Barry2717
Computer
- Jan 27, 2014
- 5
Hello,
I'm curious what the advantages and disadvantages are to winding a solenoid coil using copper foil, as opposed to wire. I'm also curious what the difference is for calculating the inductance of a foil coil. Is the inductance roughly the same, if you wind the coil full of wire or foil?
For instance, lets say I have a bobbin that's got a 0.75" hole in it, with a total bobbin diameter of approx. 1.25", so the area to hold the wire coil is approx. 0.25" x 0.25". (not really taking into account for bobbin material thickness) If I use a wire that has a 0.01" diameter, I should be able to get 25 windings on each layer, and 25 total layers, for a total of 625 turns.
Using the inductance calculator located here: Plugging in these values:
Turns = 625
Diameter = 1.25" (Their definition of coil diameter is hard to decipher from the picture)
Length = .25"
Depth = .25"
Units = Inches
Coil = Multi-layer, multi-row
I get an answer of 14,361 microhenries.
Now, what if I wrapped the same bobbin with copper foil instead? If the foil was 0.01" x 0.25", I should be able to get approx. 25 turns on the bobbin. For the inductance calculation, is each layer of wire (approximately) represented by one layer of foil, or does the lowered number of foil turns greatly affect the inductance?
It's my understanding that using foil can help the coil to dissipate heat better. It's also my understanding that a foil coil will have a lower DC resistance.
How do the ampere turns translate into a coil using foil?
What would be the difference between the magnetism generated between my two example coils?
Does magnetism power relate more to the total watts pumped into a coil, or is it more closely related to just the amps?
Thanks for your time!
Barry
I'm curious what the advantages and disadvantages are to winding a solenoid coil using copper foil, as opposed to wire. I'm also curious what the difference is for calculating the inductance of a foil coil. Is the inductance roughly the same, if you wind the coil full of wire or foil?
For instance, lets say I have a bobbin that's got a 0.75" hole in it, with a total bobbin diameter of approx. 1.25", so the area to hold the wire coil is approx. 0.25" x 0.25". (not really taking into account for bobbin material thickness) If I use a wire that has a 0.01" diameter, I should be able to get 25 windings on each layer, and 25 total layers, for a total of 625 turns.
Using the inductance calculator located here: Plugging in these values:
Turns = 625
Diameter = 1.25" (Their definition of coil diameter is hard to decipher from the picture)
Length = .25"
Depth = .25"
Units = Inches
Coil = Multi-layer, multi-row
I get an answer of 14,361 microhenries.
Now, what if I wrapped the same bobbin with copper foil instead? If the foil was 0.01" x 0.25", I should be able to get approx. 25 turns on the bobbin. For the inductance calculation, is each layer of wire (approximately) represented by one layer of foil, or does the lowered number of foil turns greatly affect the inductance?
It's my understanding that using foil can help the coil to dissipate heat better. It's also my understanding that a foil coil will have a lower DC resistance.
How do the ampere turns translate into a coil using foil?
What would be the difference between the magnetism generated between my two example coils?
Does magnetism power relate more to the total watts pumped into a coil, or is it more closely related to just the amps?
Thanks for your time!
Barry