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I'm trying to use de IC ULN2003 to logically drive the current of different LEDs from a logic input. I have searched for the data sheet but i don't understand how the IC works. I know that there is a Darlington pair for each Input-Output set but I don't understand what should be connected in the COM input of the IC. Can somebody explain me this?

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Take a look at this firstly: -

enter image description here

It shows the common pin connected to all the individual outputs via individual diodes. Now see this diagram - they have used common but also they've used external 1N4004 diodes: -

enter image description here

Don't ask why they did it this way - it seems to me that the internal diode should be as good as the 1N4004. Anyway the diodes catch the "back emf" from the inductance of the motor or relay or solenoid or electromagnet when the transistor is turned off.

If you are just driving LEDs or non-inductive loads you can use it with common tied to the highest voltage or leave it open.

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    \$\begingroup\$ The internal diode is "as good" as the external one only if the relay is physically close to the driver. The schematic lacks an important detail: potentially long wires between ULN2003 and the diode+relay combo. In absence of the external diode, and if the relay is far away, the flyback current will flow through the loop formed by the diode on the chip, the wires going to the relay, and the relay coil. Such a loop is an EMI concern. By placing the diode right next to the relay, the flyback current loop's area is minimized. \$\endgroup\$ – Kuba Ober Apr 24 '15 at 14:11
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COM is a COMmon rail for flyback voltages from inductive loads or for input protection, if you are driving LED's you can leave it disconnected.

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If there is any inductance on wire leads to LEDs , it can result in a dangerous reverse voltage on LED, thus the common clamp diodes are tied to V+. LED's can only handle -5V, or Vcc +5V, which can be exceeded when L di/dt exceeds this during rapid shut-off such as if L is > 1 uH and di/dt>5*10^-6. Although load capacitance can also limit this rise time, so not a big issue. Circuit cable determines this characteristic.

enter image description here

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  • \$\begingroup\$ Not true. If a LED is connected to a ULN2003 and is activated, the current flows from anode to cathode thru lead inductance. If the transistor open circuits, the action of the inductor is to maintain the same current (and flow direction) until the energy stored in the magnetic field is exhausted. This, I believe force current to go in the opposite direction thru the LED therefore it cannot become reverse biased. \$\endgroup\$ – Andy aka Apr 14 '14 at 20:23
  • \$\begingroup\$ Are you sure about that? \$\endgroup\$ – user38637 Apr 15 '14 at 14:15

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