ULN2003A maximum input voltage/current

[Nogtail]

I have a product that uses a ULN2003A to drive a LED matrix. I am not sure of the input voltage, however I can assume it is larger than 7.5 volts due to the dropout voltage of a regulator on the board. Most of the components are rated for at least 15V however the ULN2003A is a 5V part which seems a strange choice. I expect the input for the board to be 9-12V and it appears to run without problems on 12V.

What is the maximum input voltage of the ULN2003A? It has no resistors on its inputs and is being driven by a shift register that can supply a maximum of 10mA per output. Due to the 2.7K inbuilt base resistor it should draw around 4.5mA at 12V which seems a bit high.

Sorry in advance about the noobiness of this question, the answer is probably staring me in the face but I cannot find it for the life of me.

[flynwill]

Only takes a few seconds to google and get the TI datasheet for the part.

The ULN2003A was (perhaps still is) a very popular part for driving higher voltage loads from logic.

In any case to answer your question:  The inputs are rated for up to 30V and the outputs are rated for up to 50V.

[KerryW]

You could also consider using ULN2004 instead.

[Nogtail]

You could also consider using ULN2004 instead.

It is a second hand product that I was trying to get working - not something I had made. I just thought it seemed a strange choice when, as you say, there are better parts suited to the task.

[mariush]

Well, they used it because it's a simple IC, just 7 darlington transistors , with resistors on the bases of each transistor (on the UL chips ending in A). It saves PCB space as you don't need to use additional resistors.

There's no 5v, the transistors inside can work with up to 30v or something like that, check the datasheet to be sure.

There's several versions in the UL2* series, the differences being in the maximum voltages the individual transistors can handle and the value of the base resistors (and if there are base resistors built in or not) ...  depending on the base resistors built in, some UL2* series chips will work with 5v signals on the base pins, some would need at least 8-10v or something like that.

There's also UL2803a series which has 8 darlington npn transistors .. and there's also chips which have pnp transistors but don't know the names.

[danadak]

The part input current should ~= [ Vin - (Vz + 2 Vbe) ] / 10.5k =~ .34 mA.
Room temp kind of calculation.

See attached schematic of part.

Note the Rs in the part may have very large variation. Also the "typical" input graph
does not correlate well to the R values shown on the schematic. Also would venture
a guess Vz part to part varies a lot.


Regards, Dana.