SCOPE-3

From: Paul Harden, NA5N (na5n@rt66.com)
Date: Mon Dec 06 1999 - 14:44:46 EST

OSCILLOSCOPES - BASIC USE AND MEASUREMENTS
by Paul Harden, NA5N

PART 3 - LET'S MAKE SOME MEASUREMENTS
Amplifier gain and insertion loss
----------------------------------------------------------------------

NOTE: This is a text version of an article appearing in the Summer
1997 issue of "QRPp." The article contains numerous illustrations and
photos of oscilloscopes displays, which unfortunately can not be
included in a text file.

AMPLIFIER GAIN
The gain of an amplifier can be measured in terms of VOLTAGE GAIN,
which is simply Av = Vout/Vin. For example, if the input to an
amplifier is 1Vpp, and the output is 4Vpp, then the amplifier has a
VOLTAGE gain of 4.

GAIN IN DB is often more useful and is how the gains of amplifiers
are usually expressed. With dB (decibels), everytime you DOUBLE
the AC voltage, you ADD 6dB of gain. It is the RATIO of output to
the input, and this RATIO is easy to measure on a scope, even for
signals that exceed the cited bandwidth of your scope to some
extent.

Say you just built a single transistor amplifier to boost the audio
signal before the final audio amplifier (usually an LM386). It is
Inject an audio signal into the amplifier (transistor base). Place the
scope lead on the transistor COLLECTOR, and set the scope so the
output waveform is exactly 4 divisions peak-to-peak. Do not disturb
oscope settings.

Now move the scope leads to the amplifier INPUT, the transistor's
base. You will of course have a much smaller signal, and the ratio
of the input to the output will be the gain in dB. For example, say
the input signal is 2 divisions peak-to-peak. This would be 6dB of
gain, since you are DOUBLING the signal in the amplifier. Everytime
you DOUBLE the voltage, it is 6dB of VOLTAGE gain. If the input
signal is 1 division peak-to-peak, then the amplifier gain is 12dB.
(With the output still at 4Vpp or 4 divisions). Going from 1 division
to 2 division is 6dB gain; going from 2 divisions to 4 divisions is
6dB gain. Therefore going from 1 division to 4 divisions
is 12dB (6dB + 6dB).

This illustration shows how 0dB---|------|------|------|------|
to measure Amplifier GAIN | | | | |
on an oscope. FIRST, adjust | | | | |
the scope so the amplifier 3dB---|----- | -----|------|------|
OUTPUT is 4 DIVISIONS peak- | | | | |
to-peak. THEN, switch to | | | | |
the INPUT, and position the 6dB---|--**--|------|--**--|------|
waveform so the NEGATIVE | * * | | * * | |
peaks are on the bottom |* *| |* *| |
division, as shown. Where 12dB--*----- * -----*------*------*
the POSITIVE peaks of the 18dB__| |* *| |* *|
input occur will be the GAIN | | * * | | * * |
in dB by reading the scale ---|------|--**--|------|--**--|
on the left. You may want Amplifier INPUT displayed
to make such a scale and
attach it to the side of your CRT screen for making quick measure-
ments in dB's.

IF YOU WANT TO DO IT MATHEMATICALLY ...
Vout = 4Vpp
Vin = 1Vpp
Therefore, voltage gain Av = Vout/Vin = 4v/1v = 4
and gain in dB is:
dB = 20log(Av) = 20log(4) = 20(0.602) = 12dB

OR AS SHOWN DIRECTLY ON THE O-SCOPE AS DESCRIBED ABOVE.
Since this is a relative measurement (a RATIO), the absolute value
of Vin or Vout does not need to be determined.

INSERTION LOSS.
In some circuits, such as filters or attenuators, the LOSS in dB
needs to be determined, and this is called the INSERTION LOSS. It
is determined the same way as amplifier gain, except the INPUT will
be GREATER than the OUTPUT since there is a LOSS involved.

For example, with a signal generator connected to your QRP rig
antenna input, you want to measure the insertion loss of your IF
crystal filter. At the filter input, you can just barely squeek
out 2 divisions of input signal on your scope at its most sensitive
setting. (Perhaps due to exceeding the scope's bandwidth). The
output from the crystal filter is 1 division, or a 50% reduction.
The insertion loss would be 6dB (since if the power were HALF, or 50%,
it would be a 6dB LOSS).

OR MATHEMATICALLY USING SCOPE DIVISIONS:

Insertion loss (dB) = 20log(Vin/Vout) = 20log(2 div./1 div.)
= 20log(2) = 20(.30) = 6 dB

If the output were 1.5 divisions,
Insertion loss (dB) = 20log(2 div/1.5 div) = 2.5 dB

Again, you are determining the insertion loss of a circuit element
from the RATIO of input to output. You do not need to make absolute
measurements. So if the frequency is beyond the bandwidth of your
scope, as long as you can get enough vertical deflection to measure
it's magnitude in some terms of divisions, and able to see the signal
either get smaller or larger, you can estimate the gain or loss in
dB fairly accurately. If the signal DOUBLES, it is 6dB; if it is
about half of doubling, it is about 3dB; if the change is barely
noticeable, it is around 1dB. This is usually sufficient for
determining if circuit elements are working properly. For example,
using the insertion loss of an IF crystal filter as above, if you
determine the loss to be a few dB, the crystal filter loss is
acceptable. If it is much more than around 6dB, you may have a
problem. And if you can't see any output, you have a real problem.
(Loss is about 1 to 1.5 dB per crystal in filter). Same with checking
the gain of amplifiers. It's not important if the gain is 6.2 dB or
6.5 dB, but whether the gain is ABOUT what you'd expect. If the
output of an amplifier is ABOUT DOUBLE the input, you have 6dB of
gain. If the output is just a bit larger than the input (or the
same) ... then you've got a problem (no or little gain occuring).

With a little practice on your o-scope, you will learn to recognize
approximate gains and losses in dB's very quickly from the oscope
display.

You might want to go through your QRP kit with the circuit and measure
the gains through different stages. If your rig has an MC1350 IF
amplifier, what is it's gain with a strong signal vs. a small signal
to see if the AGC is working properly. What is the gain of the LM386
audio output amplifier? With larger bandwidth scopes, check the gains
of the RF driver transistor and output PA transistor. Knowing what
these gains are could help troubleshoot the circuit later should a
problem develop.

END OF PART 3

72, Paul Harden, NA5N
NA5N@Rt66.com

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