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Question about ULN2003A

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Newbie level 5
Aug 11, 2009
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how does the ULN2003A works

and how it used in stepper motor????:|

uln2003 example

The ULN2003 and 2802 family of ICs are just lots of power transistors in one package. They are not specially for stepper motors, you can use them wherever you need to switch current on and off.

The reason you see them used frequently to drive stepper motors is that they have 8 or more outputs (8 for the 2003, 9 for the 2803) and as steppers need several current feeds, having them all in one device makes the circuit easier to build.

The only difference between each channel of a 2003 and using a single darlington transistor is that each of the output pins has a diode connected to it. The diodes are normally not conducting and are all combined to one pin. They are to 'absorb' the spikes you get from switching inductive loads so the IC and other connected equipment is protected. You don't have to use them, leave the pin open if they are not needed.


uln2003a works

thanks for this information

but i like to give me details about the COM port
where did you connect it???

on ground ar VCC?
plz give me details

and thank you again

uln2003 comport

connect to VCC after diode katode first.

Added after 12 minutes:

sorry, I mean here :
-COM pin---to--->Anode of Diode
-catode of same diode ---to--->VCC

uln2003 voltage drop

DO NOT connect it to VCC unless the load also is powered from the same VCC line !!!

Normally it is connected to the same supply that provides power to the load. For example if a motor ran from 12V, connect it to the 12V line. This applies even if you are driving the ULN from a 5V powered circuit.

Imagine running a 12V motor with the common pin connected to 5V, the diode would conduct the 12V line through the load into the 5V line and likely cause damage.



hello again

on protus program we conect the com pin to vcc and its work

but in actcual we connect it to ground

plz give me some details if you know

and thanks

uln2003 seems not to work

Pin 8 is ground, it is the return path for the current being switched by all the driver stages. It is actually the emitter connection of all 7 output transistors.

Pin 9 is the "com" or "common free wheeling diodes" pin. If you connect it to ground the IC will not work, the diodes will conduct the load current to ground all the time so the driver will always be 'on'.

The idea of the diodes is that they are normally not conducting so they must be connected to a voltage equal to or higher than the load being driven. They are present because in many applications, the 2003 is used to drive inductive loads (relays or motors for example) which can produce a reverse voltage across their coils as they are switched off. The diodes safely clamp the voltage to 0.6V (diode forward voltage drop) so the 2003 and possibly other things connected to it are protected. The manufacturer of the 2003 had a pin left over on the package so they wisely used it for diode clamps to make the device more versatile.

If you look at the data sheet for the ULN2003 it shows the way the diodes are connected inside the package. It may give a clue as to why grounding the com pin makes no sense.



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