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Over-voltage & over-current Protection

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Discussion in 'Electronics Homework Help' started by mrkbuddy, May 18, 2013.

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  1. May 18, 2013 #1
    mrkbuddy

    mrkbuddy

    3
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    May 2, 2013
    Please help me out to calculate the values of R1 & R2. what is the formula or how to derive it. i am attaching the circuit design. thanks

    Design task:
    Target specification for Current Limiting circuit
    (i) Io [max allowed current] = 2A
    (ii) Isc [fold back current] = 0.5A
    (iii) Regulated output voltage Vo = 12V

    d. Calculate the value of Rs.
    Io = 2A
    VbeM = 0.6A

    From Ohm’s Law
    Rs = VbeM¬¬ / Io
    Rs = 0.6 / 2
    Rs = 0.3Ω

    e. Calculate the values of R1 & R2 to achieve the above data?

    [​IMG]
     
  2. May 18, 2013 #2
    duke37

    duke37

    5,364
    769
    Jan 9, 2011
    Please tell us how the circuit should work.
    As drawn, R1 and R2 just dump power. Is this what is wanted?

    It looks as if Qm is fighting the output of the op-amp.

    There is a connection between op-amp- and Qm base, both of these are inputs so where is the signal coming from?
     
  3. May 18, 2013 #3
    (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,452
    2,812
    Jan 21, 2010
    A very round about one. It's confusing. At first I thought the voltage divider was connected to it.
     
  4. Jun 12, 2013 #4
    Merlin3189

    Merlin3189

    250
    69
    Aug 4, 2011
    Maybe R1=R2 and the base of Qm and the inverting input of op amp should be connected to this point.
    Then the op amp is comparing 6V with half the output: so output =12V.

    Rs is a current sense for short circuit of the output, developing a Voltage across the base-emitter of Qm. If the current drawn is enough to pull the emitter of Qm 0.7V below the base, then Qm switches on and reduces drive to Q1
     
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