7.2
HOW TO INTERPRET SCOPE DISPLAYS OF RECOVERY TIME
This sub-section provides guidance, through examples, of how the information obtained from an
oscilloscope display can be used to estimate recovery time, without using a setup where the
amplitude of a small pulse following a large pulse is actually measured.
Examination of two cases will allow us to draw some informative conclusions. In both cases, we
assume a detector with a K of 700, and that we need an accurate log transfer at input powers of -
36 dBm and higher. In the first case, let us assume that the unit is set to start logging at -36 dBm,
and that pin 9 has been tied to ground (which kills L1, the first log stage). In the second case, we
will assume that even though we only need to log accurately above -36 dBm, we have set the unit
to enter logging at -42 dBm by leaving the gain of A1 unchanged and de-grounding pin 9, which
re-activates L1.
Case 1. For the first case, we choose a gain of 6.4 for A1; the gain of A2 is 4, and the gain of A3
is 16. With an input power of -36 dBm, a detector with a K of 700 will provide an output of 175
V,
which leads to an output at A3 of 6.4 x 4 x 16 x 175
V = 72 mV. Since L1 is killed, this
corresponds to an input of 18 mV into L2 [see Fig. 1(a)]. If we assume the unit is set for a logging
slope of 50 mV/dB, we have a gain of 1.625 between the output of A3 and the main output,
(7)
so
we start logging with an output voltage of 72 mV x 1.625 = 117 mV. The question we need to ask
is: "How big can a tail be, and still have a superimposed -36 dBm pulse give an output of less than
117 + 50 = 167 mV?".
(8)
All amplifiers in this example are still linear, so we can work backwards
as follows:
167 mV
V
IN
= ---------------------------- = 250
V
(9)
1.625 x 16 x 4 x 6.4
i.e. you could tolerate a tail of 250 - 175 = 75
V, which translates to a voltage of 50 mV at the
output. (This is actually obvious, since all amplifiers are still linear.)
Case 2. Now consider the second case, where logging starts at -42 dBm. Here we still have A1,
A2 and A3 gains of 6.4, 4, and 16, respectively, but now, since L1 is active, we have a gain of 6.5
between the output of A3 and the main output. At -42 dBm, the
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(7)
The gain in this case is 1.625 because L1 has been killed. Normally, this gain would be
6.5.
(8)
This is actually a slight oversimplification, because the local slope at the start of logging is
less than 50 mV/dB, since there is a peak in the "wave" of the log transfer curve in this region.
The example still correctly illustrates the intended point, however.
(9)
Note that 250/175 corresponds to about 1.5 dB, not 1 dB, for the reason mentioned in
footnote 8, above.
-21-
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