DC voltage regulator after the load

Discussion in 'General Electronics Chat' started by bharesc, Oct 8, 2010.

  1. bharesc

    Thread Starter New Member

    Oct 8, 2010
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    I have 20 illuminated three terminal rocker switches that will have their input from a 12V bus bar. When switched on the internal lamp of each switch is fed from this 12V supply and is grounded through the third switch terminal. I want to dim the lights without using resistors on each of the switches. With a volt drop of about 5 volts across the lamp I get the correct illumination. I found it easy to drop the voltage using a fixed positive voltage regulator (National Semiconductors LM2940CT-5.0) before the load but can't think of a similar solution for after the load.

    Any suggestions.
     
  2. tom66

    Senior Member

    May 9, 2009
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    I don't understand why you don't want to use resistors when the linear regulator is effectively an "active" resistor which wastes power as heat to get the correct voltage drop.
     
  3. KMoffett

    AAC Fanatic!

    Dec 19, 2007
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    Use an LM317L as a 7V regulator. Put the "ground" terminal of all the switches to the (+) output of the regulator. 12v-7V=5V :)

    Ken
     
  4. bharesc

    Thread Starter New Member

    Oct 8, 2010
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    Thanks for your replies.

    Tom66. The application is for a yacht's instrument panel and I didn't want 20 resistors connecting to a bus bar as well as the normal illumination ground bus bar. The back of the instrument panel needs to be pretty rugged in case of repair on the high seas.

    Ken. The LM317L is an adjustable voltage regulator. I presume the input terminal is fed by the 12v and the circuit would look like fig 1 in the National Semiconductors LM317 data sheet. If I was happy with 4.5V on the lamps I guess I could use a 7.5V fixed voltage regulator.

    Bob
     
  5. bharesc

    Thread Starter New Member

    Oct 8, 2010
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    Ken
    Just tried it with a fixed voltage regulator and it worked fine. However I needed to have a load across the output and ground (a LED) to effect this. I presume it was needed to trigger the regulator. The regulator was an LM2940CT. Wrong voltage but proved the point.
    Bob
     
  6. cumesoftware

    Senior Member

    Apr 27, 2007
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    I don't understand why using a low-dropout regulator if you have lots of headroom to use a regulator like the LM7805. The LM2940 brings only advantage in situations where the input voltage is very close to the targed output voltage. In your case, using a LM2940CT would not bring any advantage since it is lots more unstable than a regulator like the LM7805 and requires hard to find low ESR capacitors. On the other hand, the LM7805 doesn't require any special components. And I think the LM7805 is cheaper. ;)

    By the way, how many amps do you need? Are you planning to use a regulator for each lamp, or one regulator for all lamps? :eek:
     
  7. eblc1388

    Senior Member

    Nov 28, 2008
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    Not really, no.

    A LM317 regulator, or any common positive output voltage regulator can only *SOURCE* current but not sink current.

    You'll need a circuit(shunt regulator) that sinks enough current to maintain its terminal voltage at seven volt.
     
    Last edited: Oct 9, 2010
  8. retched

    AAC Fanatic!

    Dec 5, 2009
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    Some voltage regulators require a "minimum" load before they regulate to the published voltage. This can often be overcome by simply using an "indicator LED". That will provide the draw on the regulator to provide a happy voltage.

    Would a typical diode drop after the switched event, give the required voltage drop to offer the wanted illumination?

    And also, You said they have a lamp... Are they incandescents or LEDs?
     
  9. cumesoftware

    Senior Member

    Apr 27, 2007
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    You are right, I didn't knew that the lamps were all connected to a common (+).

    How about using an LM7905? It will work if your load is directly connected to the plus terminal, and if you need to regulate the voltage from there. All you need to do is to connect the common of the regulator to (+) and the input to ground.

    After all all you need is 4.5V going to the lamps, right? Using a 7V regulator to subtract 7V from 12V is a bad idea, since the voltage going from the battery will vary. A current regulator is not a good idea either.
     
    Last edited: Oct 9, 2010
  10. Wendy

    Moderator

    Mar 24, 2008
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    Actually, this is not true.

    [​IMG]
     
  11. eblc1388

    Senior Member

    Nov 28, 2008
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    That works. Excellent simple idea by cumesoftware.

    @bharesc, this is how you would connect up the circuit.

    [​IMG]
     
    Last edited: Oct 9, 2010
  12. eblc1388

    Senior Member

    Nov 28, 2008
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    Oh, dear.

    What I've meant was the output pin of a positive voltage regulator can only *Source* current but cannot sink current.
     
  13. cumesoftware

    Senior Member

    Apr 27, 2007
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    Exactly what I had in mind. Thanks for posting!

    Now make sure that the LM7905 has a proper heatsink, and that is it! :)
     
  14. eblc1388

    Senior Member

    Nov 28, 2008
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    Missed the back feed condition via the load. :eek:

    1N4001 diodes added accordingly.
     
  15. bharesc

    Thread Starter New Member

    Oct 8, 2010
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    I only used the LM2940 because that was all I had. I had stability problems but this seemed to be solved by sticking a 37μf capacitor across the output.
    Thanks for the circuit diagram, ebls1388, and the idea from cumesoftware.
    I can source a L7905CV locally, will this be ok?
    The current for each lamp appears to be around 25 mA and I don't know whether they are incandescent or LEDS.
    It might be useful to be able to vary the illumination. What would be the circuit for using an adjustable voltage regulator in place of the 7905?
    Bob
     
  16. eblc1388

    Senior Member

    Nov 28, 2008
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    Yes, that will do.

    You can use the LM337 to replace 7905. That would allow you to tune the voltage from about 2 volt to 10V across the lamp/LED.

    [​IMG]
     
    tom66 likes this.
  17. bharesc

    Thread Starter New Member

    Oct 8, 2010
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    Thanks again LC. I will build the circuit when I have the bits.
    Bob
     
  18. KMoffett

    AAC Fanatic!

    Dec 19, 2007
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    I agree with L.Chung. A shunt voltage regulator makes the most sense. Though Bill is correct...it only sinks in the constant current mode. Since the OP will be switching on and off several different parallel LED/Resistor sets the shunt regulator will sink all need current for any combination, while maintaining the 7V.

    I don't know what I was thinking. :(

    Ken
     
  19. bharesc

    Thread Starter New Member

    Oct 8, 2010
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    Success. I have only connected one switch up as yet and it works perfectly.
    One further question. I buzzed out the switch and couldn't understand the contact arrangement so I flicked it apart. Its quite clever as the line voltage is isolated from both the load and the lamp in the off position and connects to them both in the on position. There should therefore be no back feed through switches that are not on. Do I need the IN4001 diodes?
    Bob
     
  20. eblc1388

    Senior Member

    Nov 28, 2008
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    I'm glad the circuit works.

    No. You don't need the diodes if there are no back feed from the load.

    [​IMG]
     
    Last edited: Oct 12, 2010
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