I have to supply 80 Raspberry Pi's with power, so I began looking into it and found a power supply of 12V 20A (might not be enough current to supply 10 Pi's that typically draw 1.5A - 2A though).

This meant that I would need to drop 12V to 5V, which I was originally planning to do using resistors for a voltage divider. However with a bit of brief researching I found you should never use voltage dividers as a power supply to loads due to the power that needs to be dissipated.

So I discovered using a voltage regulator was the way to go, however what I don't understand is, if a schematic of a voltage regulator component shows it is comprised of many resistors and transistors wouldn't this suffer the exact same problems as a voltage divider? Wouldn't the resistors internally burn up due to the power dissipation?

Or is it due to the heat sink attached to the component, which would mean hypothetically if you could attach a heat sink to a resistor it'd be fine to use a voltage divider to supply a load?

The reason I ask is trying to work out the best way to efficiently and safely supply 5V at 1-2A to all 80 Pi's I'm aware a few computers is a smarter option however the project involves 80 Pi's and that'sthe way it must remain

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    \$\begingroup\$ you should never use voltage dividers as a power supply to loads due to the power that needs to be dissipated That only applies to applications where the power exceeds a certain value. There is a more important reason why voltage dividers can only be used under certain conditions. The output voltage isn't constant, it depends on the load current, see my comment below. So in your case: put all your Pi's in standby (low current) => They will get a higher supply voltage and will be destroyed. Put all Pi's in active mode (high current) => Supply will drop and all Pi's will reset. \$\endgroup\$ Nov 27, 2019 at 14:44
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    \$\begingroup\$ That's not what your question asks. You asked "which would mean hypothetically if you could attach a heat sink to a resistor it'd be fine to use a voltage divider to supply a load?" The answer to that is in the linked question, and is pretty much the same answer as the ones you are getting now. It isn't (just) that voltage dividers get hot, it is that they don't regulate the output. \$\endgroup\$
    – JRE
    Nov 27, 2019 at 15:02
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    \$\begingroup\$ This is obviously an XY Problem. I think it's time for you to tell us what you're really doing, so we can really help you. \$\endgroup\$
    – TimWescott
    Nov 27, 2019 at 16:22
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    \$\begingroup\$ Your best bet may be to use one or a few 12V power supply of sufficient power rating; these can be had new or easily obtained on the surplus market. Then use one 12V to 5V switching power supply module of sufficient rating for each Pi, co-located with each Pi. This will spare you trying to supply clean power over long wire runs from a regulated supply. Keep in mind that the modules aren't 100% efficient (but way more efficient than a linear regulator), so you have to account for both the heat generated and the extra input power. That's how I'd do something like this, at least for a 1-off. \$\endgroup\$
    – TimWescott
    Nov 27, 2019 at 18:04
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    \$\begingroup\$ @Explorex - Thank you for accepting my answer. At the time I wrote it, I thought I had seen something which other answers didn't address at that time. However, since I wrote my answer, the question has developed into areas not covered by my answer. Some people will not read questions which already have accepted answers, since no further answers appear to be needed in that case. If you want further answers to your question, I suggest you unaccept my answer and when you finally reach a conclusion, you can then choose an answer (mine or another) which helped the most, to finally close the topic. \$\endgroup\$
    – SamGibson
    Nov 28, 2019 at 6:29

6 Answers 6


No [it's not a duplicate of "When would I use a voltage regulator vs voltage divider?"] because I know why they are used differently and their application, I wanted to know the physics of why the resistors inside the [linear voltage regulator] component wouldn't burn in comparison to a voltage divider.

OK, I think I see the question you are asking, and the answer is fairly simple:

  • With a voltage divider, comprising only resistor components (which is typically what people mean when they talk about voltage dividers in this situation) the current for the whole load goes through the "upper" resistor. One of the effects of this (as well as poor regulation) is that the resistor has to be able to dissipate all the heat caused by passing that load current.

    In this type of circuit, the resistors have to be comparatively low values, to reduce the effect of the load current on the voltage divider's "output" voltage. However using low resistor values increases the overall current flowing through the voltage divider to ground, and so increases the power dissipation in those resistors.

  • Using a linear voltage regulator IC, whether its feedback resistors are external or internal to the voltage regulator itself, the load current does not flow through those feedback resistors. Instead, the load current goes through what is called a "pass element" e.g. a transistor.

This difference means that the feedback resistors for a linear voltage regulator (and I'm addressing just your question above, about the resistors) only dissipate a small power since they only pass a tiny current, which is not related to the current required by the load. Those feedback resistors can be comparatively much higher in value, than the resistors in a "simple resistor-only voltage divider".

For example, in page 1 of this datasheet for the old Signetics 7800 series, R19 and R20 are the feedback resistors (shown as 0.25kΩ + 5kΩ) so the current through them is just under 1mA at 5V output. The point is that this small current through those resistors stays approximately constant (and so does their power dissipation), no matter what the load current is.

(There is also this interesting webpage from Ken Shirriff, where he reverse-engineers a 7805 regulator. On that 7805 schematic, the feedback resistor divider is labelled R20 + R21.)

The pass element (e.g. BJT or FET) in a linear voltage regulator behaves like a variable resistor, under the control of an "error amplifier" (see below) and dissipates the same amount of power as the "upper resistor" in the equivalent voltage divider scenario.

Wouldn't the resistors [inside the linear voltage regulator] burn up due to the power dissipation?

No, it's the pass element (e.g. BJT or FET) which can dissipate significant power (and is designed for this, with heatsinking added by the circuit designer where necessary) - not the feedback resistors for the linear regulator, which don't dissipate enough power to "burn up".

That pass element can be internal to a linear voltage regulator IC (typical these days), or external to it, or a combination of both, depending on the regulator IC and the circuit designer's choices.

In case it helps to see it, here is a block diagram of one type of linear voltage regulator. The load is connected to the VO terminals:

block diagram of LDO linear voltage regulator

(Image source: From "Figure 1 LDO block diagram" of Linear Low Dropout Voltage Regulators, from Analog Devices ADALM1000 Active Learning Module)

The series pass element (in the diagram above, it's a P-Channel MOSFET) still dissipates a power related to the load current (P = (VI - VO)·IO approximately). The feedback resistors are termed "Sampling Resistors" in that diagram. As I explained, the load current IO does not flow through those sampling (feedback) resistors.

The "Error Amplifier" (measuring the difference between the reference voltage VR and VS which is the output voltage via the divider formed by sampling / feedback resistors R1 and R2) varies the effective resistance of the pass element, as the output voltage (and therefore VS) changes (whereas the reference voltage VREF and therefore VR, would be stable in an ideal regulator).

Does that explain what I think you are looking for in the question above, about why the resistors in a "pure resistor voltage divider" get hotter than the feedback resistors in a linear voltage regulator?

As the question has developed after I originally posted this answer, it's clear that a good approach to the whole problem is unlikely to involve a linear voltage regulator (or pure resistor voltage divider) at all. Instead, it may involve a buck-mode switching regulator (e.g. 12V to 5V) - perhaps several of them (e.g. one per RPi, or per several RPi boards).

There are advantages & disadvantages of using one or more 12V PSUs (and additional buck regulators down to 5V) or using one or more 5V PSUs, depending on various factors (e.g. voltage drop over the DC power cabling). This has been explained in another answer.

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    \$\begingroup\$ Still the transistor dissipates as much power as the upper resistor in the divider! This answer should neither be upvoted nor accepted! The regulator will become (almost) as hot as the divider! \$\endgroup\$
    – yar
    Nov 27, 2019 at 23:41
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    \$\begingroup\$ @yar - Please look at the OP's words I quoted. I never tried to address your point, because that is not what the OP asked in the text I quoted! The OP asked specifically about the resistors in the voltage divider vs. linear voltage regulator cases. That is what I noticed hadn't been addressed in other answers, at the time I posted this answer, and so I tried to help the OP by explaining what I had noticed. If you choose to focus on a different part of the question and not upvote this answer, that's fine with me. Feel free to write your own answer too! Thanks. \$\endgroup\$
    – SamGibson
    Nov 27, 2019 at 23:51
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    \$\begingroup\$ "resistor" is a function, not a technology \$\endgroup\$ Nov 27, 2019 at 23:55
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    \$\begingroup\$ @SamGibson Ah, I see. The current state of the answer is much better, thank you! \$\endgroup\$
    – yar
    Nov 28, 2019 at 0:04
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    \$\begingroup\$ @Circuitfantasist - I've been told not to write over-long answers, so I won't go into linear regulator design in my answer, as it wasn't part of the original question :-) My thoughts about your suggestion are (a) it wouldn't save 2 discrete resistors in most cases; those resistors are already integrated within many linear regulator ICs & so have minimal cost (and zero PCB size penalty); (b) it would require \$\small{V_{REF}}\$ to be equal to \$\small{V_O}\$. That sounds harder than altering the R1/R2 values for different \$\small{V_O}\$ during production - see the Ken Shirriff page I linked. \$\endgroup\$
    – SamGibson
    Nov 28, 2019 at 19:33

All input so far seems to miss your main point.
Using ANY linear means of reducing voltage at the same current will waste the power difference between power in and power out.

This SE EE answer of mine from 2011 explains what happens when a linear regulator is used - and shows why you do not want to use one here - except perhaps to drop a small fraction of a volt at a Pi input if a local bus slightly above 5V is used.

Here for divider OR linear regulator, Iin = Iout (apart from a small amount used in regulator control).

Pin = Vin x Iin
Pout = Vout x Iout
Power lost in linear regulator or divider = (Vin-Vout) x Iin


Input 12V, 20A
Output5V 20A
Pin = 12 x 20 = 240 Watts
Pout = %v x 20A = 100 Watts
Power dissipated (wasted) 240 - 100 = 140 Watts
Efficiency = Pout/Pin = 100/240 = 42%
Power (actually energy) loss = 58%

You need either:

An active regulator (switch mode supply) that converts 12V to 5V efficientlt


A power supply system that gives you what you need (~= 5V) directly

More can be said but we need some feedback from you.

Important: A linear regulator is "just"a voltage divider with a degree of intelligence.


A Pi ["Raspberrby Pi"] is designed to be powered by a 5V power supply and, unlike most other small microcontroller boards, has no onboard regulator.
The "proper" way to provide power is to provide a good quality 5V supply of adequate current capacity AT the 5V input pin.

Circuit diagrams:

The model 2 (and perhaps earlier) has a more interesting power input circuit but the same observations apply.

Some sources suggest a voltage of 5.1V to allow for internal drops under heavy loads. While this will probably cause no great problems it is already notionally exceeding specifications as there is an on board 5V voltage suppressor (usually SMNJ-5.0A) which has a 5.0V 'standoff voltage'. This is designed to ensure that power supply input voltages above 5V "have their wings trimmed". At 5.1V it is (again, notionally") 'starting to take interest', but probably not excessively so.

If the Pis are in clusters with small distances between them then supply 5V (or 5.1V) from a suitable power supply over short distances using suitably heavy wiring can be acceptable.

If clusters or individual Pis are distance separated such that voltage drops of more than say 0.1V would occur at their power input pins worst case, then a much better solution is to reticulate a higher voltage to 5V out switching regulators located immediately adjacent to individual Pis or perhaps small clusters.

Once reticulation plus switching regulators is used the Vin can be whatever is most suitable. Using eg 12V reduces power losses in the distribution cables by a factor of 5+ for the same cable. Using 20C reduces energy lost by a factor of about (20/5)^2 = 25 tims for the same size cables or allows smaller diameter cables.

Small switching regulator modules are available at low cost on eg ebay or from Aliexpress or ... . As ever, buyer beware. Check specs,. Check functionality. Consider an input fuse (and hope or design things so that the surge suppressor will blow the fuse if a supply output fails high.
Note that while eg the Pi model 3 has an input "PTC resettable fuse" (typically MF-MSMF250/X) in series with the input, it has been (naughtily) omitted from the model 4 and perhaps other versions - see circuit diagrams.

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    \$\begingroup\$ This is the only right answer so far! I'm shocked, that it still is the one with the least number of votes! As long as you don't use a switching regulator you will always waste the same power for heating. \$\endgroup\$
    – yar
    Nov 27, 2019 at 23:37
  • \$\begingroup\$ @yar: As Sam's answer points out, with a voltage divider you can waste extra power by pulling extra current through the upper resistor. But maybe you were only counting useful / correct ways of powering a time-varying load that needs actual regulation. (which would rule out voltage dividers.) \$\endgroup\$ Nov 28, 2019 at 9:28
  • \$\begingroup\$ @PeterCordes I don't see how what you say adds anything. A linear voltage regulator is identical to a divider thermally, ellectrically and efficiency wise. All input energy not used at the output is wasted. \$\endgroup\$
    – Russell McMahon
    Nov 28, 2019 at 10:36
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    \$\begingroup\$ @RussellMcMahon: I was only nitpicking that not all alternatives to switching waste the same power. You can waste extra power with a voltage divider because you create extra load, as well as wasting all the (Vin - Vout) * I_load power. \$\endgroup\$ Nov 28, 2019 at 10:42
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    \$\begingroup\$ @PeterCordes yes, that's true, especially since when using a divider as a power source, the load current has to be negligible small compared to the not loaded divider current! I was basically thinking of a series resistor, which you can use for a drop of a constant voltage at a constant current \$\endgroup\$
    – yar
    Nov 28, 2019 at 20:44

A regulator maintains its output voltage for a wider range of current draws.

Whereas a voltage divider is (often) built with the assumption that current draw is 0. And as soon as the current draw changes so does the output voltage.

linear voltage regulators use a transistor in the linear region to provide the resistance. This indeed gets hot and you need to make sure it stays cool enough.

If you want to supply a high power 5V load from a 12V source you should use a switchmode regulator or a buck regulator. These will waste less power in the conversion.


The difference between a voltage regulator and a divider essentially is that the regulator does what it says..... it regulates the voltage. For the majority of linear regulators, you will find the internal schematic has an 'Error amplifier' which is how it stays in regulation.

Here is a very simplified schematic of the linear regulator:

enter image description here

The resistors Rf1 and Rf2 are called "Feedback resistors". This works just like a voltage divider. These can be internal for a fixed output regulator, or they can be external components for an adjustable regulator. The Vref is almost always internal, and the value of this will be in the regulators datasheet. The amplifier will adjust the base current of the pass transistor, so that the output voltage is less than the input. That is where the feedback resistors are important. The amplifier will continue to change the base current of the pass transistor, until the 'output' of the feedback divider (Vfb) is equal to Vref. This way, if a change in load occurs, and the output voltage suffers, Vfb will now be changed too. The amplifier will then adjust the base current of the pass transistor, adjusting the output voltage until Vfb = Vref. Thus, it is self regulating.

Most datasheets will have recommended resistor values for at least one of the feedback resistors. They are usually chosen so that the current through them is small, which means less heat dissipation. For fixed output versions, the internal resistance will be high again, minimising current so they don't dissipate too much heat. The use of the pass transistor also means that the load current will not flow through the resistors.

  • \$\begingroup\$ It would be useful for understanding to say that the so-called "linear regulator" is the well-known "non-inverting amplifier". Its input is driven by constant voltage and its output is buffered by an emitter follower. \$\endgroup\$ Nov 27, 2019 at 22:23
  • \$\begingroup\$ Also, it is interesting to explain the need of the voltage divider Rf1-Rf2. Can't we connect the inverting op-amp input directly to the transistor emitter? So we would save two resistors. \$\endgroup\$ Nov 28, 2019 at 18:59
  • \$\begingroup\$ @Circuitfantasist it isn't a non inverting amplifier. And no, you can't connect the input directly to the emitter, Simulate it, see what happens \$\endgroup\$
    – MCG
    Nov 28, 2019 at 23:27
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    \$\begingroup\$ @Circuitfantasist ahh, that's my bad, for some reason I had inverting amplifier in my head when typing that! By the way... You know you can type your own answer if you want to? That would be better than adding things in the comments of an answer. \$\endgroup\$
    – MCG
    Nov 29, 2019 at 8:11
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    \$\begingroup\$ @Circuitfantasist the reason I said to write your own answer is because people don't tend to read comments very often. I don't need the circuit explained to me, and people that do will be looking at the answers. \$\endgroup\$
    – MCG
    Nov 29, 2019 at 9:57

The problem with using voltage dividers for any moderate or changing load is that the voltage at the load will change due to resistive loading.

Consider this circuit:


simulate this circuit – Schematic created using CircuitLab

Now this is a gross example, but it gives you an idea of the problem.

Let's arbitrarily choose Vin = 6.6V with a goal of 3.3V to the load; that will hold true (roughly) for loads above about 1k (so <= 3.3mA) (the load voltage will actually drop to 3.14V at a 1k load) - this is due to the voltage divider not being stiff enough.

To maintain the voltage within 10%, the lower divider resistor needs to be no greater than 1/10 of the effective load resistance.

For a 1A load from 3.3V, the effective load resistance is 3.3\$ \Omega\$ so the divider resistors must be much less than this (0.33\$ \Omega\$).

That is only the first problem; the upper resistor must also be low enough to let that amount of current pass.

Then we have load transients - a load that varies between 100mA and 2A will have significant power rail transients that would be difficult to effectively filter.

Voltage dividers are used for reference voltages with very small load currents but as the load currents increase, the (current) divider effect of the load with the divider itself leads to errors in the effective reference voltage.

So the biggest problem is not necessarily heat in the divider, but inaccuracy (which can be very large) of the actual output voltage with any varying load.

Note that this issue is still somewhat problematic even when using a zener as a simple regulator.

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    \$\begingroup\$ This is half of the right answer. To make it the complete answer, you need to show how the linear regulator replaces the upper resistor with an auto-adjusting resistance typically made of some sort of semiconductor element. \$\endgroup\$ Nov 27, 2019 at 23:58
  • \$\begingroup\$ Exactly... There is a need to reveal the "philosophy" of such a "dynamic voltage divider" vs the ordinary "static voltage divider"... \$\endgroup\$ Nov 28, 2019 at 6:31
  • \$\begingroup\$ @PeterSmith, another +1 about noticing the current divider R2||R3. \$\endgroup\$ Nov 28, 2019 at 16:26

The reason I ask is trying to work out the best way to efficiently and safely supply 5V at 1-2A to all 80 Pi's

The best way is to use a switching regulator! Both, the linear regulator and the divider will dissipate 7W-14W (1A-2A) - which is not efficient.


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