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I accidentally bought 12 V LEDs, so an LED with a built-in resistor (you can see it as a black dot on the anode). Now I measured the voltage and get after the LED still a voltage of 11.3 V, when the LED is supplied with 12.5 V from a power supply.

I find this strange, because usually the resistor will drop the voltage so far that the LED gets its 1.7 - 2.0 V. So after the LED should be only 1.7 V or am I wrong?

Where is my thinking error or are the LEDs with series resistor internally connected in parallel?

Shouldn't that be the same circuits, just one had the resistor inside the LED and the other one has the resistor before the LED?

Edit: Of course I'm measuring this way, I'm sorry

Enter image description here

Enter image description here

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  • \$\begingroup\$ Your question is unclear. When you say, "Now I measured the voltage and get after the LED still a voltage of 11.3V" what EXACTLY does that mean? Please include a schematic drawing. \$\endgroup\$
    – jwh20
    Apr 11 at 10:33
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    \$\begingroup\$ The photo seems to have the voltage meter in series with LED. As extremely small current flows, there is an extremely low voltage drop over LED component and therefore almost all of the drop will be over the multimeter. \$\endgroup\$
    – Justme
    Apr 11 at 12:14
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    \$\begingroup\$ Looking at the circuit you're measuring from negative to negative. That should yield a 0V reading. \$\endgroup\$ Apr 11 at 18:46
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    \$\begingroup\$ I agree with @ThomasWeller if you really measure what you have drawn you have always 0V (since your multimeter is basically measuring the same point). Probably you are measuring something else. \$\endgroup\$
    – lalala
    Apr 11 at 19:07
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    \$\begingroup\$ This schematic can't be correct. If you did have things wired up like that, you'd be measuring 0 volts no matter what. \$\endgroup\$
    – Hearth
    Apr 11 at 20:05
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Your battery, LED and voltmeter are all in parallel.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. You're not measuring the voltage "after" (or "before") the LED. You are measuring the voltage across the LED-resistor combination.

In this circuit it is the same as measuring the supply voltage.

To measure the LED voltage you would need to do this:

schematic

simulate this circuit

Figure 2. To measure the LED voltage in your 12 V LED would require some delicate surgery.

Now I measured the voltage and get after the LED still a voltage of 11.3V, when the LED is supplied with 12.5V from a power supply.

That means that your 12 V supply is drooping. To confirm this connect your voltmeter to the battery and monitor while you switch the LED on and off.

schematic

simulate this circuit

Figure 3. Measuring the voltage in series with the LED gives no useful information.

In this situation the voltmeter presents a series resistance of (typically) 10 MΩ in series with the 1.2 kΩ of the LED. Since the meter's resistance is 10,000 times that of the LED resistor almost all the voltage is dropped across it rather than the LED.

If you wish to do something useful then switch your meter to mA and use Figure 3 to measure the current through the LED. Remember to switch back to V when finished. If you forget and connect it up as shown in Figure 1 you will pass a high current through the meter and blow the fuse if it has one and blow the meter if it hasn't.

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  • \$\begingroup\$ I'm sorry, of course your right and I'm measuring with your method... \$\endgroup\$
    – 0x30
    Apr 11 at 11:20
  • \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Apr 11 at 11:50
  • \$\begingroup\$ I understand now. So I can measure the amperage and then calculate with this. \$\endgroup\$
    – 0x30
    Apr 11 at 12:01
  • \$\begingroup\$ Yes. You'll then know the current, you will have to estimate the LED voltage drop (typically 2 to 2.2 V for green) and then you can work out the internal resistor value. You might find what I've written on LEDnique.com useful. \$\endgroup\$
    – Transistor
    Apr 11 at 12:06
  • \$\begingroup\$ Thank you. But why is the voltmeter showing this 11.3V if it's connected in series (e.g. my picture in first post from my bench). Should it not show like 0V because of the internal resistor? \$\endgroup\$
    – 0x30
    Apr 11 at 12:17
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On picture voltmeter connected in series with LED and internal resistance. Voltmeter input resistance is high, usually around 1 Mohm. So you restricted LED current and diode drop voltage smaller then normal. Voltmeter shows drop voltage on itself. So the rest voltage 12.6-11.24=1.34V is drop voltage on diode and internal resistor.

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