Voltage scaling

Discussion in 'Homework Help' started by jrekha, Apr 11, 2010.

  1. jrekha

    Thread Starter New Member

    Mar 31, 2010
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    I have voltage measured in range 3.2V -4.2V.
    I need to convert this to range 0-10V using opamp circuit.Can anyone help me realise this.
     
  2. Jony130

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    Feb 17, 2009
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    So input voltage change from 3.2V to 4.2V or from - 4.2V to 3.2V ?
     
  3. jrekha

    Thread Starter New Member

    Mar 31, 2010
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    input will be 3.2V to 4.2V

    Actually my original circuit is like i ve 32V to 42V which need be scaled to 0-10V. I used a differential opamp circuit to reduce the voltage to 3.2V-4.2V range.Now i need to change to 0-10V.

    I need to use opamp, and in the first stage i had to use differential opamp .so did like that. anythng wrong?or is ther an alternate way?
     
  4. Jony130

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    And what is the supply voltage
     
  5. jrekha

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    Mar 31, 2010
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    15V is the supply
     
  6. Jony130

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    So you need this circuit:

    [​IMG]

    Resistors R3 and R4 need to be very accurate.
     
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  7. jrekha

    Thread Starter New Member

    Mar 31, 2010
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    can i use lm324?
     
  8. Jony130

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    LM324 is a quadruple version of a LM358.
    So yes you can use it.
     
  9. jrekha

    Thread Starter New Member

    Mar 31, 2010
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    is ther any steps to arrive at these Resistor values?i mean design, so that i cud know the circuit more.
     
  10. jrekha

    Thread Starter New Member

    Mar 31, 2010
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    how do u come with those Resistor values, please tell me too..
     
  11. Jony130

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    Feb 17, 2009
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    The voltage gain mus be equal :
    Au = (10V - 0V) / (4.2V - 3.2V) = 10V/1V = 10[V/V]
    So gain of a non-inverting amplifier is equal:
    Ku = 1 + (R1/R2) then R1/R2 = 9
    so R1 = 9*R2 so i start with R2 = 20KΩ
    R1 = 9 * 20K = 180KΩ.

    And we wont for Vin = 3.2V Vout = 0V and this can only be done when non-invent input "+"V is equal inventing input voltage "-"V.
    So we have this situation:

    [​IMG]

    So voltage divider voltage must be larger then 3.2V by voltage drop on a R2.

    Vd = 3.2V + I * R2

    I = 3.2V /180KΩ = 17.77uA then

    Vd = 3.2V + 17.77uA * 20KΩ = 3.2V + 0.355V = 3.555V

    So we must design voltage divider that deliver 3.555V

    Voltage on R3 = 15V - 3.555V = 14.445V
    14.445V/ 3.555V = 3.2194 = R3/R4
    R3 = 3.2194 * R4 = 470Ω * 3.2194 = 1.513K = 1.5K

    I start with low value for a R4 because current that is flow through R3 and R4 must be much large then current that is flow through R1, R2.
     
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    Last edited: Apr 12, 2010
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