If you had a circuit hooked up like this:


simulate this circuit – Schematic created using CircuitLab

it would be using a voltage divider, as Vin(R2/(R1+R2)) is the voltage divider rule. That outputs 3.3V which powers one of my LEDS.

My question is: what is the advantages of using the above circuit that is just using a 120 Ohm resistor directly from Vin to limit Vout to 3.3V?

  • 2
    \$\begingroup\$ Hint: How much current is being drawn from the source in either case? Where is that power going? I.e., how much power is being dissipated by each component? \$\endgroup\$ – Dave Tweed Oct 13 '13 at 11:48
  • \$\begingroup\$ I can think of none (unless you include negative advantages, commonly called disadvantages: the voltage divider draws 1.7 A from your 5V source!). \$\endgroup\$ – Wouter van Ooijen Oct 13 '13 at 11:48
  • \$\begingroup\$ Not with these resistors, but if you use larger ones, you get something that's actually useful. \$\endgroup\$ – John Dvorak Oct 13 '13 at 11:51
  • 4
    \$\begingroup\$ It helps to keep you warm in winter. SCNR. \$\endgroup\$ – starblue Oct 13 '13 at 12:04
  • 2
    \$\begingroup\$ Its a good example of why mathematics is not the same as engineering \$\endgroup\$ – JIm Dearden Oct 13 '13 at 12:41

At first, you should never power a LED from a voltage source. (Voltage source is such source of electricity that have constant or near constant voltage on the load attached. Voltage sources have very low internal resistance)

You always must power the LED from a current source. (Current source is such source of electricity that have a constant or near constant current through the load attached. Current sources have very big internal resistance).

Now, the presented schematic is more "voltage source" than "current source". So, it is not good idea to use it in order to power LEDs.

Another problem is that this schematic has very low efficiency, because the most of the current will flow through the resistors, not through the LED.

On the other hand, the serially connected resistor, will turn your voltage source (battery for example) to be more current source that voltage source (high resistance) and this way will make it suitable for powering LEDs.

Another talk is why LEDs should be powered by current sources. That is because the current through the LED highly depends on the voltage. So, even very small changes of the power voltage (or of the ambient temperature), the current will be changed by high value and can become smaller than needed or big enough to burn the LED. When the LED is powered by a current source, the source is that fixes the current through the LED and this current does not changes so much.


what is the advantages of using the above circuit that is just using a 120 Ohm resistor directly from Vin to limit Vout to 3.3V?

I can't think of an advantage of using a resistor divider over a single series resistor, only a disadvantage.

Recall that, to a good approximation, the voltage across the LED is constant over a wide range of diode current. Thus, given the nominal LED voltage \$V_D \$ and the desired operating current \$I_D \$, there is a need for just one series resistor with resistance:

\$R_1 = \dfrac{5V - V_D}{I_D} \$

What happens if you add an \$R_2\$? Remember that \$V_D \$ is (approximately) constant and thus, the current through \$R_1\$ is unchanged. Thus, the effect of \$R_2\$ is to reduce the current through the LED without reducing the power required from the source.

Why would one want to do this? If you desire less diode current, simply increase the value of \$R_1\$ with the benefit of reducing the power required from the source.

In other words, adding an \$R_2\$ wastes power without benefit.


You make some mistakes here.

A LED produces a certain forward voltage over it when it conducts the right amount of current. This voltage depends on the type of LED. A typical red LED has something like 2.2V over it when conducting.

As such, your objective isn't to provide the LED a specific voltage, just enough current to allow it to operate.

The easiest way is just to place a resistor in series with the LED, and attach this series connection directly to the voltage source. For example:


simulate this circuit – Schematic created using CircuitLab

Here the LED forward voltage is 2.0V typically. The current going into the LED is determined as follows: I = (5V-2.0V)/120Ohm = 25mA.

This is quite high, you probably would do fine even with a 1k resistor.


According to Thévenin, all you do here is:

  • Create an equivalant voltage source of:

$$V_{TH} = \dfrac{R_2}{R_1+R_2} \cdot V_{IN} = 3.3\text{V}$$

  • With an ouput impedance of:

$$R_{TH} = R_1|| R_2 = \dfrac{R_1 × R_2}{R_1+R_2} = 660\text{m}\Omega$$


simulate this circuit – Schematic created using CircuitLab

An LED is entirely different from a light bulb. For argument's sake, consider the LED to always drop 3V (reasonably typical for a white LED). No matter how much current you feed into the LED, it will drop these 3V. That is until it releases the magic blue smoke, which it will at $$I_{LED} = \dfrac{V_{TH}-V_{LED}}{R_{TH}} = \dfrac{3.3-3}{0.660}= 450\text{mA}$$.

Now what happens if your LED has a slightly higher or lower "internal" voltage? This is very common, even two LEDs from the same batch may differ quite a lot.

  • $$V_{LED} = 2.9V \Rightarrow I_{LED} = \dfrac{0.4}{0.660} = 606\text{mA}$$
  • $$V_{LED} = 3.1V \Rightarrow I_{LED} = \dfrac{0.2}{0.660} = 303\text{mA}$$

With this slight variation in LED's, you can imagine the difference in brightness.

So what happens if your LED just happens to have a slightly higher forward voltage than your proposed 3.3V voltage? It will not light up at all!

On a side note:

At the expense of roughly:

$$P = \dfrac{V_{IN}^2}{R_1+R_2} = 8.5\text{W}$$

dissipated in the two resistors.

How to properly connect an LED

Check the datasheet for:

  • its forward voltage \$V_F\$;
  • the typical current for the device \$I_F\$.

You probably want to derate the LED current to 80% or lower of the maximum value to extend lifetime of the LED considerably.

Calculate the series resistor \$R_S = \dfrac{V_{IN}-V_F}{I_F}\$


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