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enter image description here

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?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Can you try and cover the LED, so that no light gets in it and see if the reading changes (in the open circuit case)? \$\endgroup\$ – Arsenal Aug 26 '15 at 13:22
  • \$\begingroup\$ Welcome to EE.SE, new user! Your question is clear, but it would be much better if you posted a better schematic. Our site has a built in schematic editor that you can bring up by pressing CTRL+M. Why don't you give it a try? Your circuit is simple enough, so I'm sure you'll be able to do it. Then your question will be much better received here. Cheers! \$\endgroup\$ – Ricardo Aug 26 '15 at 13:24
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    \$\begingroup\$ 10M ohms input resistance of meter with 0.345uA flowing thru it produces 3.45 volts on the display. \$\endgroup\$ – Andy aka Aug 26 '15 at 14:00
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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: -

enter image description here

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.

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    \$\begingroup\$ Thanks for the reply. This makes sense. I guess I need to use a second DMM to check for a voltage drop across the LED when the first DMM is connected across the open switch. \$\endgroup\$ – user84102 Aug 26 '15 at 14:29
  • \$\begingroup\$ See this document about the diode equation and how it pertains to LEDs: lumileds.com/uploads/225/AB20-3A-PDF \$\endgroup\$ – Andy aka Aug 26 '15 at 14:35
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    \$\begingroup\$ @user84102 a meter across the LED is going to shunt the LED current and make the original meter voltage a wee bit higher. Not easy to measure. \$\endgroup\$ – Andy aka Aug 26 '15 at 14:49
  • \$\begingroup\$ Wow that's a nice datasheet, typically they end somewhere in the single digit mA region. \$\endgroup\$ – Arsenal Aug 26 '15 at 15:07
  • \$\begingroup\$ @user84102: If the current injected by the first DMM and its internal resistance are sufficient to influence the measurements and make them appear "irregular" at first sight, then adding a second DMM of the same kind will only produce more "unexplainable" distortions. Your DMMs are apparently too intrusive for this circuit. \$\endgroup\$ – AnT Aug 27 '15 at 5:24
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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.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I understand what you're saying but I'm still missing something :-( If the meter is loading the circuit and causing a voltage drop, why does the load (voltage drop) not show up if the diode is removed and replaced with a wire? I get the full 5V reading across the open switch when the LED is shorted with a piece of wire. \$\endgroup\$ – user84102 Aug 26 '15 at 14:24
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    \$\begingroup\$ That is exactly what you would expect, if you replace the LED with a short then with the loading effect of the multimeter you end up with the voltage divider of 200 ohm and the multimeter impedance which results on a voltage drop of 5V*(1MOhm / (1MOhm + 220Ohm)) = 5V*.99978 or 5V. so the voltage drop across R1 is very insignificant that is negligible \$\endgroup\$ – Kvegaoro Aug 26 '15 at 14:52
  • \$\begingroup\$ Got it - Thanks!!!! I didn't realize the diode would conduct that much voltage at such low current. Makes sense now. \$\endgroup\$ – user84102 Aug 26 '15 at 15:00
  • \$\begingroup\$ Conduct current, not voltage. Really, really, really don't confuse the two! \$\endgroup\$ – Graham Aug 26 '15 at 16:48
  • \$\begingroup\$ I was always told to think of diodes (including LEDs) as basically constant voltage drops, regardless of current (again, within reason). So it makes sense it took it's 1.5 volts or so. This is why LEDs need resistors to limit current, as they are unable to do that for themselves. \$\endgroup\$ – Jiminion Sep 3 '15 at 12:15
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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).

schematic

simulate this circuit – Schematic created using CircuitLab

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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.

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  • \$\begingroup\$ So, is the 1.55V volts being dropped across the diode when the meter bridges the switch? If I remove the diode and replace with a wire, and then measure the voltage across the open switch, I get the full input voltage (5V) on the meter. \$\endgroup\$ – user84102 Aug 26 '15 at 13:55

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