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I have been messing around with TLC5917 LED Constant Current Sink Drivers and I don't fully understand how they work.

Because of how I was taught how Voltage/Current works, to my understanding, a voltage of 5V should be able to be measured between an active pin on the driver and a 5V source going through an LED.

What I measure however appears the be the forward voltage of the LED. It is my understanding that current is drawn and voltage is "applied".

Since the driver is constant current, the behavior makes sense but how exactly does the driver supply the correct voltage?

Or is my conceptual understanding of how voltage/current work misguided?

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3 Answers 3

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If you look on the datasheet you'll see that the external voltage applied to the LEDs is on the anode of the LED. The cathode of the LED is connected to the pin of the TLC device.

enter image description here

In this case the TLC is sinking current. To reiterate, the driver doesn't supply the voltage, the voltage is applied externally. The driver does, however, ensure the correct current flows through the LEDs.

It should also be noted that the driver takes up any difference between supply voltage and what is dropped across the LED. So you have to take care that you don't have a combination of external voltage and current drawn that would damage the chip. The higher VLED the higher the power dissipated by the the TLC.

The image below gives you a very basic internal schematic of the IC.

enter image description here

Remember that the LEDs' cathodes are connected to OUT0, OUT1 etc, with the anode connected to the external voltage. The little circles with the down pointing arrows are representing constant current sink. It is this that ensures the current drawn by the LEDs is the same and constant.

It might be that you're not familiar with the difference between a current source and a current sink, so it might be worth doing a little reading up of the difference between them if that is the case.

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  • \$\begingroup\$ Thankyou for this thorough explanation. So the voltage is being applied externally and the driver provides a current sink creating a potential difference. The TLC presumably uses some resistance mechanism to limit current and any residual voltage is absorbed by the chip. So if I have 5V going through an LED, the LED drops 3V, that means 2V is somehow disipated through the TLC Chip? \$\endgroup\$
    – Zane Carter
    Jun 12, 2018 at 14:19
  • \$\begingroup\$ That's pretty much it. So, you can see immediately that if the LED drops 3V and you apply 6V, then the driver has 3V to take up. As P = VI there's an immediate increase in power dissipated. The datasheet will tell you maximum values for each pin and the chip as a whole. It'll be a transistor of some flavour inside that is taking up this "slack". en.wikipedia.org/wiki/Current_source shows some nice, simple examples. \$\endgroup\$
    – DiBosco
    Jun 12, 2018 at 14:24
  • \$\begingroup\$ So if this TLC Chip takes up all this slack does this negate the need for current limiting resistor completely as current is already controlled and voltage is handled for you? These chips are awesome! \$\endgroup\$
    – Zane Carter
    Jun 12, 2018 at 14:29
  • \$\begingroup\$ Well, Rext, sets the output current for them all. You can tweak each output, the datasheet goes into that in more depth. When I've used them in my applications I've never needed to do that. Some of these types of devices also allow you to have PWM control on each individual output, so you can set the brightness of them all separately in that case. They are, indeed, awesome! :D \$\endgroup\$
    – DiBosco
    Jun 12, 2018 at 14:43
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The clue is in the name of the device. It is a current sink which means instead of sourcing the current, it provides a place to sink current. This means that the LED cathode must be connected to the device pin, not the anode. Standard LED current limiting resistors are NOT required. The voltage must be supplied from the power rail, the device is basically providing a ground.

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  • \$\begingroup\$ LED drivers such as the TLC5917 do not require external series resistors, that's the whole clue about the current regulation. \$\endgroup\$
    – po.pe
    Jun 12, 2018 at 14:13
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The question in this case is more how does a current source or in this case a current sink work rather than how does the TLC5917 work. As you noticed, when connected to the driver chip, you measure only the LED's forward voltage between the source and the driver pin.

What the driver does is, it puts the voltage level of the sink pin high enough so only the specified current will flow through the LED - by regulating its forward voltage. enter image description here

Usually it works the opposite way that you have a current source instead of a sink and it generates an output voltage high enough to have the desired output current.

It may helps to understand if you take it as a variable internal series-resistor to the LED. If you configure it to draw more current, the resistor will become smaller and if you configure it to draw less current the resistor will become larger.

schematic

simulate this circuit – Schematic created using CircuitLab

Notice, this is just an explanation why you're not measuring V_LED between the LED's anode and the input pin of the TLC5917, it's not how the chip actually works.

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