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I am looking to control a 12v LED light off of a zwave appliance module. Here is my issue, the zwave module outputs about a 5v pulse every few seconds while in the off state, which is enough to momentarily turn on the lights. I put a relay inline, but the 5v is enough to charge the coil and switch the relay on. How can I create a buffer so only when 10-12v is applied will the circuit be complete?

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  • \$\begingroup\$ how long is the pulse? whack a capacitor there to reduce the pulse spike voltage, and only a good strong DC 12V for intended operation will make it go on? \$\endgroup\$ – KyranF Jan 16 '15 at 1:42
  • \$\begingroup\$ otherwise a comparator (which is a simple op-amp setup) with a 10V threshold that turns on a pass element (PFET as a high side switch, or a relay) will work. \$\endgroup\$ – KyranF Jan 16 '15 at 1:43
  • \$\begingroup\$ Pulse is less than a second. Will the capacitor discharge fast enough or will the constant pulses build up a large enough charge to trip the relay? \$\endgroup\$ – BlurrVT Jan 16 '15 at 2:09
  • \$\begingroup\$ What size cap should I use and would I have to worry about heat buildup while the light is on? \$\endgroup\$ – BlurrVT Jan 16 '15 at 2:10
  • \$\begingroup\$ You could try to replace the relay with a potentiometer to limit the current so that 5 V will not turn on the LED but 12 V will. What's the datasheet for your LED? Is it something like this? digikey.com/product-detail/en/SSL-LX5093GD-12V/67-1099-ND/… But now that I think of it, the typical LED won't change much for a between a current of I and a current of roughly 2*I. \$\endgroup\$ – FullmetalEngineer Jan 16 '15 at 3:01
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You might be able to find a 12v relay that has a pull-in and drop-out voltage greater then 5v. That way a 5v signal will not turn it on or hold it on.

If such a relay can not be found you can create one.
Use a relay with a 5v coil. Place a 6v zener diode in series with the relay coil, (not in the wire going to the LEDs). The zener will need to be rated with the proper reverse current and wattage to withstand the on current of the relay's coil and the expected power dissipation, (Vz x I).

With the above setup a 5v signal will not be enough to force any appreciable current into the relay coil. At 10 or 12v you should get 4v to 6v across the relay coil. You will need to check the relay spec's to be sure that voltage range is acceptable. If the LED line voltage is consistently closer to 12 then you might try using a 6.8v zener.

Note that zener voltage parameters often have a wide tolerance so you may need to do some testing to find the best part. Also remember to place a reverse standard diode across the relay coil to reduce any flyback voltages from getting onto the line.

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  • \$\begingroup\$ If you start with a 6v zener and later determine that you really need a higher Vz you could just place a standard silicone diode in series with the zener to get up to 6.7v. \$\endgroup\$ – Nedd Jan 16 '15 at 15:32
  • \$\begingroup\$ Thanks for the help. I was able to solve the problem by using a different power supply. I initially hooked this up because I wanted to power both lights off of one plug and this one is rated at a higher amperage. I noticed when I turned on the lights, there was a slight delay and when I turned them off, there was some residual dimming. I'm guessing there is enough capacitance or regulation in the new power supply to buffer the 5v pulses. Not the most technical solution, but it works. \$\endgroup\$ – BlurrVT Jan 17 '15 at 16:48
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Try a 6v 1W zener diode in series with either a 5v relay or the LED. You may need to use a lower value resistor with the LED to get sufficient light output.

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