Question about Pode Plotter reading interpretation?

Discussion in 'General Electronics Chat' started by samy555, Jun 20, 2013.

  1. samy555

    Thread Starter Active Member

    May 24, 2010
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    hello
    When the plotter read voltage gain = +3.979 dB @ 100MHz for the L-impedance matching circuit shown
    [​IMG]
    How much power from the source is on the R2 (the load resistance)? Thanks
     
    Last edited: Jun 20, 2013
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You've already placed a watt meter [XWM1] to monitor the load [R2] power. That power appears to be about 250mW or half the source power.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    BTW. If you are wondering why this equates to a voltage gain of 3.979dB remember that the "gain" would be the RMS voltage across load R2 divided by the input RMS voltage of 10V. The RMS voltage across R2 would be given by √(P[Watts]*1000[Ω]) ....
     
  4. samy555

    Thread Starter Active Member

    May 24, 2010
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    Thank you t_n_k
    It seems that I asked the question in the wrong way
    I ask you to give me a part of your time and try to understand my problem, because English is not my first language.

    I made some experiments on L-network Z matching, first I try the circuit without any match (R1 is source impedance and R2 is the load impedance), and got the following result:
    [​IMG]
    The voltage gain = -0.828 dB @ 100MHz
    I understand from the above result that:
    dB power gain = 0.5 dB voltage gain
    Our pode plotter gave us that the dB voltage gain = -0.828,
    so the dB powe gain = -0.414, and that’s mean P_load/P_source = 0.909
    So 90.9% of the source power transferred to the Load, and that's was OK.

    To make sure of the validity of the results I used this link
    http://www.daycounter.com/Calculators/Decibels-Calculator.phtml
    [​IMG]
    Now I'll introduce the L-impedance match (Which was calculated by someone else)
    [​IMG]
    The voltage gain = +3.979 dB @ 100MHz
    To do as before:
    dB power gain = 0.5 dB voltage gain
    Our pode plotter gave us that the dB voltage gain = +3.979,
    so the dB powe gain = +1.99, and that’s mean P_load/P_source = 1.58
    [​IMG]
    So 158% of the source power transferred to the Load and that is imposible.
    The wattmeters show 50% was transfred.
    Where the error in my concepts?
    Thank you
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    This illustrates the potential pitfalls of naively using a tool without thinking through the issues.

    The decibel calculator works fine if you are using a common impedance base for the power calculations - which is not the case in your problem. So the reference link used has no value as a validity check for your calculations.
     
  6. samy555

    Thread Starter Active Member

    May 24, 2010
    116
    3
    Okay, I completely agree with you
    Can you tell me how do I calculate the dB power gain from the value of the measured voltage gain by pode plotter
    thanks


    Note: I'm thinking about this problem a week ago, I could not find any solutions
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It's Bode plot by the way - not Pode plot.

    If the voltage gain is 3.979dB then we may write

    \text{Gain[dB]}=3.979=20\text{*}\log(\frac{V_2}{V_1})

    enabling one to find V2 as

    V_2=V_1 \text{*antilog}(\frac{3.979}{20})=10 \text{*}1.581=15.81 \ \text{Volts} \  \( \text{rms} \)

    The power in R2 is then simply

    P_2=\frac{V_2^2}{R_2}=\frac{15.81^2}{1000}=0.25 \ \text{W}

    Since the source is now matched to the load it will see an effective load of 100Ω + 100Ω or 200Ω. The source power will therefore be

    P_{\text{source}}=\frac{V_1^2}{200}=0.5 \ \text{W}

    One could notionally use these two power values to propose a power gain value using the relationship

    \text{P}_{\text{gain}}=10*\text{log}\(\frac{P_2}{P_{\text{source}}}\)

    However, I have an ongoing concern about such an expression where the impedance base is changed but this may be "OK" (with the aforementioned reservations in mind), given the LC matching network is interposed between the source and the load.
     
    Last edited: Jun 21, 2013
    samy555 likes this.
  8. t_n_k

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    Mar 6, 2009
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    As a further comment I am inclined to regard what you are doing as very much like homework. If it is in fact homework, I suggest you use the Homework forum to post in future and observe the rule regarding demonstration of some personal effort in attempting solve the problem.
     
    samy555 likes this.
  9. samy555

    Thread Starter Active Member

    May 24, 2010
    116
    3
    To the esteemed Mr. t_n_k
    Thank you very much and I wish you a happy life
     
  10. samy555

    Thread Starter Active Member

    May 24, 2010
    116
    3
    Came to my mind two questions if you allow me:
    1 - When the source is fully matched to the load, is it always true that p2/P_source = 0.5 i.e -6 dB? and when matching is done, maximum power transfered to the load = 0.5 P_source?
    2- In the above calculations, I am surprised that V2 =15.81 Vrms where V_source = only 10 Vrms? Where this increase come from? Is not the power reserved?

    thanks
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    1. Yes - but it would be -3dB

    2. The matching network provides an impedance transformation. So one sees different parameters depending upon which port of the network one is looking into. There is no discrepancy with respect to the power distribution in the circuit. My earlier post provides the supporting data for this.
     
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