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The circuit consists of following:

schematic

simulate this circuit – Schematic created using CircuitLab

And video on this link.

Summary: When there is no current flow on the circuit, supply voltage is about 16 volts, and the voltage difference between V_SUPPLY and VD is 0 volts. However, the voltage on the drain of constant current LED driver is 13 Volts. How could this be? Does the multimeter itself messes the results? And how does LEDs behave in this condition - when led driver is turned off?

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  • \$\begingroup\$ @EugeneSh. Sorry for the confusion, I should have placed a current source instead. \$\endgroup\$ – C K Aug 10 '18 at 15:45
  • \$\begingroup\$ I actually realized that you referred to Vd. Right? Vd is 13V as you say? \$\endgroup\$ – Eugene Sh. Aug 10 '18 at 15:45
  • \$\begingroup\$ @EugeneSh. exactly \$\endgroup\$ – C K Aug 10 '18 at 15:49
  • \$\begingroup\$ Why are you using a voltage source and a current source in the same loop? \$\endgroup\$ – schadjo Aug 10 '18 at 16:06
  • \$\begingroup\$ @schadjo the current source is the constant current LED driver, not an ideal current source. I need it to regulate the current that flows through LEDs \$\endgroup\$ – C K Aug 10 '18 at 16:08
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There probably is a very small current leaking through your current source or your multimeter (something in the order of microamps or nanoamps). You can model this as a very high parallel resistance. This would result in almost no voltage across R1, but a considerable voltage across the LEDs according to the diode equation.

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  • \$\begingroup\$ one question, would measuring VD and (V_SUPPLY - VD) at the same time - with different multimeters - give a consistent result? I don't have the circuit right now to try, but would like to hear your opinion \$\endgroup\$ – C K Aug 10 '18 at 16:10
  • \$\begingroup\$ The LEDs are forward biased...16V at V_SUPPLY and 13V at V_D. Can you explain what you mean by "considerable voltage" given that you said there was a "very small current"? \$\endgroup\$ – Elliot Alderson Aug 10 '18 at 16:46
  • \$\begingroup\$ @ÇetinKöktürk yes, I expect consistent results if you measure everything \$\endgroup\$ – FrancoVS Aug 10 '18 at 20:03
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I guarantee the voltage do add up. No matter how you wired the circuit it will not defy the laws of physics.

If you have another volt meter (or scope) measure the voltage between VD and GND while measuring the voltage across the LEDs. See theory below.

At 23 mA the voltage across each LED will be about 2.0V.

enter image description here
Source: LTL-307EE Datasheet


The voltage across the 10Ω resistor will be 0.23V, totaling 4.23V from V_SUPPLY to VD. This would leave 11.77V across the input and output (VD to GND) of the TLC5925.

When the TLC5925 is not enabled there is a leakage current of about 1 µA.

enter image description here Source: TLC5925 Datasheet


The minimum Vf of each LED is about 1.5V.
The voltage across the 10Ω resistor will be about 0.00001 V.

This would leave 13V at VD.


If you plan to use more than 4 outputs of the TLC5925 you will need to increase the value of the resistor. With your current circuit you are dropping 0.27 watts on the one output.

At 23 mA a 400 Ω resistor will drop almost 10V. This will not have any effect on the current. The current will continue to be set by the TLC5925. This will lower the voltage across the TLC5925, and alleviate thermal stress on the TLC5925 when (if) more outputs are used.


And how does LEDs behave in this condition - when led driver is turned off?

The LEDs only need the minimum Vf to turn on. LEDs do not have a minimum current. Vf will follow the the datasheet's IV curve. Some LED datasheets (e.g. OSRAM) will show the Vf at very low (µA) currents.


Possible theory.

The Rdynamic of the two LEDs is about 3 MΩ at 1 µA.

When you put the meter across the LEDs some current is going to flow through the meter. This will reduce the current flowing through the LEDs which will increase the LEDs Rdynamic. When the Rdynamic increases more of the current will flow through the meter. This repeats until all current is flowing through the meter and none through the LEDs. Similar to thermal runaway.

The voltage is not likely still 3.0V but the meter may not be reading the voltage due to the current flowing through the meter rather than the LEDs. The voltage may be anywhere between 16V and 0V. If you have another volt meter measure the voltage between VD and GND while measuring the voltage across the LEDs.

Instead of measuring directly across the LEDs try measuring between the LEDs to GND. The voltage between the LEDs to GND should read 14.5V.

This is only a theory. You would have to run some empirical experiments to verify.

If you do not have another meter, try adding a 10 MΩ resistor in series with the meter's lead. What happens to the voltages?

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  • \$\begingroup\$ But why do İ measure 0V between supply voltage and current driver? Shouldn't it be about 3 Volts? \$\endgroup\$ – C K Aug 12 '18 at 10:24
  • \$\begingroup\$ If V_SUPPLY is 16V and the voltage between VD and GND is 13V, the voltage between V_SUPPLY and VD must be 3V. No exceptions. Why you read 0V is beyond the scope of my answer. You made a mistake. \$\endgroup\$ – Misunderstood Aug 12 '18 at 10:33
  • \$\begingroup\$ check the video on question \$\endgroup\$ – C K Aug 12 '18 at 10:39
  • \$\begingroup\$ Sorry at first look I did not get past the first minute without getting dizzy. What is the input impedance of your volt meter? At 1 µA Rdynamic of the two LEDs is about 3 MΩ. Replace the LEDs with a 3 MΩ resistor. Can you measure the tiny voltage across the 10 Ω resistor? Try increasing the resistor to a few hundred Ω. At 23 mA a 400 Ω resistor will drop almost 10V. This will not have any effect on the current, it will lower the voltage across, and alleviate the stress on the TLC5925. You then may be able to measure the voltage across the resistor to calculate the current flowing. \$\endgroup\$ – Misunderstood Aug 12 '18 at 11:30
  • \$\begingroup\$ About the theory; yes I thought of that too. However I think the resistance stabilizes at some point, if internal res. of multimeter is 10Mohms(source: fluke datasheet), it will draw 0.3 microamps, and 0.7 microamps will be left for LEDs, equating to (1.5+1.5)/(1- 0.3 uA) = 4.28 megaohms. However I also think multimeter may be affecting the measurements. I will update soon when I measure with 2 multimeters. Thank you for your effort \$\endgroup\$ – C K Aug 12 '18 at 12:32

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