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Amplifier circuit problems

  1. Jan 4, 2007 #1
    1. The problem statement, all variables and given/known data

    One of the problems question is: Determine the analitical expression that relates the resistive value of the NTC thermistor with the tension Vt.I've come to an expression but I'm not sure if it's correct.Also, the other question of the problem is: Make an analitic description of the circuit.And I'm not sure what they meant to say with "analitical discription of the circuit".

    2. Relevant equations



    3. The attempt at a solution

    For my first question the expression,which results on the attempt to the solution is: Vt = 5 + R*(-I3) + NTC*(-INTC)
     

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    Last edited: Jan 4, 2007
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  3. Jan 5, 2007 #2

    berkeman

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    How come R1 and R2 do not show up in your solution for the output voltage Vt?
     
  4. Jan 5, 2007 #3
    So let see if I get this:

    Since it's negative feedback Va=Vb and the equations for the circuit would be:

    IR + INTC =0
    IR1 = IR2
    Vt - Vb =INTC*RNTC
    5 - Vb= R*IR
    5 - Va=IR1*R1
    Va - 5=IR2*R2

    So, the final equation would be Vt=5+R*((5-Vb)/R) +INTC*RNTC
    Is this right?
     
  5. Jan 5, 2007 #4

    berkeman

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    Sorry, I'm having a hard time tracking your work. Maybe call the input voltages to the amp Vp and Vn to make it more intuitive, and be careful about the current directions and signs.
     
  6. Jan 5, 2007 #5
    Sorry that it looks so confusing!So my equations with Vp and Vn should be:

    IR + INTC=0
    I1=-I2
    Vt-Vn=RNTC*INTC
    5 - Vn=R*IR
    5-Vp=R1*I1
    Vp-5=R2*I2

    INTC=The current on the NTC resistance
    IR=The current on the R resistance
    I1=The current on the R1 resistance
    I2=The current on the R2 resistance

    Am I on the right track?
     
  7. Jan 5, 2007 #6

    berkeman

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    It still looks to me like you are switching direction definitions on the currents sometimes between different equations, but I could be wrong. Just be sure to draw current arrows and be consistent, and you'll be fine. You're on the right track.

    I have to bail for a few hours. I'll try to check back later.
     
  8. Jan 5, 2007 #7
    I don't know why,but with these equations,the value of Vp and Vt is 5v!
     
  9. Jan 5, 2007 #8
    I keep getting instead of an expression for vp and vt the value 5v!
     
  10. Jan 5, 2007 #9
    I have another question: IS the voltage in the point where there's a wire connected to the positive input 5v?
     
  11. Jan 5, 2007 #10

    berkeman

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    No. There is a voltage divider with R1 and R2 that presents a lower voltage than 5V to the + input (Vp), and the gain of the amp will hold Vn at that same voltage.

    So start with more traditional KCL equations and voltage divider equations like these:

    (1) [tex]\frac{V_n-5V}{R} = \frac{V_n-V_t}{R_{ntc}}[/tex]

    (2) [tex]V_n = V_p = \frac{5V * R_2}{R_1 + R_2}[/tex]

    Then combine these and solve for Vt in terms of the other variables.
     
  12. Jan 5, 2007 #11
    I'm not quite understanding how did you get that expression for Vp...The current that goes through R1 and R2 should have the following expression:
    I(R1+R2) =5(R1+R2) and the voltage on Vp=R2*I2

    Is this correct?
     
  13. Jan 5, 2007 #12

    berkeman

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    Check your units. How in the world could a current be equal to a voltage? Vp is determined by the stiff voltage source V1 and the voltage divider with R1 and R2.
     
  14. Jan 5, 2007 #13
    So, we have a current that goes through the resistance R and leaves through R1 and through the voltage source.That same current that goes through R1,goes thourgh R2 too.So, for the current that goes through those resistances we would have the formula: 5-0/(r1 + r2)=Ir1 + r2 .And what about the current that goes through the voltage source?
     
  15. Jan 6, 2007 #14
    I now know how to find the value for Vp and how berkeman found it:

    IR1=(5-Vp)/R1
    IR2=Vp/R2
    IR1=IR2 and solve for Vp

    Still I'm not understanding why the expression IR=Iv1 + I1 (the currents entering the node equals currents leaving the node and Iv1 is the current that goes through voltage source) isn't considered!
     
  16. Jan 6, 2007 #15

    berkeman

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    Sorry, I'm not understanding what you are asking in the last part. Do you see where I got my equation #1? Since you know Vp and Vn now, equation #1 should be enough for you to solve for Vt.
     
  17. Jan 6, 2007 #16
    I don't understand why there isn't an expression for the current entering the node where it(the current) divides to the voltage source and to R1
     
  18. Jan 8, 2007 #17

    berkeman

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    You can go ahead and write an equation for that, but I don't think it will help in solving for the output voltage. V1 is shown as a stiff voltage source, so it alone defines the voltage across R1 and R2. And that voltage divider R1 R2 alone defines the Vp input voltage. And since this is an ideal opamp, the Vp voltage alone devines Vn. And Vn along with R and Rntc alone define the output voltage.

    You need to learn to look at a circuit like this and see what the defining things are, and not worry about other items. (Assuming that I haven't missed something, of course.)
     
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