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Looking at the datasheets, I'm not sure their specifics match;
TIP127 and ULN2003A

My intention is to do something like this, but with the ULN2003A, that I have already:

wirings

The reason why is that I want to remove the provided driver/power supply (5V 12V, driving 6 LEDs 0,25W each) and use PWM with the RPi to dim the strip's leds (using an external 5V power supply). I've already done this to regulate a 12V fan (but I understand this is another story).

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    \$\begingroup\$ How much current does that RGB thingy draw? (And connecting TIP120's without base resistors is not a good idea.) \$\endgroup\$ – Wouter van Ooijen Sep 4 '14 at 16:53
  • \$\begingroup\$ Your question title and graphic references TIP120's, but the datasheet link you provided references TIP127's. I'm assuming TIP120 is correct because that's a NPN transistor. The TIP127 is a PNP, which will definitely NOT work in this circuit. \$\endgroup\$ – Dan Laks Sep 4 '14 at 17:08
  • \$\begingroup\$ @WoutervanOoijen I'm not sure how much current my strip drowns. It's a 6 x 0,25W LED strip (1,5W tot.) using 5V. Amperage should be written on the power supply but I don't have them now. \$\endgroup\$ – dentex Sep 4 '14 at 17:18
  • \$\begingroup\$ @DanLaks The pdf contains data also for the TIP120. \$\endgroup\$ – dentex Sep 4 '14 at 17:19
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For the most part, yes, the ULN2003A will work in this circuit, but with some caveats.

The VCE(sat) is slightly higher for the ULN2003A vs the TIP120, which means the LEDs will have slightly less voltage across them (and so slightly dimmer). But that may not matter for your application.

My biggest concern is the total current through the ULN2003A chip when all three LEDs are on. Figure 14 of the datasheet shows the maximum current you can drive through each of the Darlington pairs. As you can see, if you have all three LEDs on at 100% duty cycle, each color can only draw a little more than 150mA. That's 450mA total. Any more than that and the chip will overheat. It's important to get a handle on how much current your LED strip pulls. enter image description here

If, as you said in your comment, the strip is 1.5W with a 12V power supply, then presumably that translates to 125mA total. If that's true, you're probably ok then.

EDIT: You indicated you're using a 5V power supply. Since your LED strip is sized for 12V, the LEDs may not even light up. Especially the blue LED. If they do light up, they'll be significantly dimmer than they were designed to be.

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  • \$\begingroup\$ See the edit to my answer about the use of a 5V power supply on this circuit. \$\endgroup\$ – Dan Laks Sep 4 '14 at 17:39
  • \$\begingroup\$ So is the LED strip a 12V strip or a 5V strip? Is the image you used in your picture a photograph of the actual strip or something you grabbed off the internet? \$\endgroup\$ – Dan Laks Sep 4 '14 at 18:01
  • \$\begingroup\$ Ok. I checked. The strip's power supply is rated 12V at 350mA. Sorry for the mess ;) \$\endgroup\$ – dentex Sep 5 '14 at 6:40
  • \$\begingroup\$ Accepting the answer, you were right. Although the power supply it's 350mA, the measured current across the strip it's about 100mA. Thanks. \$\endgroup\$ – dentex Sep 7 '14 at 6:16
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VI(on) (VCE=2V, IC=300mA): 3V

Achievable by a 3.3V CMOS output. Lower currents have a lower VI, which would be a good thing.

VCE(sat) Max (IC=350mA): 1.6V

Your strip would need to be okay with 10.4V. Lower currents have a lower voltage drop (also good).

The common won't interfere, and actually won't matter since you're not driving an inductive load.

So once you know what current and voltage each component requires, you can decide if it will work for you.

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  • \$\begingroup\$ Thanks for the answer, but due to my level of skill in this field, I don't understand it. I can tell that my strip uses a small driver/power supply of 5V and it's composed of 6 LEDS 0,25W each. I'm going to edit the Q to explain better what I want to do with it. Thanks again for your help. \$\endgroup\$ – dentex Sep 4 '14 at 17:22
  • \$\begingroup\$ I think that in the future you need to qualify "like this", since what you've described is not like that in very many ways. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 4 '14 at 17:24
  • \$\begingroup\$ Do you mean qualify myself as a beginner? Sure, it seems appropriate. \$\endgroup\$ – dentex Sep 4 '14 at 17:28
  • \$\begingroup\$ Qualify that the issue isn't really "like this", since the components are very different. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 4 '14 at 17:32

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