I used a power supply the other day to power an op-amp. I was told to set it up in the following configuration, however I do not completely understand why this works. The shorted cable acts as ground??
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3\$\begingroup\$ Who or what told you that? \$\endgroup\$– Long PhamAug 29, 2018 at 5:14
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3\$\begingroup\$ Does it have isolated outputs and you've been instructed to connect them in series to produce a center tap? It's clear you've connected the + from 1 to the - on the other, but what are you depicting with the black lines at the bottom? \$\endgroup\$– K HAug 29, 2018 at 5:27
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2\$\begingroup\$ Just to make it clear - the left + isn't shorted to the left -, and the right + isn't shorted to the right -, but the right + is shorted to the left -. I assume the black lines indicate connectors, not shorts. \$\endgroup\$– user253751Aug 29, 2018 at 5:27
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1\$\begingroup\$ Related: electronics.stackexchange.com/questions/392660/…. \$\endgroup\$– TransistorAug 29, 2018 at 6:06
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\$\begingroup\$ Watch out for limitations on how high you can float the individual power supplies. Check the manual. \$\endgroup\$– Jon CusterAug 29, 2018 at 22:31
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4 Answers
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You can pick anything you want and call it ground.
This power supply box has two power supplies in it. Shorting the + of one to the - of the other puts the two power supplies in series. That gives you a so-called "split" power supply.
You can call the middle connection ground and then you have +15V, 0V and -15V wires.
Or you can call the right connection ground, and then you have +30V, +15V and 0V wires.
Or you can call the left connection ground, and you get 0V, -15V and -30V wires.
Note - some dual power supplies have the - of both supplies connected, inside the box. In that case, shorting the + of one to the - of the other would short out one of the supplies. This one doesn't have them connected.
answered Aug 29, 2018 at 5:34
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4\$\begingroup\$ "Shorting the + of one to the - of the other puts the two power supplies in series" - but only if the two supplies are originally isolated from each other, and don't share a common negative rail. \$\endgroup\$– AlnitakAug 29, 2018 at 21:15
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In most electronics, "Ground" is simply the point that we want to call "Zero Volts" - it need not have any relation to Earth (AC Power) Ground.
That power supply appears to have three independent isolated supplies. The red line you show connects two of the supplies in series. If you consider that red line as "zero volts/circuit ground" you will have a positive supply from the left-most terminal, and a negative supply from the fifth terminal.
answered Aug 29, 2018 at 6:25
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It might be easier to understand if I would just draw a schematic of how this supply is configured:
simulate this circuit – Schematic created using CircuitLab
So: only the port with the green marking and ground symbol is actually connected to earth (assuming you connected the supply properly to mains ground using the mains connector).
If we ignore V3 (which you're not using in your example) then all outputs are "floating" sources, like batteries they can be used in almost any configuration you like. If you do not connect to earth but combine the supplies (for example to make 60 V by connecting V1 and V2 in series) then that combination is still floating relative to ground.
In your example, the Earth/ground pin is unconnected so your supply is floating. V1 and V2 are in series but the potential relative to ground is still undefined.
answered Aug 29, 2018 at 6:23
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\$\begingroup\$ Except, on the DP832, V2(-) and V3(-) are connected. \$\endgroup\$– Jeroen3Aug 29, 2018 at 9:58
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\$\begingroup\$ @Jeroen3 Indeed you're right, I missed that one, updating the schematic. \$\endgroup\$ Aug 29, 2018 at 9:59
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0V differential is called a virtual ground as in Op Amp inputs, even though without connections they are high input impedance.
Any 0V reference can be called "ground" in electronics ( but not earth bonded ground).
I believe you are referring to when a dual supply has a jumper from V1- to V2+ giving a floating virtual 0V ground reference in between and bipolar outputs.
