forming a op-amp circuit formula

Discussion in 'General Electronics Chat' started by ws0822, Feb 8, 2011.

  1. ws0822

    Thread Starter New Member

    Jul 18, 2010
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    can someone show me the formula of IL in the figure [​IMG]in term of Vin??
    I formed an long formula and wanna chk whether is it correct
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    Post your work here to see if it is correct.
     
  3. AdrianN

    Active Member

    Apr 27, 2009
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    The easiest way to calculate this is to apply Thevenin's theorem across RL. After calculations, the result is:

    IL = Vin * (K2 / (K2 * R1 + RL))

    where

    K2 = (1 - K1) / (1 - K1/K), and

    K1 = R1 / (R1 + R2)

    K = R4 / (R3 + R4)
     
  4. ws0822

    Thread Starter New Member

    Jul 18, 2010
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    thx for the replies...

    according to wat i hv done, i used nodal analysis and i get a formula of

    IL =( Vin*(R2/R1)*y)/(1-x*y*RL)

    where x = 1+ (R3/R4)
    and y = R1/(R1R2+R2RL+R1RL)

    is it correct??

    bwt, adriann how u use Thevenin's theorem to solve?? u open circuit RL then find Vth and Rth??
     
  5. ws0822

    Thread Starter New Member

    Jul 18, 2010
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    another thing... how to verify that the circuit is a negative impedance converter(VDCS)? can i just use different voltage to test the circuit and measure the current then plot the graph to so than the V-I graph having negative gradient (negative resistance) to verify it?? can anyone suggests any better method??
     
  6. AdrianN

    Active Member

    Apr 27, 2009
    97
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    Yes, it is, I verified it. However, it is difficult to see how the circuit behaves, when the result is presented in this format.

    I will post the solution soon. Stand by.

    I assume you refer to the resistance seen by Vin. If the current flows into Vin, the resistance seen by Vin is negative. For this, all you have to do is prove that the voltage drop across RL is greater than Vin. In other words

    IL * RL > Vin.

    In this case, R1 current flows into Vin. To see this, let’s look at the K, K1 and K2 ratios in the result I posted earlier. This is the reason why I like to leave formulas with resistor ratios like K and K1. That way I can better see what is going on as resistors change. Here is how I would analyze the result I posted earlier, aside from plotting I1 versus Vin.

    1. Assume that R2 is infinity (not connected). In that case, K1 is zero and K2 is 1, so IL becomes Vin / (R1 + RL) and the whole circuit is a regular non-inverting amplifier.
    2. Assume that R2 is connected. As R2 decreases, K2 increases and IL*RL increases.
    3. This circuit presents a negative resistance as R2 decreases furthermore. As R2 decreases K1 (which is sub unity) increases and, at some point 1-K1/K becomes negative, since K is a constant. As a consequence, K2 becomes negative, and IL*RL becomes greater than Vin. Therefore, R1 current flows into Vin and increases as R2 decreases. From Vin point of view, the circuit is a negative resistance.
    4. As R2 decreases more, at some point Vo saturates to the op amp supply and the current supplied by op amp through R2 into Vin cannot increase any more.

    I do not know your resistor values and what you are trying to accomplish. This was just a sample of the analysis you can make. Whenever you see a difference at a denominator (like K2) you should assume that you will see some instability for some values.
     
  7. ws0822

    Thread Starter New Member

    Jul 18, 2010
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    my lecturer wan us to verify the circuit capability(it is to produce voltage dependent current source?) and its limitation(for practical only) with using circuit analysis and experimental results.

    Any idea??
     
    Last edited: Feb 12, 2011
  8. AdrianN

    Active Member

    Apr 27, 2009
    97
    1
    I posted the solution here:
    http://masteringelectronicsdesign.c...eorem-to-solve-a-negative-resistance-circuit/

    I think you have plenty of info now to come up with an analysis.
     
  9. ws0822

    Thread Starter New Member

    Jul 18, 2010
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  10. AdrianN

    Active Member

    Apr 27, 2009
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    Again, you can analyze it yourself. Look at the IL expression. As long as it depends on RL, the circuit is not a current source. It can become a current source when K2 is large. In that case IL = Vin / R1. So, yes, at RL connections, the circuit becomes a voltage dependent current source. For example, if R1,2,3,4 = 10k, and Vin = 1V, IL is 0.1mA and does not depend on RL.
     
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