TGSL Experiments

[golfnut]

For Q

Hi for Q, simply.

Q= wL/r

w=2 *PI *freq


r in ohms

so if you got a 5ohm coil, L is 6mH, frequency of 14kHz

Q= 2* 3.14 *14x10^3 * 6x10^-3 / 10 (Q has no units)

Steve

[mikebg]

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.

[Qiaozhi]

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

[mikebg]

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.

[Qiaozhi]

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.

[golfnut]

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

[simonbaker]

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.

[der_fisherman]

Colpitts Oscillator Frequency

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

[simonbaker]

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

[der_fisherman]

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

[simonbaker]

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

[mikebg]

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

[der_fisherman]

Dust on the brain

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

[salih]

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

[golfnut]

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