I am designing a IR base home light control. for this i use 6 relays and i try to driver it through uln2003. I connect the GND and COM wire with 12 volt power supply and input pins are connect to microcontroller which gives 5 volt supply but OUTPUT will become constant (about 1 to 1.5 voltage) and not change on any logic 0 or 1
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\$\begingroup\$ circuit diagram? \$\endgroup\$– jippieCommented Jun 19, 2015 at 19:52
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\$\begingroup\$ You do know that a ULN2003 has open-collector outputs and will never drive them high? \$\endgroup\$– brhansCommented Jun 19, 2015 at 19:53
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\$\begingroup\$ ULN2003 is a darlington transistor array. It is expected that the output voltage won't drop below 1 volt or so. \$\endgroup\$– jippieCommented Jun 19, 2015 at 19:54
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\$\begingroup\$ The uln2003 is an open collector device. For testing just add a resistor between the output and +12Volt. Or the relay... \$\endgroup\$– auchmonoabspielbarCommented Jun 19, 2015 at 19:55
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\$\begingroup\$ Did you connect the ULN2003 ground to the microcontroller ground? \$\endgroup\$– Peter BennettCommented Jun 19, 2015 at 20:00
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2 Answers
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Here's something that should work:
answered Jun 19, 2015 at 23:52
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The circuit below takes into account the open collector operation of the ULN2003. The diode D1 is to prevent kickback when deactivating the relay. Technically this diode is built into the ULN2003, but it never hurts to have an extra!
simulate this circuit – Schematic created using CircuitLab
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1\$\begingroup\$ That's just crazy! The Darlington in the chip is a pair of NPNs, so the final emitter connection should in no way be connected to the positive side of V1, and the COM connection is the cathode of a diode designed to clamp the kickback of an inductive load to the positive rail when the inductive load is abruptly turned off. And, yet, two sheep support fantasy. \$\endgroup\$ Commented Jun 19, 2015 at 23:35
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\$\begingroup\$ You're right, I mis-remebered the data sheet and got COM and E mixed up in my mind. Fixed! \$\endgroup\$ Commented Jun 20, 2015 at 3:28