Low power relay selection

Discussion in 'The Projects Forum' started by a_hayward, Jul 28, 2011.

  1. a_hayward

    Thread Starter New Member

    Jul 28, 2011
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    Hi,

    I'm working on a project using the following capacitive sensing circuit:

    http://ics.nxp.com/products/pcf/datasheet/pcf8883.pdf (pg 6, fig 4).

    I've tuned this circuit to the particular sensor surface I wish to use, powered by a 9v battery. The output of the circuit I'd like to trigger a relay, so that it breaks the contact of another sensor, which is running on a 12v circuit at 0.02 A.

    Firstly, the Limiting value specifications on pg 9 are confusing, it states that output current at max is 50mA, and 9v max, yet total power dissipation is quoted as 100mw, not 450mw?

    Secondly, it seems I need a relay that will switch at 8V (allowing a margin for voltage drop) and 30mA that would then have to cope with 12V and 20mA. Can anyone point me in the direction of a suitable one?

    Thanks :)
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    does it have to be made with a relay? if you post a circuit of the sensor circuit you want to interrupt we could see if there are other possibilities.

    You have to consider the output voltage drop when the output is HIGH. This will be the voltage drop on the internal PMOS when it's conducting x output current. (not 9V x 50mA)

    Look at digikeys. I found THIS , there may be others.
     
  3. a_hayward

    Thread Starter New Member

    Jul 28, 2011
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    Attached is a circuit diagram for the overall circuit and a bit on how the chip works.

    I'm assuming i'll need a relay because I want the inverse of the circuit output at the moment (i.e I want the output to be closed when the sensor is low and open when the sensor is high). I think the output however is always closed and just variable voltage. I'm not sure really, a little out of my depth on this one, the most i've done in practise is a 555 timer!
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Well, trying to power just about any relay with a 9v battery means that you will be replacing the battery VERY frequently (annoyingly frequently) if the relay spends more than a few seconds a day closed.

    What you could do instead is to look at using a Schmitt-trigger inverter, such as one gate of a 4093 quad NAND schmitt trigger with its' inputs tied together, or one channel from a 40106 hex NAND schmitt trigger inverter, grounding the rest of the inputs - and use THAT output to drive an N-ch or P-ch MOSFET gate to switch your load. The one requirement is that the circuit will have to have the ground in common with your battery-powered circuit if an N-ch power MOSFET is used, or the +V in common if a P-ch MOSFET is used.

    The 4093 or 40106 will draw very little current (microampere range) and the MOSFET only requires that its' gate be charged or discharged to act as an on/off switch.

    If your load is small, you could use something like a 2N7000, which is a TO-92 packaged MOSFET rated for 200mA drain current (~50mA would be a good practical limit though), 60v Vdss, and will turn on with as little as 5v on the gate relative to the source terminal.
     
  5. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    Have a look at section 8.1 output switching modes in the datasheet. It's a switch, not an analog output voltage.;)

    You didn't post the other circuit (12V). If it doesn't has to be galvanically isolated I suggest you consider Wookies recommendation.
     
  6. a_hayward

    Thread Starter New Member

    Jul 28, 2011
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    Thank you both for your replies.

    praondevou, before posting initially I put a voltmeter across the output terminals. When the sensor was low, there was 0v, when high there was just under 9v. I ran a continuity test on the multimeter when it was low and it showed that the circuit was closed and not open (didn't beep). Does that mean that the output will not pass any current even though the "switch" is closed rather than open?

    If it is infact act as an open switch when the sensor is high, then all i'd need to do is invert the output, using a transistor output like http://www.kpsec.freeuk.com/images/trinvert.gif ?

    SgtWookie, when you say is the load small, do you mean the circuit I intend to switch on and off with this one? If so, it is, 0.02A and 12V. How would I wire the 2N700 in with my output connection?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    A transistor used like the one in your link will indeed invert the signal; however it will be rather power-hungry.

    If you use a 4000-series CMOS IC like the 4093 or the 40106 like I mentioned, it will invert the signal for you, and not wear down your battery like a transistor will - as the standby current is next to nothing (on the order of a microampere or two).

    The output of the PCF8883 is a P-ch MOSFET with an open drain. This means that it will raise its' output to near whatever its' positive supply voltage is (say, 9v), which means the load needs to be between the output and ground. When the output turns off, it's basically disconnected.

    If you need to invert that signal, you will need to use a resistor to ground so that when the output turns off, the resistor will pull the output back towards 0v.

    Then you can connect up one of the 4093 gates, a 40106 gate, the gate of a 2n7000 N-ch MOSFET, or if you insist, the NPN transistor's base via a resistor.

    In all of these cases, the ground of your sensor must be connected to the ground of your unit that you are attempting to "close" a connection with.

    It's not obvious by looking at what you've posted thus far exactly what this output is supposed to connect to.
     
  8. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    from the datasheet:
    "Pin OUT is an open-drain output capable of pulling an external load Rext (at maximum current of 20 mA) up to VDD. The load resistor must be dimensioned appropriately, taking the maximum expected VDD voltage into account. The output will be automatically deactivated (short circuit protection) for loads in excess of 30 mA. Pin OUT can also drive a CMOS input without connection of the external load.A small internal 150 nA current sink Isink enables a full voltage swing to take place on OUT, even if no load resistor is connected. This is useful for driving purely capacitive CMOS inputs. The falling slope can be fairly slow in this mode, depending on load capacitance."

    So the output is an open drain P-MOS with a current sink of 150nA. Either passes current from + through the IC to the load and back to GND or the output is pulled down to GND by the internal 150nA sink.
     
  9. a_hayward

    Thread Starter New Member

    Jul 28, 2011
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    Sorry, to clarify, I wish to place this capacitive sensing circuit in another 12v circuit, which only has a bulb. I wish to use this capacitive sensing circuit as a switch, an on/off switch. I don't wish to power anything with it, just to use the circuit as either a broken or closed contact. The capactive circuit is low when there is no contact and high when there is contact. However, I want the bulb to be high when there isn't contact and low when there is.

    I've done a little research since both your replies to try and familiarise myself with what you mean. I also managed to connect my power source in reverse polarity, so have had to order another circuit... but in the mean time;

    I've looked up open-drain outputs; am I right in saying that the open drain is connected to ground when low, but then stops current flow in the high state due to very high impedence?

    Is this diagram what you mean to do with the circuit SgtWookie?
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    No, like the attached.

    Note that the output of the 4093 only connects to the gate of the N-channel MOSFET.

    Note that the source terminal and the battery negative must be connected to the same ground that the 4093 is.

    You should also connect the unused 4093 INPUTS ONLY to ground or +V so that they don't oscillate and cause problems. Never leave a cmos INPUT "floating"; that means without a current path to ground or the positive supply.
    Leave the unused OUTPUTs disconnected.
     
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