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How would one go about using a 42V DC power source to power something which needs 36 V DC using resistors?

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    \$\begingroup\$ One would NOT! I can't find the reference but this is a duplicate from other questions where people think they to change the output of a power suply just using resistors. \$\endgroup\$ – Oldfart May 31 '19 at 14:19
  • \$\begingroup\$ @JRE That is the one I was looking for! \$\endgroup\$ – Oldfart May 31 '19 at 15:28
  • \$\begingroup\$ 46V in the title and 42V in the question body, which is it? Is the 46/42V regulated? Does the 36V need to be regulated against changes in the 46/42V supply ...for example If the 46/42V is a battery then it's voltage drops as SOC drops. Does the 36V need to be regulated within a spec, for example 36V +/-0.1V And last but most important, how much current required at 36V (both maximum and minimum). \$\endgroup\$ – Jack Creasey May 31 '19 at 16:22
  • \$\begingroup\$ @Oldfart Diodes, however, can be used for a quick and dirty voltage drop.. \$\endgroup\$ – Barleyman May 31 '19 at 16:58
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Not at all.

Resistors by definition cause a voltage drop proportional to the current draw.

So, you can't use resistors to make a supply at 36 V for varying current draw.

What you're looking for is a voltage regulator:

  • a linear voltage regulator converts the voltage difference (10 V or 6 V, depending on whether I believe your title, or your question text) to heat, and acts like a "self-adjusting resistor". It's not made of resistors, but semiconductor devices (transistors, mainly).
  • a switch mode supply instead charges the magnetic field in an inductor and releases just enough energy from that to its output that the voltage is stable at the target voltage.

Without knowing how much or how little current you'll draw, no definite recommendation can be made, but at 6 or 10 V drop, chances are you want a switch mode power supply, which is relatively complex in theory, but can in practice be built from a controller IC, inductor, a diode and a few capacitors.

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This is a really bad idea, but can be done if you know exactly how much current the device draws, and if this is a fixed, constant value over time, temperature, and input voltage.

If the device draws 100 mA of current at 36 volts, you need a resistor that develops a voltage drop of 42-36=6 volts at 100 mA. Thanks to Ohm's law (R=U/I) this equals 6/0.1 = 60 Ohm in series with the device.

To calculate the power dissipation in this poor resistor you can get that by the voltage drop and the current directly: 6 * 0.1 = 0.6 W.

These calculations are pointless if your device changes the amount of current it draws, and completely impractical if it draws a lot of current in which case you need a beefy resistor.

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    \$\begingroup\$ Please emphasize "and if this is a fixed value". \$\endgroup\$ – WhatRoughBeast May 31 '19 at 18:37
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No.

After you figure out if it's actually 42, 46 or perhaps 48V PSU, you could make a string of diodes to drop off the voltage. Voltage drop needed divided by 0.7v = number of diodes needed. 42 => 36V, 6 volts off, 6V / 0.7V = 8 diodes needed. And so on. You could use a zener diode for a larger drop but depending on your current drain it'll get mighty hot. And speaking of hot, depending how many amps the charger pulls, those diodes will be piping hot. Don't use small dinky types. Or Schottky diodes.

It would also be possible to use a MOSFET to generate a fixed voltage drop but one might just as well get a standard LM317, TO-220 3-pin regulator and a heatsink if you're going in that direction. Hot.

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I see a pattern:

36V is 3 x 12V

42V is 3 x "12V" lead acid batteries in series.

If the device was designed to be powered from the lead-acid batteries it will probably work just fine, if not it will probably break.

if you need 36V Then you need a voltage regulator. LM2596HVS is one example.

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One usualy would not do it that way, using resistors that is. Special casses exist, but without knowing what is soupose to draw the current, it is impossible to answer your question. Here is a link to a wiki article that explains input impedance, link output impedance, and a link explaining voltage division with resistors. If you read these articels, it is clear that the resistor voltage dividers output voltage is dependant on the input impedance of the next stage, and that the current draw is limited with the resistor that connects to the input voltage node.

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Like this (limits the output current to 0.006 amps)

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Those are some funny looking resistors... \$\endgroup\$ – pipe May 31 '19 at 14:25
  • \$\begingroup\$ Drop the current limiter or set it to some reasonable value and it'd be actually useful. \$\endgroup\$ – Barleyman May 31 '19 at 17:01
  • \$\begingroup\$ To increase the allowed current ( 0.6/100 ohm== 6 milliAmps), the collector dissipation budget of 0.2 watts needs a dramatic increase. Consider using a TO-220 NPN, perhaps even a Darlington; be sure to heatsink the tab of the TO-220 package. Be aware the TAB may be at the collector potential (42 volts) and you must either float the heatsink or insulate the tab from the heatsink. \$\endgroup\$ – analogsystemsrf Jun 2 '19 at 3:07

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