Difference between revisions of "Talk:Halogen Lighting"
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| − | + | ==Propose move== | |
| + | to Halogen, with this article becoming a link to there. It'll make life significantly easier in time. | ||
| + | [[User:NT|NT]] 09:02, 1 March 2009 (GMT) | ||
| + | |||
| + | == to add == | ||
| + | |||
| + | since google seems to be getting flaky on groups searches, I thought it work keeping this post from Andy Wade on transmission line losses to be integrated at some point: | ||
| + | |||
| + | |||
| + | There's a current loop of finite size, so the inductance is finite, not | ||
| + | zero. Whether or not it's vanishingly small depends on the application. | ||
| + | In 50 Hz work we're used to being able to neglect wiring inductance, | ||
| + | but it ceases to be negligible at higher frequencies, or even at 50 Hz | ||
| + | for large cables (50mm^2 upwards, say). In this case the frequency is | ||
| + | over three orders of magnitude above mains frequency. | ||
| + | |||
| + | > How do you get your 0,9uH figure? | ||
| + | |||
| + | Any text book on E-M theory /transmission line theory will give you the | ||
| + | following expression for the inductance per unit length of a | ||
| + | parallel-wire line | ||
| + | |||
| + | L = (mu / pi) * ln (s/r) [for s >> r] | ||
| + | |||
| + | where | ||
| + | |||
| + | mu is the permeability, in this case = mu_0 = 4*pi*10^-7 H/m | ||
| + | (so mu / pi = 0.4 uH/m), | ||
| + | s is the spacing (between centres) of the conductors, and | ||
| + | r is the radius of each conductor. | ||
| + | |||
| + | For 1.5 mm^2 T&E cable the wire diameter is 1.38 mm, so r =~ 0.7 mm. | ||
| + | I guessed s = 6 mm. | ||
| + | |||
| + | Substituting these values, the ln() term evaluates to 2.15, so | ||
| + | L = 0.4 * 2.15 = 0.86 uH/m. QEF. | ||
Latest revision as of 03:11, 4 September 2009
Propose move
to Halogen, with this article becoming a link to there. It'll make life significantly easier in time. NT 09:02, 1 March 2009 (GMT)
to add
since google seems to be getting flaky on groups searches, I thought it work keeping this post from Andy Wade on transmission line losses to be integrated at some point:
There's a current loop of finite size, so the inductance is finite, not
zero. Whether or not it's vanishingly small depends on the application.
In 50 Hz work we're used to being able to neglect wiring inductance,
but it ceases to be negligible at higher frequencies, or even at 50 Hz for large cables (50mm^2 upwards, say). In this case the frequency is over three orders of magnitude above mains frequency.
> How do you get your 0,9uH figure?
Any text book on E-M theory /transmission line theory will give you the following expression for the inductance per unit length of a parallel-wire line
L = (mu / pi) * ln (s/r) [for s >> r]
where
mu is the permeability, in this case = mu_0 = 4*pi*10^-7 H/m
(so mu / pi = 0.4 uH/m),
s is the spacing (between centres) of the conductors, and
r is the radius of each conductor.
For 1.5 mm^2 T&E cable the wire diameter is 1.38 mm, so r =~ 0.7 mm. I guessed s = 6 mm.
Substituting these values, the ln() term evaluates to 2.15, so
L = 0.4 * 2.15 = 0.86 uH/m. QEF.