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  • Author Author: oldmanraskers
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CPLD Controlled Door Lock - Output Circuit

oldmanraskers

The weekend is over and I'm back to my design... I don't have much to say about the output circuit, except that I want to drive a bunch of 24V solenoids and provide an indication (via LEDs) when the output should be active. Below is the circuit I have come up with and hopefully there is nothing more to be done image Next I will be moving on to look at providing the CPLD with an external oscillator circuit and dealing with the JTAG connector image

 

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  • You should wire your LEDs so that the resistor is connected to 24Vand the LED cathode connected to OUTx - that way the LEDs will be on when there is power in the solenoids. The way your circuit is now the LEDs will come on when the output driver is off and will draw their current though the load. 560R will give you about 40mA though each LED when it's on and that seems like a lot - obviously it depends on how bright you want the LEDs but I'd be inclined to limit the power in the dropper resistor to 100mW max which would give you a max LED current of about 4mA and imply a series R of about 5k6, although I'd go for less current (about 2mA) and use 10k.

     

    MK

  • in reply to michaelkellett

    I don't really  understand Michael - I think what your saying is, that the way I have my LED connected means that the LED will be on when the solenoid is off (because the CPLD must turn the output on for the solenoid to be off and vice-versa) - but that is where you lost me image Can you please offer me more explanation when you have the time?

  • in reply to oldmanraskers

    Here's a diagram showing one stage of ULN2803 (except with normal transistor instead of the darlington it actually has (don't use a  BC184 in real life  !!)), the load and both ways of driving the LED.

     

    image

    L1 is the solenoid.

    When the input is high Q1 turns on, pulls OUT low and current flows through the solenoid and LED2. There's about 1V at OUT for  a real ULN2803 so LED1 will not light.

    When the input is low Q1 is off, OUT will be pulled up by current through the solenoid (which is like a 200R resistor at DC) so LED1 will be on. The current through LED1 also flows though the solenoid and unless it's quite low it might keep the solenoid pulled in.

    So the LED2 arrangement is better because the LED is on when the load is on and the LED current doesn't flow through the load.

     

    Hope that helps - ask again if not.

     

    MK

  • in reply to michaelkellett

    Ah thank you, I see now what you are saying Michael - In my original post I had intended to have the LED between the output pin of the CPLD and the input pin of the ULN2803 - in my schematic OUTxx is the output of the CPLD pin; the outputs of the ULN are the terminal connections labelled X6-7 etc... Do you think it's better then to have the LEDs on the output of the ULN as apposed the outputs of the CPLD? I was thinking along the lines of our previous discussion about wasting power in the current limiting resistors.

  • in reply to oldmanraskers

    It's a trade off, small amount of wasted power if LEDs on output but you are indicating much more directly the drive state of the solenoid. Probably doesn't matter much which way you go. Now that I look again I see that your labels did indicate what you meant - they are in rather pale grey on my monitor and I didn't look carefully enough. Sorry.

     

    MK

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