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I need a simple circuit to control a IR device. I'm quite a newbie with electronics so I would like to know if this circuit is correct because seems is not working.

enter image description here These are the datasheets of IR LEDs and ULN2003A:

Thanks for your help!

FIXED: wrong LED direction and number of leds, resistor now to 8ohms.

This way I power leds with ~110ma.
Thx for your help!

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  • \$\begingroup\$ Your LEDs are backwards and your 5 volt supply won't light 3 series LEDs even if they were the correct way round. \$\endgroup\$ – Andy aka Dec 11 '18 at 12:06
  • \$\begingroup\$ @Andyaka Your nice summary appeared about 40s before (the system says) my answer - which says the same thing. Great minds / fools seldom ... :-) ... \$\endgroup\$ – Russell McMahon Dec 11 '18 at 12:08
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    \$\begingroup\$ @RussellMcMahon you deserve the rep increase because you've been flagging recently hehe. Nice to hear from you. \$\endgroup\$ – Andy aka Dec 11 '18 at 12:11
  • \$\begingroup\$ Fixed LED direction! \$\endgroup\$ – Robert Casanova Dec 11 '18 at 13:59
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    \$\begingroup\$ @Andyaka Absent - yes. Flagging - technically, no, probably. Otherwise - some areas - yes. \$\endgroup\$ – Russell McMahon Dec 15 '18 at 10:17
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Look at the ULN2003 datasheet, page 12.
This shows you that the darlington drivers sink current to ground when on.
The 1st requirement is to reverse the LED directions.

Now look at the Vce saturation voltage on page 5.
This is about 1V - 1.2V depending on LED current.
So VLED_string can be about Vcc-Vcesat = 5V - 1V_ = 4V or less.

Now look at LED datasheet Vf values on the 3rd page. These are TYPICALLY 1.2V - 1.4V at 20-100 mA and 1.5V-1.8V worst case.
So at even 10 mA typical Vf x 3 = 3 x 1.4V = 4.2V = more than the driver will supply.

Try TWO series LEDs.
2 x 1.4V = 2.8V at 100 mA TYPICAL. Vresistor = 4V or less - 2.8V = 1.2V.

With R = 7 Ohms I = V/R = 1.2/7 ~= 0.017A ~= 15 - 20 mA.

Rework for desired currents.

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