic, and voting to close since I stopped there and don't know what is being asked. You wouldn't hand in homework this sloppy. We deserve at least the same respect as your teachers. \$\endgroup\$ Mar 10, 2016 at 13:15
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    \$\begingroup\$ @Passerby Cleaned up the labels. Sorry about how bad it looked, rush prototype :) \$\endgroup\$
    – 0xDBFB7
    Mar 10, 2016 at 16:42

2 Answers 2

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You're probably running into the LT3663's maximum duty-cycle limit as well as other internal losses.
The datasheet specifies the duty-cycle limit as: typical 92%, minimum 80%.
So if you hope for the typical value and assume 100% efficiency everywhere else, the best you could get is 13.8V
But since efficiency is never 100%, you have a diode drop, and you can't bet on getting the typical value, you shouldn't expect as much as 13.8V out.
Taking into account a lower max duty-cycle, some diode drop and other losses (transistor switch, inductor, current-sense, etc), your 10.4V is not surprising.

EDIT:
I'm surprised somone didn't spot this sooner, but you should really have a look at the top of page 2 of the datasheet - particularly where it lists the Absolute Maximum Ratings for the Vout pin (amongst others) as 6V !!!
That IC is apparently not intended to be used for anything much above 5V output - even the adjustable version.
A check on Linear's product selector grid, with VOut Max column enabled shows 5.5V as the maximum output voltage.
None of the application circuits show anything above 5V either.
I hope for your sake that LT has a pin-compatible substitute ....

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  • \$\begingroup\$ Whoa, min duty cycle is 80%? I'm getting 70. That's wierd. \$\endgroup\$
    – 0xDBFB7
    Mar 9, 2016 at 22:47
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    \$\begingroup\$ Linear's website and datasheets suck. I had to use the product selector for the category and show the hidden VOUT max column to find the max recommended VOUT. linear.com/parametric/… TI doesn't have the problem, they have nice grids right on the product page. \$\endgroup\$
    – Passerby
    Mar 11, 2016 at 15:03
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    \$\begingroup\$ Came here to say this. Yep, this part is not meant to output more than 5.5V really. It has over voltage protection on the INPUT, but not the output. It would seem that one can make this chip regulate the output to the point of it's own damage, and whatever the failure mechanism was, its burned something up and now it's permanently locked to 10.4V regardless of the feedback. Certainly an interesting failure mode. But yeah, this is one of those unfortunate 'datasheet gotchyas'. \$\endgroup\$
    – metacollin
    Mar 11, 2016 at 15:29
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    \$\begingroup\$ @DC177E it happens more than you'd expect. \$\endgroup\$
    – Passerby
    Mar 12, 2016 at 0:52
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    \$\begingroup\$ @Passerby You and metacollin solved it with the edit, thanks so much everyone! And no, there's no drop in replacement, naturally, so revision 9 of this dang board required... \$\endgroup\$
    – 0xDBFB7
    Mar 15, 2016 at 0:21
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You have a diode on the OUT pin that is causing your voltage to drop. Remove the diode. The datasheet shows no diode in their sample circuit.

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  • \$\begingroup\$ Unfortunately, I can't; there's voltage on the output most of the time and it'll backfeed into the regulator. \$\endgroup\$
    – 0xDBFB7
    Mar 9, 2016 at 22:32
  • \$\begingroup\$ The diode only has a drop of about 0.6v, however, so it seems like that wouldn't affect it this much? \$\endgroup\$
    – 0xDBFB7
    Mar 9, 2016 at 22:32
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    \$\begingroup\$ @DC177E It could have something to do w/ how close the output voltage is to the input voltage. Try an experiment, bring the input voltage to 20V and run all of your testing. If you are looking for 13.8V out, you have 15V - diode drop - switch drop, you might actually be at or near 100% duty cycle and, thus, have no real control of the output. \$\endgroup\$ Mar 9, 2016 at 22:40
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    \$\begingroup\$ Actually, after a bit of thought, I believe that - in charging this huge capacitance - your circuit is basically going into current-limit (the switch is turning off) before you can fully charge your output circuit. Definitely remove the large cap. \$\endgroup\$ Mar 9, 2016 at 23:12
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    \$\begingroup\$ I'm stumped. If you have a blank board, start checking the connections with an ohmmeter. Maybe try a new IC in case that one is damaged. \$\endgroup\$ Mar 10, 2016 at 1:12

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