[time-nuts] Trimble Thunderbolt question, splitting its output.
Charles Steinmetz
csteinmetz at yandex.com
Sat Mar 28 15:55:37 EDT 2015
Dave wrote:
>??? How do you get to that 45 ohm figure. The input has a pair of
>100R in parallel, so I can't see how it can be 45R input impedance.
Also in parallel with the 2x 100 ohm resistors (R2 and R3) is 10k
(R5) in series with 100nF, plus 475 ohms (R1) to a pair of (parallel)
back-to-back diodes in series with a 100nF capacitor (C5). Both
100nF capacitors essentially sit at the half-supply bias voltage
(there is a very small AC component on C3, ~1mV, which can be
ignored). So, for peak input voltages of less than a diode drop, the
input impedance is ~49.75 ohms (100||100|10k). But for peak voltages
greater than a diode drop, the input impedance is 100||100||475||10k,
or about 45 ohms. (There is a slushy transition zone of ~100mV as
the diodes turn on.)
The TBolt puts out nominally +13dBm, or 1Vrms (2.8v peak-to-peak,
1.4v peak) into 50 ohms.
So, the TBolt sees a nonlinear load of ~50 ohms for the first ~600mV
(plus and minus) of voltage excursion, then ~45 ohms from 600mV to
1.4v (plus and minus).
I'd be inclined to change R2 and R3 to 113 ohms for use with sources
that put out +4dBm or more, although the practical effect in most
cases is probably minor. Note, however, that if one feeds the
divider and another instrument using a simple BNC "T", the
nonlinearity of the divider's input impedance will raise the
distortion floor of the sine wave seen by the other instrument to ~
-40dBc even if the source is perfectly pure.
Best regards,
Charles
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