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I have the following schematic:

enter image description here

Now I want to know the (more or less) exact current flow in my circuit. For this I did the following calculations:

I = V / R

I = 5V / 330 Ohm

I = 0.015A = 15mA

So with what I now know I expect approximately 15mA to flow across all the wires. However, when I try to measure the current flow by placing my multimeter (in series) between the LED and the resistor I measure I = 9.4mA.

This difference of 5.6mA between the calculated 15mA and the measured 9.4mA seems too large to be due to manufactering deficiencies of the parts of the circuit. Therefore I assume I am making some error in my assumptions/logic, but I cannot figure out what. Can someone help me with this?

Thanks in advance.

edit

As per comments/answers I measured the voltage over the LED and the resistor:
- Resistor: 3.18V
- LED: 1.94V
Also the resistor's actual resistance:
- 325 ohm

Now the calculation indeed does make sense: 3.18V/325 Ohm = 9.8mA Which is close enough to the measured 9.4mA.

Thanks everyone for helping out!

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    \$\begingroup\$ remove the LED from your circuit, then the equations will be more accurate \$\endgroup\$
    – jsotola
    Commented Sep 1, 2019 at 20:00
  • \$\begingroup\$ If you are going to be quite precise about currents in the mA range, you should measure the resistor value and the supply voltage for their exact values before doing calculations. It is easy to place the wrong resistor in without checking. \$\endgroup\$
    – David777
    Commented Sep 1, 2019 at 20:01
  • \$\begingroup\$ Your calculation ignores the fact that there is a LED in the circuit which usually contains a forward voltage drop of around 0.7V. A better way to calculate this would be: (5-0.7) / 330 = 13mA \$\endgroup\$
    – David777
    Commented Sep 1, 2019 at 20:03
  • \$\begingroup\$ @David777 Actually 0.7V is the forward voltage for a regular signal diode; LEDs have a larger forward voltage, at least 1.5V for visible light LEDs. See answer from Marko Buršič below for details. \$\endgroup\$
    – Mr. Snrub
    Commented Sep 1, 2019 at 20:39
  • \$\begingroup\$ @Mr.Snrub Oh of course, thanks for correcting that. \$\endgroup\$
    – David777
    Commented Sep 1, 2019 at 22:12

2 Answers 2

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$$I=\dfrac{V_{CC}-V_{F}}{R} = \dfrac{5V-1.8V}{330\Omega} = 9.7 mA$$ As example the LED forward voltage drop is 1.8V.

An easier approach is to calculate the required resistance:

$$R=\dfrac{V_{CC}-V_{F}}{I_F}$$ For example: I have orange LED and I want 10 mA current through: If=10ma -> Vf=1.7V enter image description here

Image source

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You just forgot that there is a voltage drop over the LED too. Ohm's law says that in your case there is about 3.1V over the resistor when 9.4mA flows. Which means there is about 1.9V over the LED and so that adds up to 5V.

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    \$\begingroup\$ "Ohm's law says that in your case there is about 1.9V over the LED" ... Although I know you know better this sentence could be intepret as if it implies you can apply Ohm's law to LED's. This answer is skipping steps, so newbies on electronics fail to follow. \$\endgroup\$
    – Huisman
    Commented Sep 1, 2019 at 20:59

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