Why does my potentiometer make smoke?

Discussion in 'General Electronics Chat' started by booboo, Jun 15, 2015.

  1. booboo

    Thread Starter Member

    Apr 25, 2015
    165
    2
    Hi fellas
    I'm trying to change this circuit to detect IR and get a 3v3 at output(instead of the LED) to connect it to my MCU:

    [​IMG]
    My circuit is like this but it doesn't has 1k resistor(What's the usage of this resistor?) and supply is 7.5v DC and transistor is bc547. for LED I use a red LED and 320ohm resis' instead of 100ohm resis'. I used a pot to change the voltage that the IR sensor(diode) draw. this was just for a test (I know it's not wisely). when I turn on the circuit and decrease the pot, it makes smoke! why?

    And also any seggestion for improve the circuit?
     
  2. #12

    Expert

    Nov 30, 2010
    16,298
    6,809
    If the 1K resistor is changed to zero, you have no linear current limiters. Add 100 ohms in series so it is impossible to adjust to zero ohms.
     
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  3. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,809
    834
    What #12 said. I don't clearly get it either, so could you draw a schematic showing EXACTLY what you have wired up?
     
  4. booboo

    Thread Starter Member

    Apr 25, 2015
    165
    2
    Like me.;):D
    Wasn't my explanation enough clear? ok, let me take a photo of my circuit. there you go:

    [​IMG]
     
  5. Reloadron

    Active Member

    Jan 15, 2015
    963
    232
    #12 covered the smoke. This is your circuit as it appears on the Internet. Since you mention 3.3 volts possibly the circuit will be used to drive a uC (micro-controller). The circuit itself is rather basic and if you want a 3.3 volt output from an IR detector using an IR diode theink about using a circuit like these types. Your pot is smoking because with nothing to limit current as the R value is decreased the I value increases till something has to give so the pot sees excessive current and burns up.

    Ron
     
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  6. crutschow

    Expert

    Mar 14, 2008
    13,014
    3,234
    No your explanation was not clear enough (where did you connect the pot?) and the picture does not show that either.
    Please post a schematic.
     
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  7. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
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    ...To avoid future agro, better toss the (now) damaged potentiometer...

    TTFN
    HP
     
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  8. booboo

    Thread Starter Member

    Apr 25, 2015
    165
    2
    Thanks Ron for perfect answer.
    Ok,

    [​IMG]
     
  9. booboo

    Thread Starter Member

    Apr 25, 2015
    165
    2
    I just have IR sensor (I do not have phototransistor in stock).
     
  10. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,388
    1,605
    Where does the ground go to?

    If it goes to the negative of the 7.5V then there is one source of smoke: you have a fixed voltage across a small to zero resistance.

    Please don't make us guess any further as to your intent. Make a full and complete schematic.
     
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  11. Reloadron

    Active Member

    Jan 15, 2015
    963
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    Well if you have a link to your sensor data sheet things might go much better.

    Ron
     
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  12. booboo

    Thread Starter Member

    Apr 25, 2015
    165
    2
    Ops :oops:, sorry. yes, it goes to the negative of the 7.5V.

    Unfortunately the vendor that I have purchased this sensor from doesn't provide any datasheet for this diode.:(
     
  13. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    If your sensor is a photodiode you may not get enough transistor base current to drive a LED properly. But if you just want to provide a signal to a micro you could try this arrangement :-
    Photosensor.gif
    R1 is a pull-down resistor to reduce the chance of spurious noise turning Q1 on. R2 and R3 form a voltage divider to ensure no more than ~3V gets to the micro input (assuming the micro has a 3.3V supply).
     
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  14. booboo

    Thread Starter Member

    Apr 25, 2015
    165
    2
    Thank you but that Ground symbol confused me (below R1). for what is that?
     
  15. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
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    It merely indicates connection to the 'common return' --- For the purposes at hand you may ignore it:)

    Alternatively, you may 'erase' the horizontal line at the bottom of the diagram and similarly draft the 'lower end' of R3 and the negative line from V1...

    TTFN
    HP
     
    Last edited: Jun 16, 2015
  16. booboo

    Thread Starter Member

    Apr 25, 2015
    165
    2
    Thanks a lot for schematic.
    I wired up it and now it's working very well.;)
    Thank you Hypatia's Protege for reply.
     
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