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I am using a ULN2003 to power lamps and LEDs, and controlling them via 5v TTL.

I have some LEDs can be driven by 5v, and some lamps which require 12v. Using the same ULN2003, can I have some powered with 5v, and some 12v, or do I need to use 2 different chips? Of course, the grounds would be shared.

Thanks for your help in advance.

enter image description here

Fed

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  • \$\begingroup\$ Is it possible to make legible schematics with Fritzing? This one is bloody awful! However, I can see that you don't have any current-limiting resistors in series with your LEDs - you do need those resistors, or the LEDs will be rapidly destroyed. \$\endgroup\$ – Peter Bennett Feb 7 '17 at 22:44
  • \$\begingroup\$ Sorry about the schematic. I removed it as I would like to focus on the question rather than my schematic mistakes. \$\endgroup\$ – Fed Feb 7 '17 at 22:48
  • \$\begingroup\$ Decent schematics often aid the reader in understanding a question. This site has a nice schematic editor invoked by "Control-M" (or by an icon above the edit window) while you are editing a question or reply. \$\endgroup\$ – Peter Bennett Feb 7 '17 at 22:55
  • \$\begingroup\$ How's the new schematic. I couldn't find a lamp element, so I used an LED. \$\endgroup\$ – Fed Feb 7 '17 at 23:10
  • \$\begingroup\$ Your LEDs are backwards, but other than that, this will work, though the common needs to go to 12V, not to ground. \$\endgroup\$ – user1844 Feb 7 '17 at 23:14
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The ULN2003A does not have a power pin. It has a catch diode pin that is common for all outputs. It can be connected to the highest voltage supply that has a load controlled by the ULN2003A.

enter image description here

For example, if you have loads connected from ULN2003A outputs to +5V and to +12, connect the COMMON to +12. Of course each load will be connected between one ULN2003A output (OUT 1 ~ OUT 7) and the respective supply (either +5 or +12).

In fact if your loads are resistive (and have short leads) you could probably leave the pin open, but it does no harm to connect it to the supply as indicated. Connecting it to +5 would be very bad as it could result in overvoltage (via the +12V loads) on the 5V supply, probably breaking something.

Your schematic does not look right, but it's so messy I hesitate to delve into it on a relatively small screen.

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  • \$\begingroup\$ I am sorry about the schematic, as I am quite new to electronics. i welcome constructive criticism. Doesn't the common connect to ground, as a "lamp test" to turn on all the pins? Also, what do you mean by "leave the pin open"? \$\endgroup\$ – Fed Feb 7 '17 at 22:43
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    \$\begingroup\$ Leave the pin open means don't connect it to anything. The diode is mainly used as a flyback diode when driving inductive loads. Because you are driving lamps, it's probably not needed in your case. \$\endgroup\$ – AngeloQ Feb 7 '17 at 23:33

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