From: Jeff Liebermann
Subject: Re: AN: New Logic Analyzer, 500MHz, $166
Date: Sun, 17 Nov 2002 09:23:51 -0800
Organization: Committee to Maintain and Independent Xenix
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On Sun, 17 Nov 2002 10:44:18 +0100, "Frank Bemelman"
>"chris" schreef in bericht
>> > Input circuit is 100K in parallel with 10pF. As you would
>> > expect, the sampling logic reads this signal every 2nS (when
>> > sampling at 500MHz).
>> > /Tim
>> Hi, your input is only 3ohms a 500M, not much will drive that! At
>> 500kHz its still 3kohms so it looks like my initial guess was about
>Or is it 30 ohms at 500Mhz? Of course it's not to analyse a 500Mhz bus,
>it's just that it captures at 500Mhz, so you can expect some detail
>in the results.
30 ohms is correct if we ignore the 30cm probe lead length. However,
at 500MHz, two 30cm leads (one for signal, one for ground), yields
about 3000 ohms of reactance. That makes a nice -37dBv attenuator.
L = 2W [ ln (2W/r) - 1.00]
L = inductance in nHy
W = Wire length in cm
r = wire radius in cm
I'll assume #26awg = 0.41mm dia
L = 2 * 60 [ ln ( 2 * 60 / 0.021 ) -1 ]
L = 120 [ ln 5713 ] = 120 [ 8.65 ] = 1038 nHy
L = 1.038 uHy
The inductive reactance at 500MHz is:
XL = 2PifL = 6.28 * 500 * 10^6 * 1.038 * 10^-6
XL = 3125 ohms
So, now we have a real network. Assuming negligible probe wire
resistance, and ignoring the 100K input resistance, at 500Mhz, we get:
>---- XL= 2135 ohms------------------>
probes Xc = 30 ohms to analyzer
Since the vectors are 180 degrees apart, we can assume that they
cancel resulting in an input impedance at the end of the probes of
about 2165 ohms, and a loss of about:
30/(2135 + 30) = 0.014
dBv = 20 log 0.014 = -37dBv
In other words, a 3.3v p-p input signal at 500MHz will appear as 46mv
on the input to the A/D converter.
[Note: I'm half asleep and doing some of this in my head. please
check my calcs, and ignore my sloppy abuse of significant figures,
before assuming they're correct].
Jeff Liebermann 150 Felker St #D Santa Cruz CA 95060
(831)421-6491 pgr (831)336-2558 home