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From this datasheet: https://www.murata.com/products/productdata/8807040057374/uhe12-30w.pdf?1617679818000

"Note: The Sense and VOUT lines are internally connected through 10Ω resistors. Nevertheless, if the sense function is not used for remote regulation the user should connect the +Sense to +VOUT and –Sense to –VOUT at the DC/DC converter pins."

Yet in the section about using the trim pin it says

"A resistor connected from the Trim (pin 9) to the –Output (pin 7), or –Sense where applicable, will increase the output voltage for all models with the exception of the 1.2V models, which will decrease the output voltage in this configuration"

That "or" throws me off a little. Figure 8 shows a dotted line from -sense to -output. I've looked at some other datasheets and application notes for similar DC-DC converters and they all seem to say that the + and - outputs should be tied to their + and - sense pins.

enter image description here

One of the reasons I ask is that i've seen some applications of this converter where the sense pins are left floating. The converter seems to work but i'm curious what the drawbacks are (if any).

Edit: I'm gonna add this here from page 4 as an extra nail in the coffin.

enter image description here

Also, based on the ordering guide, there may be no sense pins at all! This explains the applications i've seen where they are not connected to the outputs. enter image description here

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    \$\begingroup\$ You can have a look at figure 8 in this application note that I wrote some years ago. You will see in the right side how the 5 pins in your schematic are wired. This is a pretty standard configuration for these dc-dc bricks. \$\endgroup\$ Commented May 4, 2023 at 20:21
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    \$\begingroup\$ Ten ohms is not really "floating" in any meaningful sense of the word. All they're saying is that you "should" short out the 10 ohms locally if you're not using remote sensing. But failing to do that, 10 ohms (x2) in the feedback path should have no measurable effect on the converter's output accuracy or stability. \$\endgroup\$
    – Dave Tweed
    Commented May 5, 2023 at 11:27
  • \$\begingroup\$ @DaveTweed I changed the question to "should" they be left floating because that's really what i'm trying to understand. Since they are internally connected, why do they even recommend shorting out the 10 ohms locally? \$\endgroup\$
    – BobaJFET
    Commented May 5, 2023 at 12:54

2 Answers 2

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Yes, while they certainly can be left floating if unused, they really should not be floating when unused, as it will be better to not float them.

The drawback is that when left unconnected, it can work in some application and cause problems in some applications.

The second part of your question about TRIM pin is basically the same thing, do what is best. If you don't use sense wires, you should have output and sense connected together at the regulator, and so the regulator can sense only the voltage at regulator, and then the TRIM just trims the voltage at the regulator output. But if you do use the sense pins, you are measuring and regulating the voltage at load, not at the regulator, and so the TRIM input will affect the sensed voltage at load, not at the regulator output.

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  • \$\begingroup\$ Are you saying that, if you need to use the trim, you should still have R2 connected to both the sense pin and the output pin? \$\endgroup\$
    – BobaJFET
    Commented May 4, 2023 at 20:38
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    \$\begingroup\$ I'd like to think the other way around. Whether or not you use remote sensing, the sense should be connected to output, at load if used, at regulator if not used. Then the TRIM resistor can be connected to SENSE because SENSE is connected. But if SENSE pins are floating, the 10 ohms internal voltage drop may be a problem so I would not connect it to a floating sense pin? but to the stable output pin. \$\endgroup\$
    – Justme
    Commented May 4, 2023 at 20:48
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The Sense and VOUT lines are internally connected through 10Ω resistors. [from datasheet]

The sense pins aren't floating, because they are connected to VOUTs through an internal 10Ω resistors. 10Ω is a stiff resistor. Consider also that the current flowing through the sense pins is small, so the 10Ω resistors introduce a negligible voltage drop.

One of the reasons I ask is that I've seen some applications of this converter where the sense pins are left floating. The converter seems to work but I'm curious what the drawbacks are (if any).

The intent behind the sense lines. There may be a voltage drop between the output of the DC-DC and the load dues to resistance in the connection. The sense lines create a Kelvin connection and sense the output voltage at the load, instead of sensing the output voltage at the converter output. The feedback look in the DC-DC adjusts (increases) the output voltage to compensate for the drop between the DC-DC and the load.

The voltage drop between the power supply and the load may be small enough by itself (current is small, or connection resistance is small). In that case the DC-DC will supply enough voltage to the load without the sense connections.

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  • \$\begingroup\$ I know that they're not really floating and connected internally. So why does the datasheet still recommend connecting them? And that still does not answer the question of, can you leave the PINS floating? \$\endgroup\$
    – BobaJFET
    Commented May 5, 2023 at 12:50
  • \$\begingroup\$ Regarding he last bit "can you leave the pins floating?" You can leave pins unconnected, if the voltage drop between the DC-DC and load is small enough for your application. So, the other applications [which you mentioned] which left the sense pins unconnected could be alright. [That's what I was trying to convey in the last paragraph of my answer.] Internally the sense pins are connected to outputs through 10Ω. So, unconnected ≠ floating in case of this DC-DC. You can't physically leave the sense pins floating , even if you wanted to. \$\endgroup\$
    – misk94555
    Commented May 5, 2023 at 13:57
  • \$\begingroup\$ I understand now. But i'm talking about the actual pins, not what they are connected to on the inside. In that respect, an unconnected pin is considered a floating pin. At least, that's how I was taught. I'll admit it's confusing. This was discussed on here before electronics.stackexchange.com/questions/345148/… \$\endgroup\$
    – BobaJFET
    Commented May 5, 2023 at 14:25
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    \$\begingroup\$ By now between the datasheet, Verbal Kint's comment [and that's a gem], Dave Tweed's comment, and my answer you have the complete analysis. You have all of the necessary information. Take it from here, @Boba . \$\endgroup\$
    – misk94555
    Commented May 5, 2023 at 17:32
  • \$\begingroup\$ Case closed. Much appreciated. \$\endgroup\$
    – BobaJFET
    Commented May 5, 2023 at 18:26

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