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This question is related to my another question here.

In crux, I would like to sink current of TLP521 via 74HC595n shift register. The TLP521 is connected with a resistance such that the current is limited to 16mA.

My question is that if the current is limited to 16mA and the TLP521 "led" consumes it, then how much is left over for the 74HC595N to sink?

I understand that if "led" was not in place (consider a short), only then the 74HC595N would see a total 16mA current and would struggle to sink it ( as it can sink only 6mA current). But in this case, the LED would "eat" (sorry but this is my analogy of current) the current any a tiny amount might be left over for 74hc595 to sink.

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  • \$\begingroup\$ The HC595 must sink all the current that goes through the LED (and all the current through all of the LEDs that are on passes through its GND pin). \$\endgroup\$
    – Spehro Pefhany
    Apr 22, 2015 at 4:35
  • \$\begingroup\$ How can HC595 drive 8 leds at once (one red led needs about 16-18 mA current)? \$\endgroup\$
    – Anuj Purohit
    Apr 22, 2015 at 4:41
  • \$\begingroup\$ @AnujPurohit - it can't drive 8 LEDs at once - supply current thru the device is limited to 70mA and at 16mA per LED this means 4 LEDs can be driven simultaneously. Read the data sheet for more information. \$\endgroup\$
    – Andy aka
    Apr 22, 2015 at 7:20
  • \$\begingroup\$ @Andyaka I agree and thought the same but cant understand that how this works : protostack.com/blog/2010/05/… [Maybe because 550 ohm would limit total current at 72 mA] \$\endgroup\$
    – Anuj Purohit
    Apr 22, 2015 at 7:59
  • \$\begingroup\$ It looks like 510 ohms to me and with 2 volts (typically across the LED) and a 5V supply, the 510R resistor would see about 3V implying a current for each LED of 5.9mA. 5.9mA x 8 = 47mA and this should not be a problem. \$\endgroup\$
    – Andy aka
    Apr 22, 2015 at 9:15

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See this graph: -

enter image description here

It describes the forward current of a typical red LED as you apply a forward voltage.

With 1.7 volts applied across its terminals, virtually no current is taken i.e. the LED looks like an open circuit. With 1.9mA the LED takes 5mA i.e. looks like 380 ohm resistor. At 2V, the LED takes 20mA and looks like 100 ohms.

So, let's say you decide on an LED current of 5mA, from a 5V supply you need to "lose" 3.1 volts and this means a series resistance of 3.1V/5mA = 620 ohms.

If you applied a more simplistic rule of thumb and assumed the LED forward volt drop (for a wide range of currents) was 2V you would need to drop 3V across the resistor and if you wanted 5.9mA, you'd choose a resistor value of 3V/5.9mA = 508 ohms. The article you linked picked 510 ohms as the current limiting resistor and this is perfectly fine.

Assuming the LED forward volt drop is 2V is a compromise on accuracy but it is a rule of thumb and not too inaccurate. Given that a typical red LED may vary its characteristics a little bit, there is little point trying to predict a resistor value more accurately.

For different LEDs, different characteristics need to be chosen to calculate the forward volt drop: -

enter image description here

The bottom-line is always read the data sheet and don't use LEDs that don't explicitly state what their forward characteristic is like.

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  • \$\begingroup\$ Thanks for the details. This is very good explanation. Its hard to find "LEDs" with datasheet in India; thus the rule of thumb sounds fine. \$\endgroup\$
    – Anuj Purohit
    Apr 22, 2015 at 9:38

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