On 3/3/21 9:53 AM, David Eckhardt wrote:
A number I've always kept in mind for AWG #18 enamelled single conductor
copper wire: 18.1 nH / inch.
Dave - W?LEV
I use 1 uH/meter, which is somewhat larger, but the right general magnitude.? Inductance doesn't vary very quickly with diameter or shape, it's all about length and whether there's coupling from one piece of the wire to another. (that is, a 90 degree bend has pretty much the same inductance as the same length of wire in a straight line).
For what it's worth the formula from Grover (which is the "standard")? says that a 1mm diameter wire 100cm long is 1.5 uH? - AWG 18 is 1.02mm, so close enough.? A 1 inch (2.54 cm) long section is 19.8 nH, 50cm is 683 nH, so it's not exactly a linear scaling
10% is close enough in these sorts of calculations.
Grover's formula
x = sqrt(L + (D/2*L)^2)
inductance = 2 * L * ( ln( ( 2*L/D)*(1+x)) -x + mu/4 + D / (2*L))
I have no idea off hand without getting my copy of Grover out whether this equation is based on "first principles" or if it's a convenient accurate fit.? I suspect the former.
On Wed, Mar 3, 2021 at 2:30 PM WB2UAQ <pschuch@...> wrote:
Just something to consider. Think about the shunt C of 6 resistors in
parallel. There might be as high as 0.5 pF per resistor. I measured
axial lead resistors with my old HP (BRC) 250 RX meter and this is roughly
what I see for them. Calculate what 3 pF will do to the impedance at the
frequency of concern and it might be significant. At 30 MHz this is 1768
ohms.
