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I'm having a brain fart. Reading 3.1V ish on the output. My +5V0 net is after a schottky diode with 5V + a little bit coming from the NCP1117, and another 5V0 coming from a USB connector. Reason being I want the regulated voltage after the diode to be 5V, accounting for the diode voltage drop.

My feedback resistors look fine, I checked the parts on the board and they look fine, and my input looks solid too.

After the +5V0 net is a 3.3V regulator for 3.3V logic stuff. That's about it. The +5V0 only powers a handful of devices which seem to be functioning correctly with the 5V0 USB going through the diode and only powering to about 4.6V.

LINK TO DATASHEET

Schematic Layout

Any clues?

Thanks in advance

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    \$\begingroup\$ What is the tolerance of your resistors ? (please link datahseet) \$\endgroup\$
    – rom1nux
    Dec 2, 2019 at 16:43
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    \$\begingroup\$ Add the missing diode to the schematic please. \$\endgroup\$
    – Aaron
    Dec 2, 2019 at 16:45
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    \$\begingroup\$ A few more measurements would be useful - what are the voltages at the three pins of the NCP1117? \$\endgroup\$
    – Kevin White
    Dec 2, 2019 at 16:53
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    \$\begingroup\$ @Punchki are both both the USB and the 12V applied at the same time? Or are they being used exclusive of each other? \$\endgroup\$
    – Aaron
    Dec 2, 2019 at 17:02
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    \$\begingroup\$ The vout equation is only true if the regulator is allowed to regulate Vref across R1, in your design you loose one diode drop \$\endgroup\$
    – sstobbe
    Dec 2, 2019 at 17:40

2 Answers 2

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You (and a lot of the answers so far) seem to think that the path from OUT to SET is a feedback loop, and that putting the diode in that path will somehow allow the regulator to compensate for the \$V_f\$ of the diode.

Nothing could be farther from the truth.

The fact is, there are a lot of things you give up when you simplify a regulator to just 3 pins, and one of those things is an external feedback loop. The only thing that such a regulator can do is maintain a fixed voltage between OUT and SET, and the feedback loop that accomplishes that is entirely internal to the chip.

The equation that you show as part of your schematic is based on the assumption that there is a fixed resistor between OUT and SET, and therefore this resistor has a fixed amount of current flowing through it, which allows you to use a second resistor to establish a voltage for SET relative to ground. Putting a diode in there completely violates that assumption, because the voltage drop across the diode will be based partly on the load current, and you can no longer assume that the current through the two resistors is still constant.

If you want a circuit that regulates the voltage at the cathode of the diode, then you need to pick a different kind of regulator (either linear or switching) that actually has an external feedback pin.

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  • \$\begingroup\$ Hi Dave, thanks for the response. I did as you mentioned and it did fix the problem. I guess the bigger problem is where I didn't notice / account for this. Do you have any recommendation where I can read up on this to not make this mistake in the future? Is the current through the diode really that jumpy that it would ruin this design? I did look at the Vout and it did have quite a bit of ripple to it. Thanks again :) \$\endgroup\$
    – Punchki
    Dec 2, 2019 at 18:41
  • \$\begingroup\$ I'm not really sure where your thinking went astray, so I'm not sure what to recommend. But in general, if you're going to attempt an "off-label" application of an IC, you need to make sure you understand how it works internally to whatever level of detail the manufacturer is willing to provide. \$\endgroup\$
    – Dave Tweed
    Dec 2, 2019 at 19:18
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According to the datasheet, this regulator needs a load of at least 5mA to regulate properly. Are you measuring the output voltage with such a load.

Also, the adjustment pin current is not zero, it is up to 120μA which will give you a significant error.

Bottom line, use resistor values for setting the voltage that will also draw much more than the adjustment pin current and will provide enough of a load for proper regulation.

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  • \$\begingroup\$ On it! I see this that the feedback resistors are a fair bit above where they should be.. Thanks, and I'll try with some lower values and see how it goes. \$\endgroup\$
    – Punchki
    Dec 2, 2019 at 17:13

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