How would one go about using a 12 V DC power source to power something which needs 4.5 V DC using resistors? Is there a way to determine how much adding a resistor would drop the voltage?


6 Answers 6


The short answer is "don't do that."

The voltage dropped by a resistor is given by Ohm's Law: V = I R.

So if you know exactly how much current your device will draw, you could choose a resistor to drop exactly 7.5 V, and leave 4.5 V for your device, when that current is run through it. But if the current through your device is changing, or if you want to make more than one system and not every device is exactly alike in current draw, you can't consistently get 4.5 V at the device using just a resistor.

Your other options include

  • A linear regulator. This is basically a variable resistor that will adjust it's value to keep the output where you want it. This is probably only a good solution if your device draws very little power (maybe up to 100 mA).

  • A shunt regulator. This means using a resistor to drop the voltage like you are suggesting, but then adding an extra device in parallel with the load to control the voltage. The shunt regulator will adjust its current (within limits) to keep the current through the resistor correct to maintain the desired output voltage.

  • A switching regulator. This uses some tricks to generate your desired output voltage with much better power efficiency than a linear regulator. This is probably the best choice if your device needs more than 10 or 20 mA of current.

  • \$\begingroup\$ @Photon: Is it not possible to "calculate" what resistor you will need in order to drop 7.5V from 12V DC without knowing the (amps) current? \$\endgroup\$
    – cpx
    Sep 11, 2016 at 11:05
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    \$\begingroup\$ @cpx No, it is not possible. \$\endgroup\$
    – The Photon
    Sep 11, 2016 at 14:33
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    \$\begingroup\$ @cpx, If you have more questions, you should post them as questions. But looking at your past questions I recommend you find a decent textbook (maybe a first-year physics text or allaboutcircuits.com) and get a basic understanding of what are meant by terms like potential difference, current, and power first. \$\endgroup\$
    – The Photon
    Sep 11, 2016 at 14:51

If these conditions are satisfied you can reduce DC voltage by (high power aluminium) resistors [>50 watt]

  • Your battery is enough to supply at least 20x (or much more) current for your load.
  • Power loss is not a problem.
  • (Over)Heating is not a problem or having good cooling mechanism for resistors.
  • Even the lowest resistance of your load is much (20x or more) higher than the aluminium resistances.

Note: 20x is only an artificial number, the actual number depends on how much % voltage variation does your load can tolerate.

enter image description here


Two resistors can be used as explained by @efox29, just the only problem with that configuration is the current going through the load, connecting a load will change the output voltage because some current will flow through the load .

A simple solution myself is a voltage follower connected to the output of the two resistors, this will provide a high input impedance and therefore:

  1. the output voltage will be constant 4.5V

  2. the opamp used as a voltage follower will try to provide as much current as the load requires.

Here is a picture of what I am talking about:

enter image description here

Connect the output between two resistors to Vin in this configuration and then the output should be a constant value and the op-amp will provide the load with the required current.

  • \$\begingroup\$ This seems like a very natural and flexible solution. What are the drawbacks? \$\endgroup\$ Jul 15, 2016 at 5:13
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    \$\begingroup\$ Main drawback is (most) op amps are not designed to source significant currents. Also, IMHO by the time you've stuck an Op Amp in, you may as well have stuck a power supply / regulator IC and have done with it. \$\endgroup\$
    – John U
    Jul 18, 2016 at 11:43

Look at electro103's schematic above. You need to know four numbers: the maximum current your device can draw, the minimum current it will draw, the maximum voltage it can withstand without vaporizing into a smelly cloud, and the minimum voltage it needs to function. Without these four numbers, you cannot design a resistive voltage divider.

Note that such an arrangement is very inefficient and can result in great heat generated in the dropping resistors.

  • 2
    \$\begingroup\$ This does not answer John's almost five year old question, to which he already have an accepted answer. \$\endgroup\$
    – winny
    Jul 6, 2018 at 8:24
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    \$\begingroup\$ I sure will check the age of any future questions before I answer them. The negative votes are cheerfully accepted. For what it's worth, I think it is time to end my engineering career and go into something useful like law or real estate. Shortage of STEM workers can be solved by rehiring some of the millions who were laid off. \$\endgroup\$ Jul 13, 2018 at 4:04
  • \$\begingroup\$ @richard1941 AND you replied an entire week later, please be more considerate and timely. \$\endgroup\$
    – Lathryx
    Feb 7 at 23:21

Simply take 7805 from market and connect pin number 1 with positive and connect pin 2 with negative and take output from positive from pin number 3 and negative from pin number 2 and keep the distance of output wire 1.5 meter from output terminal of supply to load.

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    \$\begingroup\$ Could you please explain in details about "keep the distance of output wire 1.5 meter from output terminal of supply to load" ? \$\endgroup\$
    – Aura
    Aug 2, 2017 at 17:18

You can only do it with 'just' resistors if you know your load will be a fixed current.

If that's the case (a 20mA LED, say), then you would compute a 'dropping' resistor as follows:

  • R = (Esupply - Etarget) * Iload = (12-4.5V) * 20mA = 375 ohms

If your load isn't fixed (which is almost always the case) you need to use some kind of voltage regulator.

  • For a light load (less than 500mA), use a linear regulator like the LM317, which is adjustable.
  • For a heavier load (500mA or more), a DC-DC would be better. There are adjustable modules available.

And now, a Hacktastical cheep-and-cheerful idea:

  • Go to a quickie-mart and buy a car USB charger (5V output) and a cheap USB cable. (If you want Funyuns and 12-pack of Steel Reserve to go with that, I won't judge.)
  • Hack off and strip the USB cable end. GND will be black and +5V will be red.
  • Add a silicon diode in series with +5V red. A 1N4001 or similar diode is fine. This will give you about a 0.5~0.6V drop.

That will give you 4.4~4.5V or so at 1A, with about 70-90% efficiency.


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