Hoping someone can help me to understand what I am seeing.

I have the following:

  1. A 5V DC power supply derived from a USB power supply

  2. A LED (LINK) with a Forward Voltage requirement of 3.2V (Maximum Forward Voltage of 3.5V DC), and Continuous Forward Current of 20mA (Peak Forward Current of 100mA).

  3. A 91 ohm 1/2 watt flameproof resistor

I used the popular LED Calculator to design this ultra simple circuit.

However, before connecting the LED I figured I'd use my multimeter to confirm that the resistor was properly dropping the voltage to 3.2V.

I tested this by placing the resistor first at the +5V side, and then at the GND side, of the appropriate multimeter leads -- but I am still seeing 5V on the meter.

I then place two of the same resistors in series, at both polarities, and I am still seeing the 5V on the meter.

Is there a simple reason I am still seeing 5V? Even without a load, I would have expected to see a voltage in the realm of 3.2V.

Thanks in advance.

  • 4
    \$\begingroup\$ The resistor only drops voltage when there is current flowing through it, without the LED connected you have no current flow. \$\endgroup\$ – Jack Creasey Dec 23 '18 at 22:57
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    \$\begingroup\$ Naughty Jack shouldn't have posted an answer as a comment. Now he can't win any points! \$\endgroup\$ – Transistor Dec 23 '18 at 23:28
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    \$\begingroup\$ I'm not that concerned about points, a one line answer would seem somewhat over the top. I think the idea here is help people, not seek reputation ...there are far too many of those folks already. \$\endgroup\$ – Jack Creasey Dec 23 '18 at 23:42
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    \$\begingroup\$ @blafarm - Although this wasn't your question, looking at that linked page for the LED, the Continuous Forward Current rating of 20mA is listed in multiple places, including under the heading of "Absolute Maximum Ratings". If that part of the listed specification is true, then running the LED continuously at 20mA can be expected to reduce its lifespan (that's part of the typical meaning of "Absolute Maximum Rating"). Therefore personally I would not run it at 20mA, as per your plan, unless maximum brightness is required, even though it may cause premature LED failure. \$\endgroup\$ – SamGibson Dec 23 '18 at 23:48
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    \$\begingroup\$ @Blafarm, SamGibson's explanation is not a perspective, but the correct interpretation of the table. These are the absolute maximum ratings for the LED when the ambient temperature is 25 degrees C. Running an LED at maximum rating will indeed reduce it's operating life considerably. That said, this is one of your first projects, it is simple and cheap and need not be perfect. The lower the power level you run LEDs at and the better you cool them, the more efficient they will be and the longer they will last. \$\endgroup\$ – K H Dec 29 '18 at 22:25

By putting the resistor in series with the multimeter leads, you will always see a voltage of 5V since the multimeter (with effectively infinite resistance) will act as a potential divider with the source, and compared to the 91 ohms, your meter will take up all of the voltage. This happens no matter which way you connect the meter. Whats worse is that the LED isn't even connected to begin with. The voltage dropping happens when the circuit is fully connected, and for a voltage to actually drop, a current has to flow. Connect your circuit so that it turns on, then connect your voltmeter in parallel to each source. You will observe that the voltage across the LED is 3.2V, and across the resistor is the remaining 1.8V (This is due to KVL but i wont get into that).


simulate this circuit – Schematic created using CircuitLab


When you connect the resistor with out load (here led)... the circuit is not closed (this is called OPEN CIRCUIT).

In open circuit ---

voltage = the maximum voltage applied.

lets say if you apply 10V, you will see 10 like that

if you apply n volts... you will see n volts

Current = 0;

SO, that is the reason you are seeing 5V.


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