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  • Qiaozhi
    replied
    Originally posted by golfnut View Post
    Yes, Tx and Rx frequencies are different.

    Returns from an energised target get through the Rx chain as an I.F. signal after the mixer of

    = 1.76kHz

    Bit like a superhet Radio.

    The bit I dont get is ... I assume Rx listens to signals from an energised ringing target - which re-radiates a component different to what it was energised with.

    IE we energise taget with 14kHz, we listen on 16kHz

    Any takers for this last bit?

    Steve
    A lot of people get confused by this.

    One thing you must understand is that, even though the RX coil is tuned to 16kHz, the signal received is still 14kHz. Think of the RX coil / capacitor as a filter. You put 14kHz in, and you get 14kHz out, albeit shifted in phase and a lower amplitude.

    Since the sample pulses for the synchronous demodulators (analog switches - JETs, in this case) are triggered from the TX oscillator, the RX signal needs to be phase-shifted to allow correct sampling of the in-phase and quadrature components. This is the reason for the different frequencies. It has nothing to do with creating an intermediate frequency, as per superhet designs.

    Leave a comment:


  • golfnut
    replied
    Freq tune

    I did,

    Freq = 1 / 2 PI SQRT(L Ctot) Units are Hertz (Hz) or cycles per second if you prefer.


    where Ctot is combo of C1,2

    C tot = (220x22)/(220+22) - Product / sum


    You Must :-
    work out Top row, get the number.
    Work out the bottom row, get the number.
    Then do the divide.

    C tot - Should come out as 20nF

    xlpy Ctot by L value

    gives, 120x10^-12

    take the square root = 10.95x10^-6

    times by 2 and times by PI (PI is 3.14159) = 68.8x10^-6

    So now we got all of the bottom row of the original equation done, so now plug numbers back in..


    freq = 1/68.8x10^-6 Hz

    = 14.52x10^3

    Or 14.52kHz





    For Rx coil 6.5mH with associated caps of 4n7 and 10n in parallel.

    As caps are in parallel you simply add the values together. 10 + 4.7 = 14.7nF

    Same as before...

    Freq = 1 / 2 PI SQRT(L Ctot) Hz

    = 1/ 2 PI SQRT (6.5x10^-3 x 14.7x10^-9)

    If uneasy with numbers TO A POWER like engineers use, you could do long-hand with the same result... i.e.

    = 1/ 2 PI SQRT (0.0065 x 0.0000000147)

    = 16.28kHz

    Yes, Tx and Rx frequencies are different.

    Returns from an energised target get through the Rx chain as an I.F. signal after the mixer of

    = 1.76kHz

    Bit like a superhet Radio.

    The bit I dont get is ... I assume Rx listens to signals from an energised ringing target - which re-radiates a component different to what it was energised with.

    IE we energise taget with 14kHz, we listen on 16kHz

    Any takers for this last bit?

    Steve

    Leave a comment:


  • salih
    replied
    I built Iron box for TGSL and Connected with (-)
    When I'm close to the Iron box my tgsl Becomes a weak sensitivity
    what I do?
    help
    Greetings

    Leave a comment:


  • der_fisherman
    replied
    Dust on the brain

    Originally posted by simonbaker View Post
    Just for fun, solving for Cd, a capacitor you put in parallel with C2:

    C = 1 / (4*pi*pi*L*f*f) Combined capacitance needed for your coil

    C = C1 * (C2 + Cd) / (C1 + C2 + Cd)

    .... lots of rearranging...

    Cd = (C*C1 + C*C2 - C1*C2) / (C1 - C)

    Cd is the capacitor you want to put in parallel with C2. If it is a negative number, probably means you need to solder in a smaller C2 and start again. You could also do similar calc if you want to put the trim capacitor in parallel with C1 instead.


    -SB
    Perfect Simon, saved me dusting of my Brain!!!

    regards

    Andy

    Leave a comment:


  • mikebg
    replied
    Originally posted by Qiaozhi View Post
    No ... you've completely missed the point.
    I was not commenting on the notation. It was simply that you used the frequency of 14kHz in the equation, instead of the inductance value of 6mH.
    Remember, T = L/r ... not F/r.
    OK
    At last I understand that I have made two errors: other value used at calculation and incorrect notation of the result (according SI).
    Here is the correct inscribe of calculation according SI:
    T=L/r=(6x10^-3) / 5 = 1.2x10^-3 s

    Leave a comment:


  • simonbaker
    replied
    Originally posted by simonbaker View Post
    Hi der_fisherman -

    Formula seems to be:

    f = 1 / (2 * pi * sqrt(L * C))

    f in Hz
    L in Henries
    C in Farads

    C is the series combination of C1 and C2 which should be C1*C2 / (C1 + C2)

    So you'll have to solve for the capacitors you want knowing the frequency you want, the L of your coil, and maybe fixing the value of one of the capacitors.

    -SB
    Just for fun, solving for Cd, a capacitor you put in parallel with C2:

    C = 1 / (4*pi*pi*L*f*f) Combined capacitance needed for your coil

    C = C1 * (C2 + Cd) / (C1 + C2 + Cd)

    .... lots of rearranging...

    Cd = (C*C1 + C*C2 - C1*C2) / (C1 - C)

    Cd is the capacitor you want to put in parallel with C2. If it is a negative number, probably means you need to solder in a smaller C2 and start again. You could also do similar calc if you want to put the trim capacitor in parallel with C1 instead.


    -SB

    Leave a comment:


  • der_fisherman
    replied
    Originally posted by simonbaker View Post
    Hi der_fisherman -

    Formula seems to be:

    f = 1 / (2 * pi * sqrt(L * C))

    f in Hz
    L in Henries
    C in Farads

    C is the series combination of C1 and C2 which should be C1*C2 / (C1 + C2)

    So you'll have to solve for the capacitors you want knowing the frequency you want, the L of your coil, and maybe fixing the value of one of the capacitors.

    -SB
    I posted the formula as well but the editor lost it......looked great and after I "saved" I did not check it out again.....

    Thanks for the series capacitor reminder, it must be almost 50 years ago that I learnt it!!! I will do exactly as you say, perfect!

    Have a great day.

    Regards

    Andy

    Leave a comment:


  • simonbaker
    replied
    Originally posted by der_fisherman View Post
    Hi Simon,

    you write very interesting posts.

    Please help me as I cannot calculate the 14.6KHz using the values given with the usual:-

    It appears to be important to somehow include C1 in the calculation is some manner......

    My Tx coil is not exactly 6.0 mHenries, its slightly less, so I need to calculate an extra Cap to solder across C1 and/or C2 to get the frequency right. The Rx coil is spot on.....

    .....(somewhere I have an old frequency counter, I must try and find it.......)

    regards

    Andy
    Hi der_fisherman -

    Formula seems to be:

    f = 1 / (2 * pi * sqrt(L * C))

    f in Hz
    L in Henries
    C in Farads

    C is the series combination of C1 and C2 which should be C1*C2 / (C1 + C2)

    So you'll have to solve for the capacitors you want knowing the frequency you want, the L of your coil, and maybe fixing the value of one of the capacitors.

    -SB

    Leave a comment:


  • der_fisherman
    replied
    Colpitts Oscillator Frequency

    Originally posted by simonbaker View Post
    Let's remember that the formula:
    Q factor of coil = wL/R
    is only valid when w is the self resonant frequency of the coil. 14 kHz is not the resonant frequency of the coil -- it is approximately the resonant frequency of the TX oscillator tank circuit, including capacitors C1 and C2. So everyone's formulas above are calculating the Q of the TX tank circuit -- not the Q Factor of the coil. Although the Q of the tank circuit is probably of most interest in the long run, I was trying to measure the Q Factor of the coil. But it gives me the idea that adding a capacitor might make calculating the effective resistance of the coil easier.

    I believe our coils have self-resonance in the hundreds of kHz range.

    When I did my measurements, I used a signal generator and swept for a maximum or minimum of some voltage. There easily could have been hidden factors that I was unaware of. I plan to try again.

    -SB

    P.S. What I was referring to as "skin effect" in our coils is probably better termed "proximity effect" as mentioned in another thread. So I'm trying to measure the effective resistance of our coils due to the proximity effect at about 14 kHz.
    Hi Simon,

    you write very interesting posts.

    Please help me as I cannot calculate the 14.6KHz using the values given with the usual:-

    It appears to be important to somehow include C1 in the calculation is some manner......

    My Tx coil is not exactly 6.0 mHenries, its slightly less, so I need to calculate an extra Cap to solder across C1 and/or C2 to get the frequency right. The Rx coil is spot on.....

    .....(somewhere I have an old frequency counter, I must try and find it.......)

    regards

    Andy

    Leave a comment:


  • simonbaker
    replied
    Originally posted by mikebg View Post
    However in the formula is written 10 ohm instead 5 ohm
    Q=105.6 @ 14kHz and r=5 ohm
    Q=52.8 @ 14kHz if r=10 ohm
    Q=26.4 @ 7kHz if r=10 ohm
    Q=3.77 @ 1kHz if r=10 ohm.
    Instead the Q, designers prefer to use parameter "timeconstant of coil" because it is independent from frequency and relates direct to weight of coil. The above coil has timeconstant
    T=L/r=14x10^3 / 5 = 2.8 ms.
    If we use aluminium wire, we will obtain almost twice less weight for the same coil Q or coil timeconstant.
    Let's remember that the formula:

    Q factor of coil = wL/R

    is only valid when w is the self resonant frequency of the coil. 14 kHz is not the resonant frequency of the coil -- it is approximately the resonant frequency of the TX oscillator tank circuit, including capacitors C1 and C2. So everyone's formulas above are calculating the Q of the TX tank circuit -- not the Q Factor of the coil. Although the Q of the tank circuit is probably of most interest in the long run, I was trying to measure the Q Factor of the coil. But it gives me the idea that adding a capacitor might make calculating the effective resistance of the coil easier.

    I believe our coils have self-resonance in the hundreds of kHz range.

    When I did my measurements, I used a signal generator and swept for a maximum or minimum of some voltage. There easily could have been hidden factors that I was unaware of. I plan to try again.

    -SB

    P.S. What I was referring to as "skin effect" in our coils is probably better termed "proximity effect" as mentioned in another thread. So I'm trying to measure the effective resistance of our coils due to the proximity effect at about 14 kHz. I realize now that I was wrong to use the Q Factor of the coil to get what I wanted because the proximity effect will be stronger at the coil self-resonant frequency than at our TX oscillator frequency. So I'll re-try measuring the Q with a tank circuit at about 14.5 kHz resonance and see if I can estimate the effective resistance of the coil at that frequency.

    Thanks everyone for your insights.

    Leave a comment:


  • golfnut
    replied
    Q

    Hi, I was responding to SB ..

    "It's also possible my Q measurements were affected by something else and I didn't get a real estimate. Wish I had one of those "Q" meters.

    Regards,

    -SB


    Im correct. Q is essentially a measure of the ratio of L to R. High R for a given L means poor Q.

    If you use a poor Q coil in a resonant cct the cct is lossy and has a wide bandwidth.

    Steve

    Leave a comment:


  • Qiaozhi
    replied
    Originally posted by mikebg View Post
    According SI and other standards, the correct notation is
    T=L/r=14x10^3 / 5 = 2.8x10^-3 s,
    because L should be expressed in Henry and r in Ohm.
    All dimensions should be in standard units (without mili-, micro-, pico-, mega- etc).
    My error is that I expressed the result in miliseconds instead of seconds.
    I think Steve is an engineer because he uses SI notation.
    No ... you've completely missed the point.
    I was not commenting on the notation. It was simply that you used the frequency of 14kHz in the equation, instead of the inductance value of 6mH.
    Remember, T = L/r ... not F/r.

    Leave a comment:


  • mikebg
    replied
    Originally posted by Qiaozhi View Post
    While we're correcting each other's mistakes ...
    I think you meant to write:

    T = L/r = 6mH/5 = 1.2ms
    According SI and other standards, the correct notation is
    T=L/r=14x10^3 / 5 = 2.8x10^-3 s,
    because L should be expressed in Henry and r in Ohm.
    All dimensions should be in standard units (without mili-, micro-, pico-, mega- etc).
    My error is that I expressed the result in miliseconds instead of seconds.
    I think Steve is an engineer because he uses SI notation.

    Leave a comment:


  • Qiaozhi
    replied
    Originally posted by mikebg View Post
    T=L/r=14x10^3 / 5 = 2.8 ms.
    While we're correcting each other's mistakes ...
    I think you meant to write:

    T = L/r = 6mH/5 = 1.2ms

    Leave a comment:


  • mikebg
    replied
    However in the formula is written 10 ohm instead 5 ohm
    Q=105.6 @ 14kHz and r=5 ohm
    Q=52.8 @ 14kHz if r=10 ohm
    Q=26.4 @ 7kHz if r=10 ohm
    Q=3.77 @ 1kHz if r=10 ohm.
    Instead the Q, designers prefer to use parameter "timeconstant of coil" because it is independent from frequency and relates direct to weight of coil. The above coil has timeconstant
    T=L/r=14x10^3 / 5 = 2.8 ms.
    If we use aluminium wire, we will obtain almost twice less weight for the same coil Q or coil timeconstant.

    Leave a comment:

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