Ok, here's the schematic I'm trying to use to test if my uln2003a chip is working. While checking the pinout with my multimeter, I couldn't find that pin out1 was connected in some ways to the com pin. When I short the input with 1.5v the led doesn't light up and the led never light up.

The led works when connected to the battery directly.

Is there a better way to test if my uln2003a chip actually works. I'm worried I might have damaged it while soldering it (it's a SMD part). Looking at the schematic if I check for resistance between any of the output and com, there should be no or little resistance but in my case my multimeter always shows 1 (infinite).


simulate this circuit – Schematic created using CircuitLab


Ok since leds are a bit worrying I decided to move a bit further on this without leds.

I uploaded a stepper program on an arduino board. I connected the pinout to a digital analyser and can confirm that the arduino program is fine.

I connected my inputs to the uln2003 and checked with my analyzer the output. The uln2003 is powered with a 5v power supply.

When I check the output pin with the digital analyzer, I can see that I have spikes of less than 1ms from times to times but not similar to the input.

Now things get weird when I connect only one input, then I can see the 10ms up and down of the stepper motor on all four output but only 1 input is connected.

When I measure the resistance between all input pins, I can sense around 18kohm. That said, the digital analyser will raise up even if it sense really little amount of energy passing through.

Then I tried it without actually pluggin the power supply and if only one input is connected I can still see some output on all four connected output.

Then I configured the program on the arduino to make really long step so I could measure the voltage on output pins when they're up. But it looks like it's sinking only a few mv.

I measured the resistance between inputs on a chip I didn't solder yet and can still sense around 18k ohm between inputs.

  • 3
    \$\begingroup\$ 1. You need more than 1.5 V applied to the input to ensure the output turns "on" (pulls low). 2. You need to limit current through your LED if you don't want to blow it up. 3. You probably need more than 1.5 V applied to the LED anode to be sure the LED will light up when the '2003 output goes low. \$\endgroup\$
    – The Photon
    Commented Jun 7, 2017 at 20:35
  • \$\begingroup\$ @ThePhoton I was looking at the datasheet and they don't seem to give a minimum for the input voltage. But there seems to be a maximum of 3v (I have a Texas Instrument one if that matters). For the led, I'm not worried too much if it burns. I have a load of IR led the only way to see if they're on is either through a camera or if I barely fry them it'll glow red. \$\endgroup\$ Commented Jun 7, 2017 at 20:43
  • \$\begingroup\$ That said, if I increase the source to around 2.5v I'll need the resistor or it'll burn faster than I can see. \$\endgroup\$ Commented Jun 7, 2017 at 20:44
  • 1
    \$\begingroup\$ Look at figure 20 in the TI datasheet. You need at least two forward Vbe drops (~1.4 V, plus a little bit to get some current through the 2.7 kOhm base resistor. Given the high gain, 1.5 will probably turn it noticeably on, but performance will be highly variable depending on device temperature. \$\endgroup\$
    – The Photon
    Commented Jun 7, 2017 at 22:26
  • 1
    \$\begingroup\$ On the LED side, you should always use a series resistor when powering with a constant voltage source. Never count on the forward voltage being whatever it says in the datasheet. The actual forward voltage will change depending on temperature. Also look at the Vce(sat) values in sec. 6.9 of the datasheet. With 1 V or more Vce(sat) you're never applying more than 0.5 across the LED --- not enough to turn it on. \$\endgroup\$
    – The Photon
    Commented Jun 7, 2017 at 22:29

2 Answers 2


You are misinterpreting the datasheet. It is guaranteed to be able to sink 300mA when 3.0V is applied. That's a functionality specification. Below 3.0V (while still above ground) it is not guaranteed to sink 300mA. It might do something with a couple volts in, but not likely with 1.5V and no guarantees below 2.7V for performance with input voltage- and that's with 2V drop at the output.

The maximum input voltage is specified as 13V.

Try this out with a visible LED. Use a 1K series resistor and a 5V supply. Apply 5V to the input (or 3V). You can also damage the ULN2003A by leaving out the resistor.

It is fairly unlikely you have damaged the chip by soldering to it, however you have a good chance of damaging it by leaving out the series resistor.

  • \$\begingroup\$ Is it normal that between all the input pins, I have around 18k of resistance? \$\endgroup\$ Commented Jun 16, 2017 at 19:47
  • \$\begingroup\$ Depending on your meter, how it is connected etc. it's not particularly unusual. Check out the internal schematic. \$\endgroup\$ Commented Jun 16, 2017 at 21:49

The ULN2003A output will go no lower than about 0.8 V when on. Your LED will drop something like 1.2 V to 1.6 V when conducting so you want at least 2.5 V supply. That rules out your 1.5 V battery, I'm afraid.

You also need a resistor in series with your LED current to limit the current to maybe 5..10 mA.

To your question...

Connect a resistor between the ULN2003A output and your 1.5 V supply. The resistor can be of any value between about 1 K and 100 K, with the lower end of the range better than the higher. Put a voltmeter across the ULN2003A output and GND. Then connect the corresponding ULN2003A input to 1.5 V. You should see the output voltage drop from the battery voltage down to about 0.8 V. Repeat this across all 7 ULN2003A outputs.


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