In my circuit, I have 5V power supply and I have 200 ohm resistor. When I measure the current, it's 14.8 mA. Why can't I use the equation I = V/R, which is 5/200 = 25 mA?

enter image description here

  • 4
    \$\begingroup\$ You have more than just a resistor there. \$\endgroup\$
    – Eugene Sh.
    Mar 11, 2019 at 20:10
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    \$\begingroup\$ You have to measure just the drop across the resistor. \$\endgroup\$
    – stark
    Mar 11, 2019 at 20:14
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    \$\begingroup\$ @SunnyskyguyEE75, Your comment probably will not help the OP to understand how to use Ohm's Law. \$\endgroup\$ Mar 11, 2019 at 21:08
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    \$\begingroup\$ I think Ohm's Law ought to be a prerequisitve to an EE site, don't you? @SolomonSlow \$\endgroup\$ Mar 11, 2019 at 21:24
  • 1
    \$\begingroup\$ @SunnyskyguyEE75, I don't see that specific requirement in the help center. Perhaps you could suggest it to the moderators. There certainly are other stackexchange sites that prefer participants to demonstrate some minimum level of competence before asking questions. \$\endgroup\$ Mar 11, 2019 at 21:46

4 Answers 4


Ohm's law applies to the resistor. The voltage across the resistor will be about 3V. The remaining 2V or so appears across the LED.

To a rough approximation I = Vr/R ~= (5V-Vf)/200. Vf, the voltage across the LED, stays more or less the same at different currents within the normal operating range.

So if you increase the voltage to 6V you would expect the current to increase to about 20mA (actually a bit less because the voltage across the LED will increase a bit).

From this answer, there's a graph that is similar for most red LEDs.

As you can see, at about 15mA Vf is a bit under 2V. At 20mA, Vf is just about 2V exactly. At 30mA it's more like 2.05V.

enter image description here


You can use Ohm's Law but only on the resistor.

The LED in your circuit is non-linear and so Ohm's Law does not apply to it.

enter image description here

Figure 1. The I-V (current versus voltage) curves for various colours of LED. Source: LEDnique - IV curves.

From your measurements we can calculate the voltage across the resistor as \$ V = IR = 14.8m \times 200 = 2.96 \ \text V \$. (Let's say 3 V.)

That leaves 2 V across the LED and from the I-V curve in Figure 1 we can see that at 14.8 mA we would get a 2.0 V drop across the LED for a green(ish) LED.

The important thing to learn here is that the LED has a voltage drop and that it does not act like a resistor.

enter image description here

Figure 2. The I-V curves for an assortment of resistors. Note that these are linear. Source: LEDnique - IV curves.

  • \$\begingroup\$ That first image has a random mouse cursor in it. \$\endgroup\$
    – Hearth
    Mar 11, 2019 at 20:39
  • \$\begingroup\$ Ha! My Greenshot screengrabber grabs the cursor if I forget to move it out of the way. \$\endgroup\$
    – Transistor
    Mar 11, 2019 at 20:46
  • \$\begingroup\$ I figured that was the case! It's not really a problem, though. \$\endgroup\$
    – Hearth
    Mar 11, 2019 at 20:54

Ohm's Law applies only to resistors. The voltage term must be the voltage across a resistor, and the current term is the current through the same resistor. Your 5V is not the voltage across the resistor, it is the voltage across both the resistor and the LED.


This is all about linear regression above the knee voltage, then you can use Ohm's Law.

It's only because you don't know how to model a diode. Diodes are nonlinear initially then they somewhat behave as a voltage drop with a small series resistance.

Ohm's Law or KVL analysis

\$Vcc= I_f*R + V_{f (LED)} = I_f*R + (V_t+I_f*R_s)\$ ......(1)

for knee threshold Vt and diode bulk series Rs and forward current If rated at Vf forward voltage.

From looking at the datasheet VI curve you can get a typical ΔV/ΔI =Rs in Ω (V/I slope or tangent) near the current range you expect.

As the current drops to 10% of the rated current (not Abs max.) the series resistance rises to about 10x that at rated current so it is mostly linear above 1/4 of the rated current given the wide tolerances on this Rs is +50%/-25% for an open "bin".

Here is how Ohm's Law can be used on diodes.

e.g. RED LED's Vf = 1.85V + 12*If (Rs = 12 Ω +50%/-25%)

thus using (1)
\$Vcc= I_f*R + V_t+I_f*R_s\$

or \$I_f= \dfrac{V_{cc}-V_t}{R+Rs}=\dfrac{5-1.85}{200+12}=14.3mA\$ which is well within expected tolerances due to Rs. enter image description here There is also a voltage drop with rising junction temperature which may be computed , but it's not a good idea to get it too hot to touch.

This model is far more accurate than the tolerance on parts so changing Rs by your Vdd and Rs tolerance allows anyone to use Ohm's Law for any diodes (including Zeners) with some understanding.

You can adjust the model to the Datasheet.
enter image description here

Note this LED Vf=1.85V (typ) 2.5V (max)

here the typical curve is shown ONLY or Vt=1.75V and slope ΔV/ΔI=(1.9-1.75)/30mA = 125mV/30mA = 4.2 Ω = Rs


Still If=(Vcc-Vt)/(R+Rs)=(5-1.75)/(204)=15.9mA. It is not much different than 14.3mA above. Now try 2.5V , what R gives 20mA?

20mA=(2.5V-1.75V)(R+4.3), R= 750mV/20mA-4.3 = 33 Ω
(Ohms Law says 20mA*33= 660mV across R)

so Vf=2.5V - 0.66 = 1.84V vs 1.86V above, which is what the datasheet says for TYP. Vf@ 20mA


One can use Ohm's law while including the resistance of the LED if you choose nom. values from the VI curve for the intercept, Vt and slope Rs, then you can interpolate or extrapolate and get more accurate results than using a fixed Vf.


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