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~~CUBIC
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COMMUNICATIONS
SIrAltI DIVISIDN
TRANSISTOR TROUBLESHOOTING
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150
+12V
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1.2K
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Many techniques
for troubleshooting
transistor circuits have been ev-
olved,
of varying degrees of complexity and reliability.
One system,
per-
haps the simplest and most reliable,
is outlined here for your information.
With it, you can determine, with only a few minutes thought and basic arith-
metic calculations, what the normal operating voltages on any given transis-
tor stage should be, and by comparing them with those actually measured, 10-
cate the fault in the circuit.
.
The key
to the scheme
is to remember that,
for normal operation, the
base-emitter junct ion of a transistor, must be
biased into
conduction--and
that when this is so,
the voltage dT~p across that junction will be approx-
imately
0.7
volts fdr silicon devices and about
0.25
volts for germanium.
To illustrate
how the system works, , lets
take a simple
amplifier stage
(at the right).
To start with, we know that the base current is
very small,
on the order of microamperes.
For
our purposes,
then,
we'can ignore it,
for it
would
be masked by circuit
tolerances suffic-
iently that we couldn't
measure a voltage drop
caused by it with our usual methods.
We there-
fore'treat the
bias network (the
5p
kn
and 1.2
kO resistors) as a
simple voltage divider, and
can
compute the
voltage that would
appear at
their junction (the base of the transistor). In
this case,
it works
out to
about
2.1 volts.
Since we know
that this silicon transistor has
a base-emitter drop of about
0.7
volts, it fol-
lows that
the eimtter
voltage
is that amount
less than
the base voltage,
or about
1.4 volts.
The emitter
current
cän
then be calculated by Ohm's law, and comes out to be about
9.3
mA.
We know
that bath the base current
and the collector current flow th-
rough the emitter resistor, but the base current is so small that we can say
for practical purposes that the emitter and colleotor currents are the same;
therefore, the current through the
4700
collector resistor is also about
9.3
mA, and the voltage drop across it is therefore about
4.4
volts.
This means
that the collector voltage would be 12 -
4.4,
or
7.6
volts.
Actual measured values
on a
normally operating circuit would be close
to these values,
while a málfunctioning transistor would
result in greatly
different
values being found.
Remember,
transistors do not get "weak" , as
do vacuum tubes (with the exception of some emitter-ballasted power devices)
so when they fail, "they do so catastrophically--they open or short, or bath.
Thus, the voltages
found on their terminals will be greatly different than
if it were operating normally.
For example, a shorted base-emitter junction
would result in no
0.7
volt drop between those terminals; an open base-emit-
ter junction would leave the emitter
at ground potential
(no base current
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Chapters
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Troubleshooting
