# Resistor in series, or voltage divider?

There's something I just can't get my head around. Say I have a 5 V DC source, and I need to power an LED rated at 280 mA @ 3.2 V.

Should I use a resistor in series (1.8 V drop needed, so ~6.5 Ω), or make a voltage divider (R1 - 5.5 Ω, R2 - 60 Ω, RL ~ 11.4)?

Does one have an advantage over the other? If I use a voltage divider, should I consider higher resistances?

• Have you learned how to analyze networks with series and parallel resistors? Commented Aug 25, 2022 at 21:04
• Series resistor... Commented Aug 26, 2022 at 3:56
• As a side note, using a resistor for a 320 mA LED albeit the small drop you need (1.8 V that means 576 mW) is starting to push what is reasonable to do with a resistor in my opinion. Commented Aug 26, 2022 at 9:20
• Nobody makes a 3.2V LED. All LEDs have a range of voltages. Some of yours are maybe 2.8V and others are maybe 3.6V or more. You need to test your LED to see what is its forward voltage then calculate the resistor. How will you cool the LED? Commented Aug 26, 2022 at 20:19
• They do! I'm using these Xiaoyztan 100 Pcs White Light LED Bead Chips DC 3.0-3.2V 280mA 1W Emitting Diodes LED Emitters (White Light) a.co/d/aLtasXs Commented Aug 28, 2022 at 15:08

A voltage divider is the way to go when you are trying to control voltage. A single resistor is better when you are trying to control current.

The 3.2 V figure you have for your LED is forward voltage drop of the diode (D in LED). This is not trying to say you need to feed this LED 3.2 V, rather that it's voltage drop across it will be about 3.2 V when the current through it is 280 mA. Notice I said "about". If you try to provide exactly 3.2 V, the current could be almost anything from not enough to see the LED to burning it out completely.

With LEDs, you are trying to control the current, and the voltage at that current is just an artifact of that. Therefore, you will want to use a resistor that will provide the correct voltage drop at the desired current.

The resistor divider would actually work (because it also has a series resistor limiting current), but it is wasting power by dumping current to ground rather than just a series resistor where all the current goes through the LED.

• Series resistor also wastes power. You could consider a switched current supply Commented Aug 27, 2022 at 10:55

A series resistor of 6.5 Ω is the correct way to do this. A voltage divider will change its voltage when you add the LED in parallel with the "bottom" resistor.

Keep in mind that your resistor's power is 0.282·6.5, or just over half a watt, so you will need probably a 1 W resistor to handle it comfortably and it will get warm if not hot.

Standard resistance values are almost always cheaper than the exact value, so 6.8 Ω would be an appropriate choice. Also your LED will be dissipating almost a watt itself so it should have some form of heatsink.

• You forgot the decimal point in 68 :-) Commented Aug 27, 2022 at 10:58
• Got it, thanks!
– vir
Commented Aug 29, 2022 at 17:00

We start out with the simple KVL equation which can be applied in either case:

$$V_{_\text{TH}}-I_{_\text{LED}}\cdot R_{_\text{TH}}-V_{_\text{LED}}=0\:\text{V}$$

With a simple resistor $$\V_{_\text{TH}}=V_{_\text{CC}}\$$ and $$\R_{_\text{TH}}=R_{_\text{LIMIT}}\$$. With a resistor divider ($$\R_1\$$ connected to $$\V_{_\text{CC}}\$$ and $$\R_2\$$ connected to ground) $$\V_{_\text{TH}}=V_{_\text{CC}}\cdot \frac{R_2}{R_1+R_2}\$$ and $$\R_{_\text{TH}}= \frac{R_1\,\cdot \,R_2}{R_1+R_2}\$$.

Solving for $$\I_{_\text{LED}}\$$:

$$I_{_\text{LED}} =\frac{V_{_\text{TH}}-V_{_\text{LED}}}{R_{_\text{TH}}}$$

Now comes the question of better.

Power, while a problem, isn't the bigger problem. A sensitivity analysis will help to see why. We do this by trying to examine how the LED current changes, as a percent of its nominal value, when some other factor changes by some percent of its own nominal value. This uses calculus to be precise. But it is straight-forward:

\begin{align*} &\text{Take the derivative using the chain rule,}\\\\ \text{d}\:I_{_\text{LED}} &=\frac{1}{R_{_\text{TH}}}\:\text{d}\:V_{_\text{TH}}-\frac{1}{R_{_\text{TH}}}\:\text{d}\:V_{_\text{LED}}+\frac{V_{_\text{TH}}-V_{_\text{LED}}}{R_{_\text{TH}}^{^\:2}}\:\text{d}\:R_{_\text{TH}} \\\\ &\text{Divide both sides by }I_{_\text{LED}},\\\\ \%\, I_{_\text{LED}}&=\frac{\%\, V_{_\text{TH}}}{1-\frac{V_{_\text{LED}}}{V_{_\text{TH}}}}-\frac{\%\, V_\text{LED}}{\frac{V_{_\text{TH}}}{V_{_\text{LED}}}-1}-\%\,R_{_\text{TH}} \end{align*}

And that specifies quantitative regulation with respect to each parameter.

With a voltage divider, it's not % variations in $$\R_{_\text{TH}}\$$ so much. But instead it is mostly about % variations in $$\V_{_\text{TH}}\$$, affected by resistor tolerance, that will kill you. For example, suppose you choose $$\V_{_\text{TH}}=3.5\:\text{V}\$$, which is a little bit higher than your expected LED voltage, and used 2% tolerance resistors for the divider. The resistor part alone would yield about a 2% variation in LED current, looking at the rightmost term and ignoring $$\V_{_\text{TH}}\$$. But when you look at the leftmost term and consider a 2% variation in $$\V_{_\text{TH}}\$$, you now find that there is about 20-25% variation in the LED current as a result of that.

It's still very much worse once you take into account LED voltage variations, one to another!

So you really do not want to use a resistor divider for that reason, alone. Even without considering LED voltage variations (which are also not so good.)

So this returns you to the idea of using a simple $$\V_{_\text{CC}}\$$ and $$\R_{_\text{LIMIT}}\$$. While this case is much better, the regulation still won't be all that good with so little headroom (you can plug in some numbers to see what happens with the LED voltage variations, which are large.) But you are much better off without the use of a resistor divider purely for regulation issues alone.

Sure, you'd also waste power with a resistor divider. And that's another reason not to go there. But that's not the biggest reason for avoiding a resistor divider. The biggest reason is that the LED brightness, one instance of a circuit to another, will be all over the place.

• Be aware of long stories starting with the qualifier "simple" Commented Aug 27, 2022 at 10:57
• @Roland The KVL is simple and the idea of analyzing, using variables that can either use Thevenin results or just straight values is also simple, I think. The sensitivity analysis itself is only 'simple' if you have mastered basic calculus (the 2nd semester of the 1st year.) So I'll give that. But what doesn't happen enough, here or elsewhere, is to apply sensitivity analysis in order to find out what's important and what is less so in a situation. I seem to be the only person here using it habitually. It should be seen far more often here than it is. Thanks for the comment. Helps me think.
– jonk
Commented Aug 27, 2022 at 18:43

As the other answers explained, in this case a series resistor is best if all you have are resistors. A voltage divider is intended for cases where the current load on it is light (or in cases where you can buffer the load for example by using an operational amplifier or an op-amp + a push-pull transistor configuration for even more current). If you use a voltage divider, it will create even more waste heat than a single series resistor.

Actually, for these high-power LEDs rated at watt or more, you usually don't want to use series resistor either. Your resistor would create 1.8 V * 0.28 A = 0.504 W of heat. A normal resistor can't handle that. A power resistor would but it's simply not a good solution for driving high-power LEDs. In your case, the energy efficiency of a resistor is 3.2 V / 5 V = 0.64 = 64%.

What you want is a buck converter with control mechanism for current. The buck converter lowers the 5 volt voltage to whatever is needed to create that 0.28 A of current. It will have way less losses than a series resistor. Buck converter won't have 100% efficiency but you should at least get 80-90% out of it, in any case more than 64%.

• Almost correct. But a buck converter creates a voltage, where a current source is needed. The solution is a current module, set to 280 mA, which is a kind of buck converter, but with a different control system. Commented Aug 27, 2022 at 11:01
• Well I did say "with control mechanism for current", so I didn't mean using an ordinary buck converter with voltage control, but rather one with current control. Commented Aug 28, 2022 at 9:55
• Thanks for clearing this up. It was just not clear to me from your description Commented Aug 28, 2022 at 10:10

The best way to explain circuits is to go back in time, reinvent them step by step and generalize the ideas derived. Let's do it then...

The problem here is to reduce a voltage by a certain value (in this particular case, from 5 V to 3.2 V).

The solution is to subtract some voltage (1.8 V here) as follows:

If the load is low resistive, and a significant current flows, it is enough to insert a resistor between the source and load. A voltage drop appears across the resistor and the voltage decreases with this value.

If the load is high resistive (e.g., open circuit), not enough (no) current is flowing. There is not a high enough (no) voltage drop across the resistor. So, to reduce the voltage, we need current to flow. For this purpose, we include another resistor in parallel with the load... and thus we "invent" the circuit of the famous voltage divider.

So, the lower resistor R2 in the circuit of a voltage divider allows current I to flow through the upper resistor R1 in order to cause a voltage drop VR1 = I.R1 across it.

The bottom line is that when the load has relatively low resistance, we can regulate the voltage across and current through it with just one resistor in series. When the load has high resistance, we can adjust the voltage across it by adding another resistor in parallel to the load.

In fact, there is always a voltage divider configuration consisting of two elements in series. One of them is a resistor deliberately added to reduce the voltage; the other is either the load (LED here) or another resistor. The resistance of the first resistor must be high enough compared to the resistance of the second resistor.

This begs the question, "What kind of load is an LED?" The problem with it is that it offers no constant resistance to the current flowing through it. At first, at low voltage, it is high... but at some point (in the vertical part of the IV curve), it starts to decrease sharply.

So, if we need to fix the voltage across the LED less than its threshold, we need to use a voltage divider; if the voltage is higher than the threshold, only one resistor is sufficient.

It is important to understand that a LED needs a certain current, contrary to the old incandescent lamp, or "bulb", that needs a certain voltage. Are you old enough to remember the 60 W bulb above the dining room table? Such lamp was rated 120 or 220 V, while the current was not printed on the bulb. Your bicycle had a 5 V bulb, your car 12 V bulbs. The filaments of those bulbs were hard wired to a voltage source, no drivers needed.

Now with LED, the electrical systems are still voltage systems, but really, LED's for 5, 12, 120, 230 V do not exist, they all need (internal) resistors or current drivers.

It is nice to know that the rated voltage of your LED is 3.2 V, so you can dimension the series resistor to be about: (5.0 - 3.2)V / 280mA = 6.5 Ohm. (The power rating will be P = I-square times R = 0.5 W, just buy a 2 W type or bigger)

When you implement this, you will want to take your multimeter, and check the current through the LED, this should not be much higher than rated in order to avoid burning it out. You may measure current directly, or by measuring the voltage over the resistor, if you trust its value.

Do NOT measure the voltage over the LED. The current through a diode (the D in LED is for diode) is an exponential function of its voltage, so even if you would measure an "about correct" led voltage, the current may be way off.

Because that 5 V may not be exact, and a 6.5 ohm resistor may be hard to implement, and eats power, you may look for a "current driver", usually a switched power supply chip or module, that can be easily set to your rated current value, and that has an efficiency of around 80 %, meaning that you save 80 % of the heat loss of a normal resistor.

Another option, not saving power, but to deal with a not exact or constant 5V input, is to search for how to use a voltage regulator chip as a current source, e.g.:

Voltage regulator for linear constant-current (1.5A) LED driver?

By the way, do not fall in the trap of connecting two of those LED's in series, thinking that 2 times 2.8 V is just about 5 V, and eliminating that power wasting resistor. Because of the exponential U - I curve, the current in the LED's will be way too low, and so the light output. In case the power voltage gets only slightly over 5.6 V, the LED's will draw a way too high current, and burn out.

I leave it as an "exercise for the reader" to why you should not connect old bulbs in series, or with a series resistor, or LED's in parallel . . .

Good Luck!

• Interesting observations on the behavior of these nonlinear devices... Maybe my story would be of interest to you? Commented Aug 27, 2022 at 13:28
• @Circuitfantasist Great story, but why stop there? How about the magnetics? H-B curves of iron core, V-I curves of inductors showing hysteresis and saturation of the iron core. Commented Aug 27, 2022 at 13:51
• Roland, Thanks... but I am not a physicist; I am just a "circuit thinker" :-) You can see this from my Circuit Idea wikibook. By the way, here it is accepted to answer only once (I guess for the sake of voting, maybe not to collect too many reputations). I personally do not see the point in this, also in the suppression of discussions... Commented Aug 27, 2022 at 14:22
• @Circuitfantasist - Hi, Re: "I personally do not see the point in [...] the suppression of discussions" The reason is that this site is not a discussion forum, it doesn't want to be one, and it has rules to stop attempts to use it in that way. You can see this in the comment rules mentioned here in the help center and here in the FAQ. This approach is part of what differentiates Stack Exchange from typical forums, so don't expect it to change anytime soon. You can ask more in Electrical Engineering Meta. TY Commented Aug 29, 2022 at 18:26

I answered on how to hook up the LED. Rereading your question I get the impression that you do not actually need to hook up an LED, but are just tinkering on circuit design.

Using the voltage divider is fundamentally wrong. The problem lies in the RL, the "resistance" you computed for the LED. But for the LED, the current is an exponential function of the voltage, not proportional, like a normal resistor.

That leaves the series resistor. Indeed, low power LED's like what you are thinking of are often connected with a series resistor. For 12 Volt, often 2 or 3 LED's are connected in series, with one series resistor, to reduce the power loss, while the current is stabilized sufficiently.

On tinkering, why not simulate your designs with a SPICE program? There are many free versions.