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17228 Views 6 Replies Latest reply: Nov 1, 2012 5:01 PM by circuitspecialists RSS
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Sep 25, 2012 10:29 PM

How do I connect a shift register - serial in parallel out - to lamp

I want to connect a serial shift register to a 12 volt lamp that draws about one ampere, but I don't know what to connect between the shift register and the one ampere 12 volt light bulb.  My friend said to use a 'Lamp - driver' but I don't know what that is or where to find one.  Maybe you have  a better idea... Oh, BTW could you recomend a four terminal output shift register part number to light up three lamps, - the fourth output will reset the shift register.

  • MrSpidey Novice 11 posts since
    Sep 25, 2012

    I have always used transistors to drive lamps with shift registers, decade counters etc.

     

    You could also use SSRs.

     

    Can’t really help you with choosing the register.

  • MikeWeed Novice 62 posts since
    Oct 21, 2011

    The best way to drive a lamp depends on the voltage the shift register runs at, which in turn depends on the voltage of the clock signal. If you use a CMOS shift register like the CD4015 running on 12V, then a simple

    field effect transistor like the RFP12N10L will work. Just connect the FET source to ground, the gate to the shift register output and the drain to one side of the lamp, and the other side of the lamp to +12V. This should also work with a 5V supply, since this is a "logic level " FET. A traditional TTL (bipolar transistor) shift register would probably not work because the output voltage swing is less than 5V, but any of the numerous 74xxx CMOS registers made to run on 5V only should work. A lot of logic circuits use 3.3V or less these days, and a FET cannot be used with these because the voltage swing is too small. For these, a bipolar transistor is needed, probably 2 stages to provide the 1 amp you need.

  • Shockwave Novice 1 posts since
    Oct 30, 2012

    Take a 74HC595 shift register.

     

    Each output goes to the gate of a Nchannel power Mosfet.  The gate also has a resistor (10K approx) tied to the source, and the source tied to the ground.

     

    The drain of the Mosfet connects to one side of the 12VDC light and the other side of the light is +12VDC power.

     

    This is called Low side Switching.  The lamp is connected to power and you are switching on/off the ground.

     

    As for the other request; simply write a byte out to SPI on your microcontroller and then toggle STCP pin of the shift register to latch all the bits together.  This is also extendable -- just add more shift registers after the 595 and clock out more bits if you need more in the future.  Turning them all off is a simple matter of writing out 0x00 to SPI then toggling the latch.  Turning them all on is the same by writing out 0xff instead. 

  • MarkArnold Novice 1 posts since
    Oct 30, 2012

    Go to the Parallax web site and they have a good note on using a basic stamp to control a shift register like a 74hc595.

  • J_W Novice 1 posts since
    Oct 31, 2012

    You could use the 74195 IC which is a 4 bit  shift register and at the outputs use either 2n3904 or 2n3906 transistors to handle the required 1 A load current.  The transistors will handle over 1 A if heat sinked properly.  The choice of NPN or PNP depends on your requrements as to what happens with the light vs its shift register output.  You may need current limiting resistors in the lamp cicuits to be safer, but you will have to test that and see for yourself.

  • circuitspecialists Novice 7 posts since
    Oct 19, 2012

    I use the ULN2003 darlington drivers to drive higher current loads like lamps and solenoids. It is easy to interface with the shift register either CMOS or TTL.

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