Convert a negative output to positive output

Discussion in 'General Electronics Chat' started by NeoXon, May 24, 2011.

  1. NeoXon

    Thread Starter New Member

    May 24, 2011
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    Hi,

    I am a newbie in electronics. I have a ground switched lamp. What I want is to have +12V output when the lamp is switched on, and 0V/GND when it is switched off (basically a ground switch to a positive switch).

    What I have found on the internet is that I should use a relay for this purpose.

    Now I made a small sketch of a simple circuit. What is in ... lines is the already existing lamp + the ground switch. Is it possible to attach a relay as shown, or do I need diode, snubber circuit, anything else? Note that DC12V is used everywhere, no AC in the circuit.

    Can I use a solid state relay instead of a mechanical one? I want it to consume as small power as possible, and only consume power when the light is on. This is because 99.5% of the time the lamp is off.
    If I used solid state relay instead to reduce power consumption, would I have to change something, add anything else? Which legs of SSR correspond to 30, 85, 86 and 87a legs of a mechanical relay? I know the control circuit usually has + and - on it, but does the direction matter on the load circuit? I mean if I connect fused +12V to leg1 or leg2 of the load circuit?
    [​IMG]
     
    chloride likes this.
  2. Jaguarjoe

    Active Member

    Apr 7, 2010
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    What you have there will work OK as long as the existing fuse can handle the new load, otherwise a separate fuse for pin 30 will be required.

    What are you powering with the +12v from the relay's pin 87?
    Most Bosch style relays have ~75Ω coils that consume around 150ma or so and are able to switch 30 amps, some 40 or even 50. If you do not intend to switch something that needs a lot of current like that, you can use a smaller relay that will probably have a higher resistance coil which will consume less power.

    Whether you need a snubber or commutation diode depends upon what you are switching.
     
  3. NeoXon

    Thread Starter New Member

    May 24, 2011
    9
    1
    The switched 12V+ output is connected to a voltmeter embedded in a third party microcontroller, so there's practically hardly any current flowing through the load circuit (ideally a voltmeter has infinite resistance, afaik). If that detector circuit sees +12V it will create a UDP packet and send it to my server.
    So this means that whenever a lamp is switched on, I will be notified.

    Fuse should be OK, switched output should not consume more than a few mAs, a solid state relay consumes about 150mA as you say, the lamp eats around 500mA and I have a 5A fuse or so.

    But since the whole thing runs from a battery I'd like to save as much power as possible. For this reason, solid state relays are more advisable, right?
    Do I need commutation diode and snubber circuit in this setup?
     
  4. BillB3857

    Senior Member

    Feb 28, 2009
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    Would it be easier to change the detector circuit to sense LOSS of 12 volts?
     
  5. radiohead

    Active Member

    May 28, 2009
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  6. NeoXon

    Thread Starter New Member

    May 24, 2011
    9
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    Sure it would if it was not a third party device :) I don't want to disassemble it. Not just because I am a newbie, but it would also void warranty.
    If it was possible I would just connect the output of the ground switch and everything was fine already.

    Maybe this is going to be stupid, but is it possible to use transistor instead of a relay in the above circuit? What else would I need for that? Load circuit should have only a few mA. I have seen that SSRs are generally much more expensive than relays and transistors.
     
    Last edited: May 24, 2011
  7. radiohead

    Active Member

    May 28, 2009
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    To use a transistor as a switch, you need to stay within the limitations of the transistor. For example, a 2N3904 can handle up to 200mA. Apply a base bias voltage to "turn it on"
     
  8. Jaguarjoe

    Active Member

    Apr 7, 2010
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    NeoXon likes this.
  9. NeoXon

    Thread Starter New Member

    May 24, 2011
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    Yes, this is definitely cheaper than a $20-40 SSR.
    So in the above setup if I use this 12 VDC reed relay I don't even need diode or snubber circuit. Cool.
    If this reed relay has an 1453 Ohm coil, am I right in thinking that it consumes only 12V/1453Ohm = 8mA when I close the load circuit (+12V on output, lamp is on) and consumes nothing when the load circuit is open (0V on output, lamp is off)?

    What is the drawback if I choose this and not an SSR? Does this small relay make a disturbing clicking sound during operation? Or an SSR would consume even less power when the load circuit is closed?
     
  10. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    That is a reed relay, it is inaudible.

    I don't know which SSR you would use so I can't make a comparison.

    A transistor would consume some current, about 1/2 that of the relay.

    I forgot to add a load resistor across the voltmeter. The transistor needs to have something to complete the circuit from collector to gnd and the voltmeter's resistance is way too high. 10k would be fine. FWIW, Radio Shack has transistors and resistors. Not cheap but convenient.
     
  11. NeoXon

    Thread Starter New Member

    May 24, 2011
    9
    1
    Thanks, Jaguarjoe, and also thanks for the others who helped. I will use a reed relay. That is quite cheap and it is inaudible, it is totally OK (8mA in 0.5% of time is more than acceptable).

    You meant that load resistor across the voltmeter is needed only if a PNP transistor is used instead of a reed relay. If I used a reed relay the circuit I sketched is OK, nothing else (diode, snubber circuit, load resistor) is needed, right?

    Why is that if I used a transistor it needs an additional load resistor? Is it because it can't operate if the load current is too low (voltmeter has pretty big resistance)?

    Does the reed relay care about how small current flows on the load circuit? Will it perfectly switch +12V on the output even if only a voltmeter's huge resistance is the "load"?

    Sorry for asking very basic things, but I am a real beginner in this field.
     
  12. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    As you said, the voltmeter load is infinitesimal and the PNP needs a current path, thus a load resistor.

    The contacts on the reed relay will pass the whole 12v to your other circuit. No voltage loss (just about immeasurable, anyhow).

    The relay contact has a maximum current rating (1 amp, I think) which can not be exceeded. You're far, far away from that though.
     
  13. NeoXon

    Thread Starter New Member

    May 24, 2011
    9
    1
    Hi,

    I need some further help. I couldn't find that type of reed relay you linked in any stores near me, so I eventually bought one with 1000 Ohm coil (datasheet: http://www.meder.com/fileadmin/products/en_datasheets/3212100012e.pdf

    Now I decided to put a diode to make sure the control circuit is protected from back EMF. Just for safety, because I can't know if the originally existing control circuit may later on have some transistors... Those may be killed by large voltages.
    Anyways, I could not find on the datasheet how many volts the coil may produce when deenergized. I read somewhere it could be even a few hundred volts(?). What type of diode should I use for this purpose that can surive such back EMF?

    The other thing I don't understand is how this diode should be connected. Cathode (the marked end of the diode) to pin 86 (fused +12V), the anode to pin 85 (switched GND)? (For pinout see diagram in my first post.)

    Also it is not clear why this reed relay has 8 pins, instead of 4 or 5 as shown in the diagram of my first post. I read the datasheet but it would not give any information which pins correspond to 30, 85, 86 and 87. How should I know which pins are for controlling the relay and which ones are for breaking/making the load circuit?
     
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