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I have a DC jack with three terminals. Terminal 1 is tip (+), terminal 3 is sleeve (-), and terminal 2 is a switch, my understanding of which is such that if there is no adapter plugged into the jack, terminals 2 and 3 are connected.

When I plug a DC adapter into the jack, I get the voltage drop I would expect across terminals 1 and 3, about 9V. However, there is zero voltage between terminal 2 and both 3 and 1. How is it possible that there is no voltage drop between either? Shouldn't there have to be a 9V drop between terminal 2 and 3 if there's 0V between 1 and 2?

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    \$\begingroup\$ You usually don't call it a "voltage drop" unless it's across a resistance. This would be a "voltage difference" or simply "voltage". The rest is correct, and only means that you're either measuring it wrong or that you are reading your multimeter wrong. It's probably not connected to anything and you confuse a reading of "nothing" and "0". \$\endgroup\$ – pipe Apr 21 '16 at 17:21
  • \$\begingroup\$ If there's nothing connected to that terminal, then the resistance of your multimeter will pull that terminal to the voltage of the other terminal, and the multimeter will read 0V. \$\endgroup\$ – uint128_t Apr 21 '16 at 17:23
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    \$\begingroup\$ "...my understanding of which..." Why guess? Post a link to the jack's data sheet. Then we'll have the jack's schematic diagram, after which we won't need to guess. \$\endgroup\$ – Warren Young Apr 21 '16 at 17:24
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Your reasoning about voltages is correct. What's happening when you insert a connector is that the normally-closed contact opens, and is now floating.

That means it should measure open, not 0 volt.

I suspect that your multimeter does not show the distinction clearly enough.

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No voltage is measured with the switched Pin because it is no longer connected with anything. In this diagram used for illustration purposes Pin 3, the Switched Pin, is physically disconnected when the plug is inserted.

enter image description hereenter image description here enter image description here

Pin 2 is physically moved by the barrel. Its normally made of a spring metal that can bend and return to its normal position when the plug is removed. The 3rd picture is a Phone connector with, but the concept is the same in a DC barrel jack.

The voltage seen is 9v, with +9V on the center pin/Pin 1, and 0V/Ground reference on pin 2 (for a center positive adapter).

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