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I have a sensing power supply powering a custom build LED light for video. There is an on/off switch after the sensing lines and everything works well. The setup looks like this:

enter image description here

However I'd like to sense the voltage drop after the switch. As what I have read in the PSU manuals, this configuration will cause current to flow throught the sensing lines when the switch is open and damage the PSU's resistors.

My question is, can the configuration work with a diode placed on the +SENS line to stop current when the switch is open? Like this:

enter image description here

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  • \$\begingroup\$ Could you, please, include the exact sentence from the PSU manual? The sense lines should present ahigh impedance as seen from the outside world (typically 1MOhm). It makes no sense that there is a current flowing outwards these lines, much less that the said current is capable to damage a resistor. \$\endgroup\$ – jose.angel.jimenez Dec 8 '15 at 18:47
  • \$\begingroup\$ Don't have this in front of me, but this is what it meant "We have seen applications where the user has installed a switch or fuse in series with one or both output wires. This can cause a serious problem if the remote sense lines remain connected to the load, because if the output cable switch or fuse opens, current will flow in the sense lines and cause the internal Rsense resistors to burn up." From ( edn.com/electronics-blogs/power-supply-notes/4418253/…) \$\endgroup\$ – NikoDiYana Dec 8 '15 at 18:58
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In short, a PSU sense line cannot include a diode in series; nor any other active/passive component, for what it matters. Adding a diode as in the schematic you have depicted will prevent the main current blowing up the internal Rsense of the PSU, however, this will also make also impossible for the PSU to correctly measure the remote voltage at the load.

Sense cables can have very small cross-section area (diameter/gauge), as the sensing current will be very small and the voltage drop negligible. This is always assuming that the PSU is being operated correctly. Most remote sensing PSUs does not allow the main cables to be disconnected separately from the sensing cables. This is precisely the type of PSUs the article from EDN refers to,

http://www.edn.com/electronics-blogs/power-supply-notes/4418253/Power-supply--Remote-Sense--mistakes---remedies

In any case, delving a little bit more into your root needs, I cannot understand why you want "to sense the voltage drop after the switch". Why would you like to break the power circuit but not the sensing lines? That defeats the purpose of the sensing circuit and it's explicitly forbidden in many designs of sensing capable PSUs.

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  • \$\begingroup\$ Thanks for this. My assumption is that in closed switch state current flows from Load+ back to the PSU where it drives the PSU to do the voltage compensation. So a diode put in the direction depicted wouldn't block the sensing line. However it will block the current flowing towards the Load+ when switch is open. I might be wrong of course. \$\endgroup\$ – NikoDiYana Dec 8 '15 at 21:11
  • \$\begingroup\$ Regarding "sensing the voltage after the switch" I need it because the switch drops 0.2 volts at 30amps and I want to include this drop in the compensation by the PSU. I can't find a less resistive switch for now. \$\endgroup\$ – NikoDiYana Dec 8 '15 at 21:14
  • \$\begingroup\$ Edit to my first comment: "... in closed switch state current flows from Load+ back to the SENS+ of the PSU where it drives the PSU to do the voltage compensation" \$\endgroup\$ – NikoDiYana Dec 8 '15 at 21:25
  • \$\begingroup\$ @NikoDiYana The voltage drop from the diode is not minimal, it follows the Shockley diode equation (check sources) and can be about 0.5-0.6V even for nano-micro ampere currents (in the case of a standard Silicon diode). That means your PSU will increase by that amount the output voltage at the load. The temperature coefficient is almost constant and fairly small, about 2.1mV/C, however you will have to analyze if it is small enough for your application. \$\endgroup\$ – jose.angel.jimenez Dec 9 '15 at 7:52
  • \$\begingroup\$ @NikoDiYana If your switch drops 0.2V at 30A, another solution is to simply adjust the output of your PSU to +0.2V of your desired load voltage. Many sensing PSUs also allow a small adjustment in the output voltage. \$\endgroup\$ – jose.angel.jimenez Dec 9 '15 at 7:54
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Why don't you put an on/off switch on the AC side of the power supply if you are worried about the sense lines being damaged.

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Unless the power supply is specified to function correctly without the sense lines connected, this is a bad idea. The PS may, with the sense line disconnected, try to raise the output voltage Beyond its limit and shutdown or, worse, damage itself.

Even if your idea would work, you will cause the PS to raise its output voltage by the voltage drop through the diode. This is because the diode will drop the voltage on the sense return which the supply will see as voltage drop to the target.

As @Andyaka suggested, the proper place to put a switch is on the AC input.

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  • \$\begingroup\$ Thanks. You're right on both points. Still, there is no state of the switch in which the sense line is disconnected while the supply line is connected. \$\endgroup\$ – NikoDiYana Dec 8 '15 at 18:46
  • \$\begingroup\$ Regarding the voltage drop from the diode, I assume it will be minimal since the sensing line current is low, and also constant, so it can be compensated easily by the PU. There is an AC switch, but device design for now requires DC switch too. \$\endgroup\$ – NikoDiYana Dec 8 '15 at 18:50

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