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The NØPU Big Quiz

Discussion in 'General Technical Questions and Answers' started by N0PU, Jan 9, 2003.

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  1. AE4FA

    AE4FA Ham Member QRZ Page

    I thought it would be good for me to work thru a couple ‘frinstances:

    R = 50 Ohms
    .01mH = 0.000010 or 10*10^-6
    .01uF = 0.000000010 or 10*10^-9
    ƒr = 503000Hz or 503kHz

    At resonance: Z = (31.6-31.6)+50 = 50 Ohms
    At 12V input, current would be 12/50 or 240mA
    EL = .240*31.6 = 7.584V
    EC = .240*31.6 = 7.584V
    ER = .240*50 = 12V
    Power Out = 12*.240 = 2.88W
    0W dissipated as heat by reactances

    Shifting operating frequency to 400kHz:
    XL = 25.133 Ohms
    XC = 39.789 Ohms
    Z = (39.789-25.133)+50 = 64.656 Ohms
    At 12V in, current would be 186mA
    EL = .186*25.133 = 4.675V
    EC = .186*39.789 = 7.4V
    ER = .186*50 = 9.3V
    Power Out = 9.3*.186 = 1.73W
    0.5W dissipated as heat by reactances

    Shifting frequency to 800kHz:
    XL = 50.266 Ohms
    XC = 19.894 Ohms
    Z = [50.266-19.894]+50 = 80.372 Ohms
    At 12V in, current would be 149mA
    EL = .149*50.266 = 7.490V
    EC = .149*19.894 = 2.964V
    ER = .149*50 = 7.45V
    Power Out = 7.45*.149 = 1.1W
    0.67W dissipated as heat by reactance

    So, we see that both voltage and current through the circuit both decrease – of course producing less power at the output – as the operating frequency moves away from the circuit’s resonant frequency. And, more power is dissipated as heat in the circuit. Direction doesn’t matter.
     
  2. N0PU

    N0PU Ham Member QRZ Page

    </span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td>Quote (ae4fa @ June 09 2003,17:32)</td></tr><tr><td id="QUOTE">I thought it would be good for me to work thru a couple ‘frinstances:

    R = 50 Ohms
    .01mH = 0.000010 or 10*10^-6
    .01uF = 0.000000010 or 10*10^-9
    ƒr = 503000Hz or 503kHz

    At resonance: Z = (31.6-31.6)+50 = 50 Ohms
    At 12V input, current would be 12/50 or 240mA
    EL = .240*31.6 = 7.584V
    EC = .240*31.6 = 7.584V
    ER = .240*50 = 12V
    Power Out = 12*.240 = 2.88W
    0W dissipated as heat by reactances

    Shifting operating frequency to 400kHz:
    XL = 25.133 Ohms
    XC = 39.789 Ohms
    Z = (39.789-25.133)+50 = 64.656 Ohms
    At 12V in, current would be 186mA
    EL = .186*25.133 = 4.675V
    EC = .186*39.789 = 7.4V
    ER = .186*50 = 9.3V
    Power Out = 9.3*.186 = 1.73W
    0.5W dissipated as heat by reactances

    Shifting frequency to 800kHz:
    XL = 50.266 Ohms
    XC = 19.894 Ohms
    Z = [50.266-19.894]+50 = 80.372 Ohms
    At 12V in, current would be 149mA
    EL = .149*50.266 = 7.490V
    EC = .149*19.894 = 2.964V
    ER = .149*50 = 7.45V
    Power Out = 7.45*.149 = 1.1W
    1.79W dissipated as heat by reactance

    So, we see that both voltage and current through the circuit both decrease – of course producing less power at the output – as the operating frequency moves away from the circuit’s resonant frequency.  And, more power is dissipated as heat in the circuit.  Direction doesn’t matter.[/QUOTE]<span id='postcolor'>
    Right ...
    FA: 10 points
    ------------------------

    Question 249

    How does the ideal series resonant circuit differ from a one with 'some' R?
     
  3. AE4FA

    AE4FA Ham Member QRZ Page

    An “ideal” circuit has no R – and therefore, at resonance, infinite current flows. At resonance, the reactances cancel – and, since there is no resistance, Z=Ø and I=E/Z yields infinite current. Probably not a good situation, really, given other limitations - but impossible in any case, since every circuit has resistance.

    If XL and XC are equal, say 31.6 Ohms each, the circuit is resonant. And, if R=Ø (as in the &quot;ideal&quot;), then Z=(100-100)+Ø=Ø Ohms.

    In the same situation, with 50 Ohms resistance: Z=(100-100)+50=50.Ohms

    At resonance, Z always = R, not less, not more. In other words, the value of R is the lowest impedance figure possible, and it is achieved only when the circuit is resonant.

    For demonstration, this can be plotted out in a number of ways. This would clearly show that impedance (Z) is at minimum (and equal to R) only at the point of resonance, where the XL and XC lines cross.
     
  4. GM3ZMA

    GM3ZMA Ham Member QRZ Page

    Harry is taking his time getting back to this, maybe he is looking for a bit more info?

    Lets try:

    A series resonant circuit with no &quot;R&quot; will have no losses, it will not dissipate any power.
    At resonance it will be a true short circuit, zero ohms impedance.

    The &quot;Q&quot; of the circuit will be infinite, and the bandwidth will be zero!

    Consider the formula:

    Q = (resonant freq)/Bandwidth

    turn it around

    Bandwidth = (resonant freq)/Q


    if Q is infinite, Bandwidth = 0

    Jim GM3ZMA
     
  5. N0PU

    N0PU Ham Member QRZ Page

    BOTH OF YOU ARE RIGHT...

    All hands please stand-by...

    I'm not ignoring the QUIZ but have had some serious storms here and had power loss for over 12 hours and lost 2 servers to lightning... I am buried right now and this has to wait...

    I am also not completely satisfied with the text coverage of tuned circuits and am re-writing a portion of it to include an understanding of vectors and j factor... I feel that j factor is important at this point especially to those working with antennas and IMHO the text does a lousy job of it...

    I'll be back in a few days after I scrape all the silicon gobs off the ceiling and get some other stuff fixed...

    Thanks for your patience...
     
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