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I have a device which takes 4xD batteries which I would like to instead hook up to a spare wall charger. I have looked at the voltage of the batteries and it seems most batteries are 1.5V. My device has 4 of them (seemingly in series - but with 2 down one side and 2 down the next. Ie:

schematic

simulate this circuit – Schematic created using CircuitLab

There is 1 wire (black) connected to the negative terminal on the top. A Red wire connected to the positive terminal on the top, and then an orange wire connected to the negative terminal at the bottom (although this terminal is just a plate of metal so I guess it's neither positive nor negative as they are both connected? Whereas the positive and negative at the top are separated.)

My wiring diagram seems to indicate this orange wire is ground (I guess that makes sense if both positive and negative terminals are connected then the voltage on this plate is 0 correct?

If that is the case - then am I correct in thinking 4 batteries in series this way can be substituted for a 6V dc charger? (the wiring diagram shows the terminals as -3V and 3V). Also if I substitute the batteries for a dc charger, how do I connect ground?

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Unfortunately, just a 6 V supply won't do it. You really need a ±3 V supply. Or, you can get two 3 V supplies and connect the + of one to the - of the other. That common connection will be the ground of your circuit. The remaining + and - ends are the +3 V and -3 V ends.

The latter is probably easier to source. Basically, get two cheap 3 V wall warts.

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  • \$\begingroup\$ Any reason he couldn't put a voltage divider on the 6V supply to generate the "ground" and therefore +3 V and -3 V relative to the output of the divider? \$\endgroup\$ – ks0ze Mar 6 '17 at 19:59
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    \$\begingroup\$ Because a voltage divider is not suitable for driving any significant load. \$\endgroup\$ – Passerby Mar 6 '17 at 20:01
  • \$\begingroup\$ a potential approach could be having a 3V voltage regulator generate the "virtual" ground, but there's a truckload of things about that that aren't great – basically, you'd always want your ground to be at least as "strong" as your positive and negative voltages, or else. \$\endgroup\$ – Marcus Müller Mar 6 '17 at 20:15
  • \$\begingroup\$ @MarcusMüller No one knows which way "ground" current flows, so we can't say if a +ve regulator or -ve regulator is required. It is also possible that ground current flows both ways, in which case, an op-amp would be required to make a virtual ground. And as you say, op-amp should supply peak current to be safe. \$\endgroup\$ – glen_geek Mar 6 '17 at 23:41
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You should also look at the current requirements for that device. Two small 3VDC plug-in wall transformers will work if the current they supply is adequate for your device. Since you're using a cell as large as the "D" cell it is possible to find small transformers with the correct voltage but lacking the needed current.

Not to be picky but to help with terminology, those are 4 "cells" and not batteries. All 4 cells would make a "battery." Again, my only purpose with that is to expose you to correct terminology even though cells are often referred to as batteries.

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It is entirely possible that the device needs a split supply (±3 V) and that replacing with a simple 6V "wall wart" will be insufficient.

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  • \$\begingroup\$ It is not only possible but also very likely considering the original schematic. \$\endgroup\$ – Ariser Mar 8 '17 at 9:46

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