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I ordered a few ULN2002As.

I have it wired as follows:

Pin 1 -> GPIO of Arduino Pro Mini 3.3V

Pin 16 -> 300Ω resistor in series with an LED

Pin 8 -> GND

Pin 9 (COM) -> 5V (which is also RAW for the Arduino).

Pins 1-4 were wired similarly with LEDs but I've removed 2-4 because it wasn't working and testing with 1 was simpler.

When I apply 3.3V from the GPIO to Pin 1, nothing happens. When I disconnect GPIO and just apply 3.3V with a jumper, nothing happens. If I apply 5V, nothing happens. The output just swings around 300mV.

Am I totally misunderstanding how this chip works? I thought applying at least Von (2V according to the datasheet) at Pin 1 should make Pin 16 turn on.

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  • \$\begingroup\$ Is your LED in backwards? When input is high, output for Darlington is low and visa versa with pullup. \$\endgroup\$
    – Tony Stewart EE75
    Feb 1, 2018 at 0:26
  • \$\begingroup\$ No it's definitely in correctly. If I remove the LED and resistor on the output it just sits at 300mV to 1.2V, regardless of the input. I'll maybe try another chip perhaps this one is broken? \$\endgroup\$
    – brenzo
    Feb 1, 2018 at 1:01
  • \$\begingroup\$ Test 300R resistor to ground (LED ON) then short out LED with 300R across p16 to p9(5V ) the switch should be inverted to input I assume LED Anode was at 5V not 3.3 (wrong) \$\endgroup\$
    – Tony Stewart EE75
    Feb 1, 2018 at 1:05

2 Answers 2

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If you look at the datasheet, on the first page, under the heading of description, you’ll see that the ULN2002A is designed for 14-15V PMOS Logic.

Looking at the functional block diagram on page 2, you’ll see that there is a 7 volt zener diode in series with the base of the first transistor. There’s no way that a 3.3V signal will bias that transistor.

They do offer a 5V CMOS logic version (ULN2003A), that will probably work for you, in your intended application.

Your confusion about ‘Von’ being 2V is because you were looking at the ‘test condition’ voltage and not the actual ‘parameter’, which is 13V. Look at the datasheet again. If you’re still confused, I can try to explain it in more detail.

Hope that helped!

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  • \$\begingroup\$ Thanks I totally missed that. I'll order the 2003A. What exactly does the "test condition" voltage mean in that context? \$\endgroup\$
    – brenzo
    Feb 1, 2018 at 2:19
  • \$\begingroup\$ In the “Electrical Characteristics”, it will refer you to figure 6 on page 6. The “Test Conditions” are what they give you to describe how the circuit will behave under the “Parameter” given. \$\endgroup\$
    – Nettle Creek
    Feb 1, 2018 at 3:08
  • \$\begingroup\$ Also, I just realized that we both read V(on) wrong. (I’m on an iPad.) It’s (Vion). It’s the relationship of output (collector) current to the input voltage. The 2 volts you were reading is the voltage drop across the device junctions. In other words, the lowest voltage you will see on an output pin, with respect to ground, is 2 volts. I may be confusing you even more. Sometimes, I can explain things well verbally, but stumble around when writing it out. :-) \$\endgroup\$
    – Nettle Creek
    Feb 1, 2018 at 3:37
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If you used a white LED Vf=3.1V on a 3.3V supply switched by the ULN200x Darlington which has up to ~1V rise in ON voltage.Then the LED will not turn ON.

However GPIO can drive a White LED direct from 3.3V since it has ~25R internal and some additional small R may be useful.

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