From: Bob McCowan ([email protected])
Date: Fri Jun 23 2000 - 08:12:35 PDT
Loss and size are probably the issue. There is an upper limit to the outer
diameter for a coaxial conductor. Make it too big and circular waveguide
modes can propagate. The maximum outer conductor size scales as
(epsilon)^-1/2 and as (freq)^-1. Also, for a given impedance the ratio of
the outer to inner conductor diameters are determined by the dielectric
constant of the insulator. The higher the dielectric constant the greater
the ratio. This makes for extremely small inner conductors for mm-wave
frequencies, giving a lot of loss.
Also, the wavelength is shorter in higher dielectrics. This translates to
more loss per inch due to dielectric loss. Add that to the increased
conductor loss because of the small inner conductor and you have a
hard-to-make lossy cable.
Bob McCowan
Physicist/Sr. Engineer
CPI Beverly Microwave Division
Beverly MA, 01915
(978)-922-6004 x208
[email protected]
Visit our Web Site at http://www.cpii.com/bmd !!!
.----Original Message-----
From: Ali, Hassan [KAN:0G15:EXCH]
Sent: Friday, June 23, 2000 6:37 AM
To: si-list
Subject: [SI-LIST] : Merits of low dielectric constant
I attended a presentation by a high-frequency (1GHz < f < 65GHz) coaxial
cable vendor, and the presenter claimed that their cables use a material
with a very low dielectric constant and therefore are ideal for high-speed
application as they give rise to low capacitive loading. He gave a formula
showing the capacitance (I think per unit length) decreasing as you decrease
dielectric constant. This claim, however, perplexed me as I don't know how a
cable's capacitance per unit length would give rise to a capacitive loading.
All I know from my transmission line classes, a lossless transmission line
with Z0 = sqrt(L/C) would transmit signals exactly the same way regardless
of the value of the p.u.l. capacitance C as long as the ratio L/C is
maintained. Am I missing something here?
Thanks.
Hassan.
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