Unexplained voltage loss with relay

Discussion in 'The Projects Forum' started by cocacolby, Jun 2, 2011.

  1. cocacolby

    Thread Starter New Member

    Jun 2, 2011
    5
    0
    I'm pretty new to circuits but it really seems like this should be a lot easier than it is. I'm basically just trying to use a relay to trigger a 24v solenoid valve on and off. I have the coil part of the relay attached to LabView which just generates the 5v on/off signal. My problem is that for some reason when that 5v signal is plugged into the relay, I can only seem to get a reading of 1.4 volts no matter how high I turn up the voltage. Because of this I can't ever get the relay to switch. I've verified the signal coming out is actually 5, anyone have any ideas on this? Thanks in advance for any help.

    p.s. The relay is a Gordos SM-0DC5 and I have it wired with a protective diode as well.
     
  2. cocacolby

    Thread Starter New Member

    Jun 2, 2011
    5
    0
    Quick Update:
    My 5V signal is being sent out of a National Instruments 9263 analog out DAQ module. I tried generating a 5v signal using a DC power supply instead and found that I was getting 5 volts no problem. So I guess the updated question would be why can't my computer maintain 5 volts when my DC power supply can? Can I fix this somehow or am I screwed? Thanks
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    A quick search didn't turn up the Gordos part, but there is a Crydom datasheet that has the same part number, and similar appearance.

    The link for the datasheet is on this page:
    http://www.datasheetarchive.com/pdf-datasheets/Datasheets-8/DSA-142814.html
    There is Crouzet datasheet with the same part number available on this page:
    http://www.datasheetarchive.com/Indexer/Datasheet-026/DSA00452953.html

    Do you have the polarity correct on pins 3 and 4? Pin 3 needs to be more positive than pin 4 to turn it on; the range for your device is 3v minimum to 8v maximum.
    If pin 3 is <= 2v from pin 4, it should turn off.

    The current required to illuminate the IR emitter is between 10mA and 27mA.
    It may be that your LabView device cannot source enough current. Why don't you insert a 1 Ohm resistor in the path, and measure the voltage across the resistor? That will give you the current reading in Amperes, and is much safer than trying to use the ampere function of the meter (no risk of blowing a fuse or damaging the meter).
     
  4. cocacolby

    Thread Starter New Member

    Jun 2, 2011
    5
    0
    Hey SgtWookie,
    It looks like you're right, that module can only output a max of 1ma. Again, I'm really new to this but my next step is to try and amplify the current using a transistor right? Thanks for figuring out the problem! It's been driving me nuts.
     
  5. cocacolby

    Thread Starter New Member

    Jun 2, 2011
    5
    0
    Actually, just going to buy a new relay that needs less than 1ma.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You can use the output of your module to drive the base of an NPN transistor, and switch the negative side (pin 4) of the logic input of the relay from the transistors' collector; connect pin 3 to your power supply's 5v source, and the emitter of the NPN transistor to your power supply return. Your power supply return will need to be common (connected) to your module's supply return. Use a 4.3k Ohm resistor from your module to the transistor base to limit current to 1mA, and a 0v-5v signal from your module.

    Instead of an NPN transistor, you could use an N-ch logic level power MOSFET; use the module to control the gate, sink current from pin 4 using the drain, and connect the source terminal to your supply return; as before, the power supply return needs to be common to your module.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    If you're switching DC, just use a logic level MOSFET that has a Vdss rating (max voltage from drain to source) suitable for your load, and an Id rating (drain current) several times that of your load requirement, and use the MOSFET to sink current from your load. Once switched on or off, MOSFETs don't require any current to maintain the ON or OFF state.

    But with only a 1mA drive current, you will not be able to switch it very fast, or it will spend too much time in the linear region, and will dissipate power as heat.

    Here's an IRLU8721 that Digikey carries. It's a fairly compact MOSFET design, Vdss=30v, and Id=65A. $1.28/ea if you just buy one.
    http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=IRLU8721PBF-ND
    It has a very low gate charge (13nC) and the Rdson is so low (8.4 milliOhms) that you won't need a heat sink.
     
    Last edited: Jun 2, 2011
  8. cocacolby

    Thread Starter New Member

    Jun 2, 2011
    5
    0
    Wow, thank you so much! I finally feel like I'm making progress on this again. I'll pick up parts tomorrow and give this a shot.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Beware that there is a bewildering assortment of MOSFETs out there; many thousands of different models produced by a lot of manufacturers.

    You need a Vdss of at least 30v (but don't go more than 60v or your gate charge will get very high), and you need at least 3x your current sink requirement for the Id rating. The lower the Id rating, the bigger the heat sink you will need for the MOSFET.
     
Loading...