The signal strength increase is higher than I expected. 36t/30t equals 1.20. Getting more than 1.20. I've doubled peak current on a measurement in the past and remember the signal doubled. The current profile is the same on the two coils. I was thinking signal strength is ampere turns. Am I thinking wrong?
Announcement
Collapse
No announcement yet.
Big foot style for a PI
Collapse
X
-
-
Hi Green
Attached is a Spider coil calculator that has worked for my coils. It is a different spoke configuration but appears to be close to your toothpick coil. I entered the following parameters for your coil; Inside dia = 3.9 inches, 38 turns, #28 wire = 0.014 (to calculate click outside the box) = 300.7 uH. I like the improved 4" coil. I will have to think about the x1.2 current gain.
Have another great day,
Chet
http://www.crystalradio.net/professo...piderweb.shtml
Comment
-
Originally posted by green View PostYes I use a 2mm craft foam sheet sandwiched between two pieces of cardboard. If you are interested I could put together instructions on how I'm doing it now. I'm still learning so things keep changing.
Comment
-
Originally posted by green View PostThe signal strength increase is higher than I expected. 36t/30t equals 1.20. Getting more than 1.20. I've doubled peak current on a measurement in the past and remember the signal doubled. The current profile is the same on the two coils. I was thinking signal strength is ampere turns. Am I thinking wrong?
I don't think your current ratio calculation is a reliable method for comparing different coils. When using two different inductance values you may need to calculate the energy density for each coil in milli-joules.
Best regards,
Chet
Comment
-
Originally posted by Chet View PostHi Green
I don't think your current ratio calculation is a reliable method for comparing different coils. When using two different inductance values you may need to calculate the energy density for each coil in milli-joules.
Best regards,
Chet
Comment
-
Originally posted by green View PostSomething I don't know anything about. I'm probably saying and thinking wrong. I was thinking field strength is ampere turns. If I double current, signal strength should double. If I keep current the same, signal strength should equal turns ratio. I was thinking to double signal strength you would have to double turns. If current remains the same what ratio of turn increase should it take to double signal strength? Just thinking. maybe it's ratio squared because of transmit and receive.
Comment
-
Hi Green
I haven’t done this before so I hope I didn't mess up too much.
Neglecting losses, the energy (measured in joules) stored by an inductor is equal to the amount of work required to establish the current through the inductor, and therefore the magnetic field. This should be based on the current in the coil just before flyback occurs. This is given by: 1/2 of the inductance times the current squared.
Energy examples using your coils:
30T 4” 220 uH coil @ 1 amp is .000220 H / 2 = .000110 * 1 * 1 = .000110 Joules
Notice next that adding 16 Amp-Turns doubles the Joules.
46T 4” 440 uH coil @ 1 amp is .000440 H / 2 = .000220 * 1 * 1 = .000220 Joules
36T 4” 280 uH coil @ 1 amp is .000280 H / 2 = .000140 * 1 * 1 = .000140 Joules
Notice next that doubling the Amp-Turns increases the Joules by X4.
36T 4” 280 uH coil @ 2 amp is .000280 H / 2 = .000140 * 2 * 2 = .000560 Joules
??T 6” 230 uH coil @ 1 amp is .000230 H / 2 = .000115 * 1 * 1 = .000115 Joules
Now to find the density we divide the Joules by the square inches of the coil area.
Let’s use 4.5” average diameter = a radius of 2.25” * 2.25” * pi 3.1416 = 15.9 sq. inches.
The 30T 220 uH coil @ 1 amp .000110 / 15.9 = .0000069 Joules per sq. inch.
Notice next that the added 16 Amp-Turns doubles the Density.
46T 4” 440 uH coil @ 1 amp is .000220 / 15.9 = .0000138 Joules per sq. inch.
The 36T 280 uH coil @ 1 amp .000140 / 15.9 = .0000088 Joules per sq. inch.
Notice next that doubling the Amp-Turns increases the density by X4.
The 36T 280 uH coil @ 2 amp .000560 / 15.9 = .0000352 Joules per sq. inch.
The 6” coil; use 6.5” dia. = 3.25 * 3.25 * 3.1416 = 33.18 sq. inches
The ??T 230 uH coil @ 1 amp .000115 / 33.18 = .00000347 Joules per sq. inch.
Let’s use a 21T, 12”, 300 uH coil with 2 amps; .000300 / 2 = .000150 * 2 * 2 = .000600 Joules.
Notice next what 9 additional turns does.
Now add 9T = 30T, 600 uH coil with 2 amps; .000600 / 2 = .000300 * 2 * 2 = .001200 Joules.
Divide by (6” * 6” * 3.1416 = 113.1 sq. inches) is .000600 / 113.1 = .0000053 Joules/sq. inch.
Next with the added 9 turns the density is doubled.
Divide .001200 / 113.1 = .0000106 Joules/sq. inch.
This 12", 600 uH coil is certainly not fast enough for small gold nuggets but could be used for large nuggets or relics.
Hopefully this provides some insight into the energy at the face of the coil. The losses to and from the target and target size are most important as some of your charts have demonstrated.
I hope all of this doesn't make you dizzy,
Chet
Comment
-
What I want to do. Chart the data for a 300uh coil without making a new one. I took the data for the 4 inch coil with 30t and 36t. 36/30 equals 1.2, squared equals 1.44. I multiplied the 30t data by 1.44 and it plotted on top of the 36t data. Thinking I can use your coil calculator to find turns to match inductance of existing coil, same ID and wire size. Then find turns for 300uh. Multiply existing data by turns ratio squared, same as above example. Going to do a 8 inch coil and try for 300uh and would like to compare the nuggets with 4, 6, and 8 inch coils with the same current and inductance (300uh).
Comment
-
-
A way to wind a spiderweb coil. I used materials that I already have. (1)The beads, spacers I use are 5mm long, so cut a cardboard disk with a diameter 10mm less than desired inside coil diameter. (2)Cut a piece of the foam about 1 inch larger than the disk.(tape pieces of foam together with masking tape if needed). (3)Spray one side of cardboard disk and one side of the foam(not taped side) with the contact adhesive, let dry(follow instructions on can). (4) Stick together and trim foam. (5)Cut a piece of cardboard about 1 inch larger than the disk. (6)Remove masking tape from foam if needed and repeat step (3). (7) Stick together and trim cardboard. 8 Cut some 10mm wide foam strips, enough to wrap around coil. Glue strips together with super glue(cyanoacrylate). (9) Wrap foam around coil disk and cut to length. The foam was 620mm long for the 8 inch coil I was making. For ease of marking I've been using 10, 15, or 20 mm spacing. 620/20 equals 31. Need a odd number. For some reason I used 29. The foam will stretch. Mark the foam strip, cut to length and glue the ends together. Wrap band around coil form and mark cardboard, Remove foam band and poke tooth picks in foam, about 10mm. Tape wire to disk. Put 6 or 8 beads on the tooth picks. I go under over and pinch the over tooth pick and repeat. Add 6 or 8 beads as needed and repeat. I put a mark on a piece of paper each revolution so I don't loose track of turns count. I use insulating foam at the crossover points to keep things together before removing tooth picks. I'll add instructions for how I shield the coil later.Attached Files
Comment
Comment