Is voltage divider is a good option for supply voltage to an IC ?

I am working with a 555 timer, MAX232 ,an Inverter SN74LVC and with a 12V relay.

I heard that voltage devider is not a best way to step down 12v to 5V for supply voltage.

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    \$\begingroup\$ What you have "heard" is correct. This is a terrible way to get a supply voltage. \$\endgroup\$ Jul 30, 2018 at 13:37
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    \$\begingroup\$ What happens to the voltage output of a voltage divider when the circuit load is added? \$\endgroup\$
    – Andy aka
    Jul 30, 2018 at 13:37
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    \$\begingroup\$ and many, many more questions. Also "I heard" is a strong indication that you should have researched a bit more intensely. \$\endgroup\$ Jul 30, 2018 at 13:44
  • \$\begingroup\$ What you heard is right, it's also common to power microcontrollers through a divider, if you are forced to switch over to a 3.3V MCU when everything around it is 5V. You find the worst case current for the MCU, which will be next to nothing, and figure out a divider that causes it to use a bit more than that much current all the time. Probably works fine drooped to 1.7V, in a terrible worst case. \$\endgroup\$
    – doug65536
    Dec 11, 2022 at 12:51

3 Answers 3


The problem with voltage dividers is twofold.

  1. The output voltage is dependent on the input voltage, often a nominally 12V supply will be some sort of unregulated high power supply that may vary significantly from it's nominal value.
  2. The output voltage is dependent on the output current, to keep the output voltage stable the current drawn by the divider must be significantly greater than the expected variation in output current. For a typical electronic load whose current draw can vary from near zero to whatever it's maximum demand is this means you end up wasting significaly more current down the divider than you end up delivering to the load.

As a result voltage dividers are generally only used for signal/control applications, not as part of the power supply path.

For example lets assume your electronics can draw between 0 and 10ma and can tolerate a 5% variation in supply voltage. Lets further assume that the 12V supply is a perfect 12V.

We design our voltage divider so that off-load it is at the upper end of the acceptable range, 5.25V. So the top resistor has a value of 6.75x and the bottom resistor has a value of 5.25x

Now lets put an actual value on that top resistor, the voltage across it must change by no more than 0.5V when the current drawn by the load changes from 0 to 10ma. So it's value must be at most 0.5 / 0.01 = 50 ohms. The bottom resistor will then need to be approximately 38.9 ohms.

So under no load we are wasting approximately 0.91 W in the top resistor and approximately 0.709W in the bottom resistor. Under full load we will be wasting a bit more in the top resistor and a bit less in the bottom resistor but the ballpark will be similar.

In other words we are wasting over a watt and a half to supply a load that draws a maximum of 0.05W

Reality will be even worse, our resistors will have a finite tolerance and our 12V supply will likely not be exactly 12V, so we will have to waste even more power to get a stable output.

  • \$\begingroup\$ So voltage dividers are something that only has very limited niche use? \$\endgroup\$
    – Juan Perez
    Nov 19, 2021 at 17:32
  • \$\begingroup\$ Voltage dividers are a vital building block of circuit design, but they are not generally suitable for powering stuff directly. \$\endgroup\$ Nov 19, 2021 at 17:50
Is voltage divider is a good option for supply voltage to an IC ?


At the least use a linear regulator.

If efficiency is important and your electronics skills are a bit more advanced, you might want to use a buck converter.

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    \$\begingroup\$ Why was this downvoted? The answer is correct. Sure, it could have done with an explanation as to why the answer is no, but if that is the reason for a downvote, at least leave a comment explaining! \$\endgroup\$
    – MCG
    Jul 30, 2018 at 14:19
  • \$\begingroup\$ I am new so I accepted both one and I dont know that only one answer is accepted.This answer is also accepted,but I got more explanation from other one so I voted that one.Sure next time I will leave a comment for explanation.Thanks for answer each one is helpful to me I am new in analog electronics so I learn something from each answers.Again thnaks @Olin Lathrop \$\endgroup\$
    – ARYA1992
    Jul 30, 2018 at 15:10

If you had a constant load, this would work, which none of your loads are. Voltage dividers change voltage under load, A better way is to use a zener and resistor, and the best way is with feedback in a voltage regulator. The feedback regulates the voltage no matter the current.

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Source: Art of Electronics

  • \$\begingroup\$ why is no load constant? When I only want to operate a LED that shouldn't be a problem, or will it? \$\endgroup\$
    – Ben
    Sep 30, 2020 at 7:29

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