How to activate a circuit when current flows through another given circuit ?

Discussion in 'The Projects Forum' started by 5416339, Aug 26, 2010.

  1. 5416339

    Thread Starter New Member

    Aug 25, 2010
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    The topic is quite clear ,So i was wondering how we can make such a circuit !
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Can you give us more info on the circuits?
    How much current is flowing in the "origenating" circuit?

    Bertus
     
    5416339 likes this.
  3. Wendy

    Moderator

    Mar 24, 2008
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    Actually the topic isn't clear. Don't assume, it doesn't work.

    I assume you want something like the output of a 555 to power a second circuit. Is this even close?

    Schematics and examples are your friend.
     
  4. 5416339

    Thread Starter New Member

    Aug 25, 2010
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    The originating current is about 5 A
     
  5. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Can you post a schematic of the originating circuit?
    Is it possible to connect anything to the circuit?

    Bertus
     
  6. 5416339

    Thread Starter New Member

    Aug 25, 2010
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  7. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The mobile bug you show in the link is never drawing 5 A.
    Can you give us more info.
    With the info provided we can not do anything.

    Bertus
     
  8. 5416339

    Thread Starter New Member

    Aug 25, 2010
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    Actually my aim is to amplify the PD across the LED terminals currently its just 0.1 V!

    http://www.physicsforums.com/showthread.php?p=2853797
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    A simple op amp such as the LM358 can amplify a small voltage like that, just set the gain to 10. Since your talking LEDs I am assuming DC voltage.

    PD? As in photo diode? If it is how does it apply with this?
     
  10. 5416339

    Thread Starter New Member

    Aug 25, 2010
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    Yes it is DC voltage but 10 time higher will not help 50 times is required !!
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    So change the value of the resistors. I'm assuming you know how op amps work.

    The reason suggested a LM358 is it can go very low to ground.

    A LM393 is a comparator, it can do the same thing, but it does not have an analog output. It is strictly on or off. It has an open collector output with an absolute max current rating of 16ma. Less is better, much better.
     
  12. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    I have some doubts with the circuit provided.
    The CA3130 will not work with the RF signal from the phone.
    Also the desciption in C3 is very strange, the capacitor wit its leads will never meet the frequency as given.

    [​IMG]
    Bertus
     
  13. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    This is a blast from the past! Just after the Korean War, lots of military components went on the surplus market. A man who styled himself "Madman Muntz" started putting out a line of very cheap TV's using surplus parts.

    The fun part was that many of the capacitors were old so-called "body end dot" types in phenolic packages - see the reference for "postage stamp" caps - http://en.wikipedia.org/wiki/Electronic_color_code . The individual capacitances were weenie, but could be made useful by soldering together 5 or 6 of them (possible in the days of point-to-point construction).

    Then he got carried away by finding out how many components could be left out and still have the set work. This led to the use of lead dress as a means of adjusting reactances in the set. It also meant they became effectively unrepairable, because disturbing the runs of wiring tended to make the set stop operating.

    Found a reference - http://www.freeenterpriseland.com/MUNTZ.html
     
  14. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    Then use different values in the op amp circuit to achieve the desired amount of gain.
     
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