All through my electronic career I have used OHM's law to calculate resistor values for driving a LED, 250 ohm to give 20mA at 5v etc, I have never measured the current supplied to an LED, until now, someone has asked me to produce a light consisting of 2 rows of 6 LED's. ,I built a test bed using a breadboard and the current didn't seem right so I just plugged 1 led into the board with a 5v power supply and a 470 ohm resistor near to but not connected to the led, put my multimeter (on current) across the gap between the resistor and the LED and the reading is 7.2 mAmp, it should be around 12 surely, the variable power supply giving me 5v says it is supplying 7.2 mAmps so the multimeter looks ok.
Can you see where I am going wrong?

 

You need to know how many volts the LED works at. A LED is much like a Zener diode. LED work at 1.5 to 3.6V depending on the color and temperature. Red is low voltage while White is close to 3V. Some time there are more than one LED in a package.
If you have a 1.5 Red LED and a 5V supply, then there will be 3.5V across the resistor. Do your formula for 3.5V 10mA is what resistance.

 

All through my electronic career I have used OHM's law to calculate resistor values for driving a LED, 250 ohm to give 20mA at 5v etc, I have never measured the current supplied to an LED, until now, someone has asked me to produce a light consisting of 2 rows of 6 LED's. ,I built a test bed using a breadboard and the current didn't seem right so I just plugged 1 led into the board with a 5v power supply and a 470 ohm resistor near to but not connected to the led, put my multimeter (on current) across the gap between the resistor and the LED and the reading is 7.2 mAmp, it should be around 12 surely, the variable power supply giving me 5v says it is supplying 7.2 mAmps so the multimeter looks ok.
Can you see where I am going wrong?

Why would you expect it to be around 12 mA? Even if you put the 470 Ω resistor directly across the 5 V supply, you would only expect about 10.6 mA of current to flow.

An indication of where your thinking is going wrong is that you think that a 250 Ω resistor in series with a 5 V supply will yield 20 mA. That would only be the case if the LED didn't drop ANY voltage.

Your problem is that you don't understand Ohm's Law. It relates the value of a resistor to the voltage across THAT resistor and the current through THAT resistor. You are assuming that it somehow relates the current through the resistor to the voltage across the supply. The resistor doesn't know what the voltage across the supply is, only the voltage across itself.

Let's work in backwards from your measurements.

You have a 5 V supply and 7.2 mA through a 470 Ω resistor. There's also some drop across the meter, but we'll ignore that (at least for now). That means that the resistor is dropping ~3.4 V, leaving ~1.6 V across the LED. That seems quite reasonable, especially for a red or infrared LED.

A common rule of thumb is that blue/white LEDs will drop around 3.5 V and all other colors will drop around 2 V. This is a back-of-the-envelope rule of thumb for quick estimates and results should always be reworked based on the specs of the actual LEDs that will be used.

Another thing to watch out for as you try to create this array of LEDs. It is not a good idea to put LEDs in parallel -- they don't tend to share well or play nice with each other. Each LED should have a resistor in series with it. You can put multiple LEDs in series with a single series resistor servicing them, since that resistor is in series with each of the respective LEDs.

What current are you wanting to get through the LEDs? If you want 12 mA, then one way to get close to that is to arrange your LEDs in three strings consisting of one resistor and two LEDs in each string. Then put the strings in parallel.

If your LEDs all drop around 1.6 V, then two LEDs will drop around 3.2 V, leaving a drop of around 1.8 V across the resistor. At 12 mA, that would need a 150 Ω resistor, which is a standard resistor value. The power dissipated in each resistor would be around 22 mW, so 1/8 W resistors will be more than adequate. If you are using SMD resistors, I wouldn't use resistors rated at less than 1/20 W (the common rule of thumb on power ratings is to double the expected actual dissipation at least).

 

There's also a small voltage drop when inserting a DVM into the circuit.

 

Can you see where I am going wrong?

For one thing, using your DMM to measure current. Depending on the range, it can perturb the circuit sufficiently to give erroneous readings. You'd be better off measuring the voltage drop on the current limiting resistor and calculating current.

 

There's also a small voltage drop when inserting a DVM into the circuit.

For one thing, using your DMM to measure current. Depending on the range, it can perturb the circuit sufficiently to give erroneous readings. You'd be better off measuring the voltage drop on the current limiting resistor and calculating current.

While I agree, the burden resistance of even most cheap digital multimeters (which is likely what the TS is using) are typically an order of magnitude less than the 470 Ω resistor they are using. They are typically in the 20 Ω vicinity on the ranges used for 10 mA scale currents. So while this is a perturbing factor, it is not their main issue. The main issue is ignoring the 1.6 V drop across the LED, not the <0.2 V drop across the meter.

 

Hello,

Your measurement actually looks pretty normal. The resistor isn’t dropping the full 5 V because the LED itself has a forward voltage (usually somewhere around 2–3 V depending on the type).

So the current should be calculated using (Supply voltage − LED forward voltage) / resistor value. For example, if the LED drop is ~2 V:

(5 V − 2 V) / 470 Ω ≈ 6.4 mA

That’s very close to the ~7 mA you’re seeing. The exact value can vary a bit depending on the LED and supply accuracy.

So it doesn’t look like anything is wrong with your measurement — it’s just the LED’s forward voltage affecting the calculation.

 

LEDs, 555s, Flashers, and Light Chasers

 

if you want to use Ohms law, learn to use it correctly. relationship In=Vn/Rn applies but... all three terms are for the one and same component.

I_r=V_r/R_r
I_led=V_led/R_led
I_r37 = V_r37/R_r37
etc.

only then this makes any sense.

the problem is you get too many variables and to solve it you need to know how they relate to each other.

the answer is that voltage is same for components in parallel (and their currents add up). current is same for components in series (and their voltages add up). lookup Kirchhoff's laws...

since your LED and resistor are in series (hopefully) their currents are the same (I_r=I_led) and their voltages add up (V_r+V_led=5V).
V_led can be found in the datasheet. and THEN you can calculate the rest.