Voltage drop across LED in open circuit

Question

I've searched the forums but can't find an answer for my issue. I've got a simple LED circuit powered by 5V DC. When the switch is closed, I've got 2.8V across the resistor, 2.2V across the LED, and 0V across the switch (as expected).

However, when the switch is open, I get 3.45V across the switch but no other noticeable readings (0V across resistor and 0V across the LED). Where did the other 1.55V go? If I shunt across the LED, I get the expected 5V across the open switch connections. How is there a voltage drop across the LED in an open circuit and why can't it be measured with a DMM?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

It comes down to the LED still conducting current when it is not illuminating. Take a look at this characteristic from an OMRON LED. It's very rare to see stuff like this in data sheets: -

This shows the volt drop (about 1.3 volts) across an LED when 1 uA is flowing and it's not unreasonable to expect all LEDs to behave within a ball-park or two of this. So you are seeing a volt drop across the LED due to the current through the meter's inputs when placed across the open switch. If the meter's input resistance is 10 Mohm then there is about 0.345 uA flowing and there will be probably over a volt dropped across the LED should it be the device in the picture above.

Answer 2

It has to do with the internal resistance of your multimeter. The high impedance on the multimeter input is basically creating a path to ground when you measure the open switch. So therefore when you make your measurement, it is the equivalent of adding a high value resistance( in the order of 1 to tenths of megaohms depending on your meter ) in parallel across from where you are taking the measurement in this case the open switch.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

Here is a circuit with a generic blue (or white) LED SPICE model inserted for the LED and 10M to represent your DMM.

.model QTLP690C D(Is=1e-22 Rs=6 N=1.5 Cjo=50p Iave=160m Vpk=5 mfg=Fairchild type=LED)

The simulated voltage drop across D1 is similar to what you measured at ~1.4V (with an LED of a longer wavelength such as green, yellow or red it would typically be much less than ~1.5V with 350nA flowing).

^{simulate this circuit – Schematic created using CircuitLab}

Answer 4

You could assume the multimeter is a resistor with very big resistance (several Mega Ohm).

When you measure the switch voltage, although the switch opens, you make a closed loop circuit, so you can read the voltage on the multimeter itself (3.45V).

When you measure the resistor voltage or LED voltage and the switch opens, the circuit is open. There is no a close loop circuit, so there is no current and no voltage across the multimeter.