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Circuit design, 2 equations, 3 unknowns

  1. Mar 23, 2007 #1
    I'm working with this circuit to design an ambient light meter.
    http://img201.imageshack.us/img201/6479/cktyu0.jpg [Broken]

    We were given steps to go about solving for different values of the components in the circuit.

    First we're to solve for the gain and thus V1 which I found to be

    Then we're to solve for V2, which I found by NVA:
    [tex] V_2 = \frac{R3*V1+R2*Vee}{(R3+R2)}[/tex]

    I don't think there is anything wrong with either of those equations, as I've verified that they give the correct voltages when simulating the circuit in PSpice.

    The given\known values in the circuit are Vee=-10V, R1=1k and the condition when Rphoto=400, V2 = 8V and when Rphoto=10k, V2=0V

    This leads to:
    [tex]8 = \frac{R3*\frac{-1000 \Omega}{(400 \Omega+R7)}*-10+R2*-10}{(R3+R2)}[/tex]
    [tex]0 = \frac{R3*\frac{-1000 \Omega}{(10000 \Omega+R7)}*-10+R2*-10}{(R3+R2)}[/tex]

    With those 2 equations, I am to deduce values for R2, R3, and R7, which I can't seem to do having only the 2 equations. Can anyone push me in the right direction? Thanks.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 23, 2007 #2
    Since there are two equations and three unknowns, i would derive the third equation. For instance, derive an equation that defines the relation between V2 and output.
  4. Mar 23, 2007 #3
    Hmm, which output? I'm not sure I see another equation for V2 that includes those unknown resistors.
  5. Mar 23, 2007 #4
    Hmm, now I'm completely lost. How did you derive the last two equations? Another question, you're using voltage-divider for v1, how come there are no R1 in the denominator?

    P.S I was talking about writing a equ. for the feedback loop for the second opamp.
  6. Mar 23, 2007 #5
    The last 2 equations are simply V2 with the known values plugged in. As for V1, I solved that by analyzing the left-most op-amp by itself.

    Using nva:

    [tex]\frac{0-Vee}{(Rphoto+R7)} + 0 + \frac{0-V_1}{R1}=0[/tex]
    [tex]\frac{-Vee}{(Rphoto+R7)} = \frac{V_1}{R1}[/tex]
    [tex]\therefore \frac{-R1}{(Rphoto+R7)}*Vee=V_1[/tex]

    I went on to solve for V2 in a similar manner.

    I was able to solve my original problem at school this afternoon. It turns out that R7 can be solved for directly by combining the two V2 conditional equations, and then the other values are fairly trivial to find.
    Last edited: Mar 23, 2007
  7. Mar 24, 2007 #6
    I was about to propose the same thing, but i thought that you've already tried it.
  8. Mar 26, 2007 #7


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    If you have more unknowns than equations, you can attribute an arbitrary value to one of the unknowns and solve for the other two.
    Since R7 is in series with Rphoto, it must not be much greater than the minimum value of Rphoto nor much smaller than its maximum value. A good guess would be the geometrical mean between 400 and 10000.
  9. Apr 4, 2007 #8


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    That looks like one of the designs I had to do in my engineering courses as an undergrad. What school do you go to?
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