I want to use a UDN2981 8-channel source drivers IC (see UDN 2981 datasheet) for a matrix multiplex light.

I connected the UDN2981 this way:

  • pin 9 (VCC) to 5V
  • pin 10 (GND) to GND
  • pin 6 to VCC with a 10K ohm resistor
  • pin 13 to a 3mm white LED to a resistor (220 ohm) to GND

And I connected another LED not via the UDN: from VCC -> white LED -> 220 ohm resistor -> GND

What I see is

  • that the 3mm white LED via the UDN gives only 0.1 mA (my multimeter here does not have more accuracy)
  • that the 3mm white LED directly from 5 V gives 1.7 mA

My calculation for the latter would be

V = I * R <=> (5 - 3.3) = I * 220 <=> I = 1.7 / 220 = 7.7 mA

This is still a lot more than 1.7 mA, but maybe the forward voltage is higher

However, I'm more interested to know why the LED via the UDN gives only 0.1 mA... I tried different resistors to pin 6 but this does not affect the current through the LED.

(I think it has something to do with the internal diodes of the UDN giving a voltage drop... this means I should use a higher voltage to VCC or less current; I hope I can calculate this easily).

  • 1
    \$\begingroup\$ Just off the cuff, are you aware that the UDN2981 uses a high-side Darlington output and may have a voltage drop of \$1\:\text{V}\$ or more perhaps when ON? If you are expecting full voltage to be applied, it's not a good switch to use. \$\endgroup\$
    – jonk
    Mar 30 '19 at 17:54
  • \$\begingroup\$ @jonk I was not really aware of this, but it will not be a problem, I can easily change the resistor values, and since I need a separate power supply anyway, I can as well use a 9V or 12V. \$\endgroup\$ Mar 30 '19 at 17:59
  • 1
    \$\begingroup\$ Hi, Michel. Schematics are better than words. Is this more work for the band? Give us a YouTube link sometime to see if the help we are giving is of benefit to humanity or if we should discourage you from playing! \$\endgroup\$
    – Transistor
    Mar 30 '19 at 17:59
  • 1
    \$\begingroup\$ @MichelKeijzers Then that's the approach. Just adjust the resistors. (Just measure the output voltage drop you are getting already. You can work out what the device is actually doing from that and work out the resistors. Or else raise up the supply voltage, which the device can easily handle.) \$\endgroup\$
    – jonk
    Mar 30 '19 at 18:02
  • 1
    \$\begingroup\$ Please try to reduce the impedance of the resistor from Vcc to pin6. For example 2K ohm. \$\endgroup\$
    – AltAir
    Mar 30 '19 at 20:37

I am afraid that all your observations are misleading.

First, I don't know any white LEDs that would draw 1.7 mA if solid 5V is applied. A normal white lamp should smoke out if you connect it to 5V.

The reason you have these "strange" reading is that you are using DMM in current mode. The DMM measures currents using a shunt resistor, and therefore there is parasitic voltage called "burden voltage". The burden voltage varies with measurement range, and depends on manufacturer's quality of design. This voltage "eats" you voltage budget, so the LED is exposed to a smaller voltage and hence draws much less current (check with I-V curve for LEDs).

Similar effect happens when you use the UDN2981 drivers, which output is made of Darlington and have the saturation voltage of 1.6V at -100mA, (and maybe 0.6 -1.0 V at smaller load currents, as Jonk mentioned). This voltage drop further reduces the voltage applied to V-I LED curve, which reduces the current to nearly zero.

In general, if you want to measure real current in low-voltage circuitry (under 5V), never use DMM current mode. Use instead a 0.1 Ohm resistor in series with your wires and measure the voltage drop in mV-mode. Then adjust the value of shunt resistor if your readings don't have satisfactory resolution.

  • \$\begingroup\$ About your second paragraph (First, ...), I use a 220R resistor. I wasn't aware of a DMM not able to measure mA, so I will buy some resistors (I did some calculations, and I think 2W through hole resistors of 0.1 R would do, maybe in various combinations like parallel/serial). Can you confirm this? \$\endgroup\$ Mar 30 '19 at 20:40
  • 1
    \$\begingroup\$ @MichelKeijzers, DMMs are able to measure mA, but some of them need higher overall voltage in the path such that the burden voltage doesn't affect the results much. And you don't need 2W shunt if your target range is 100 mA (which is plenty for most LEDs), 0.1*0.1*0.1 is 1 mW. Any SMD chip resistor 0805 or 1206 will do the job. If you need through-holes, solder AWG30 wires to the chip or solder the chip between 0.1" male 2-pin header. \$\endgroup\$ Mar 30 '19 at 20:51
  • \$\begingroup\$ I'm just so far checking the chip on a breadboard so I can easily add through hole resistors next to it. And eventually (and mostly for future cases). \$\endgroup\$ Mar 30 '19 at 20:55

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