Play Meter

PROGRAMMED TESTIS
transistor have been previously
discussed . Also , corrective
measures have been devised. leo
is a result of temperature
increases both external (T A) and
internal (PT) to the transistor.
Internal heat is a result of the
voltage collector to emitter (VeE)
and lc.
PT = VeE X lc
Design Procedure:
STEP 1: Find absolute value of IE:
I Vin l - I VoE I
=
Design example:
Design an emitter follower which will
drive a 6V, 150-omega lamp (the
circuit configuration is shown in fig .
5-1 .) Use a germanium transistor type
2N404.
STEP 1: f-ind l i E
Care must be taken by the
designer not to operate the
transistor beyond the manufac-
turer's stated maximum dissipa-
tion. As was the case with the
diode, the manufacturer states a
maximum dissipation on the data
sheet. Included in the collector
family curve on fig. 3-6 A (Lesson
3) and the emitter follower line of
fig . 5-2 is the curve PT . The load
line must be selected such that it
will be within the area below the
maximum dissipation curve. It
should be noted that the curve is
given for aT A of 25° C. The manu-
facturer also gives on the data
sheet the rate at which to derate
PT[max 1 at elevated temperatures .
This derating figure is the same
form as it was given for the diode
in Lesson Two, and all calcula-
tions are done in the same
fashion .
IE
The formula for lc is taken from
Lesson One, formula [2A) .
RE
The above is a general statement and
applies to all design problems. In this
course we are concerned with digital
circuits only , so Vee can be
substituted for Vin. This substitution
is allowed because the input levels for
digital circUits are either OV or Vee. If
the device being switched is in the
emitter circuit, then RL = RE ..
r1 -
=
5 · ?V
=
38 mn
1500
Since the transistor is operating close
to the saturation region , the general
rule for VBE,sat, can be used.
lo
(1 - a) IE
=
23.3V
5000
0 1d you get the ume t esult?
46.6ma
It not,
clo aot go on, Out c heck.
Vs£, Ve E can t>e /gn01ed. Then
the value I£ • 48ma . c•n be used.
3 8m a (1
0. 98)
0. 76m a
:JB ma (1
0 .99 )
0 .3 ma
What is the range in Ia?
• · --~ Is ~--
GO TO BLOCK 19
Notice that quantities of base current
have been calculated, each associa-
ted with a different value of alpha.
The two values of alpha correspond
to the high and low values of beta
given in figure 4-7 (Lesson Four). The
values of alpha were found by using
formula 6 in Lesson One and the high
beta value of 130 and the low beta
value of 50. The student should verify
the conclusions from beta to alpha.
The range in base current illustrates
the current amplifying ability of an
emitter follower.
STEP 3:
lc
Find l lc I·
=
=
37ma
a) IE
End of lesson Five
lc = 46ma or 47.5ma depending upon
which value of alpha was used. In either
case the difference will be very small.
You have completed the test for Lesson
Five. You may want to breadboard the cir-
cuit and test your results.
11 YOU ARE CORRECT!
Design an emitter with an RE of 500
omega.
Use a silicon NPN transistor with a range
in beta from 25 to 100. Make Vee = 24Vand
Vin = Vee.
a.
alE
0.98 x 38ma
4
What is IE?
0.99.
PLAY METER, May 1,1981
3
Since in rlus cue V ee n
STEP 3: Find lc.
alE
The input and output voltages are
Inverted
GO TO BLOCK 21
b.
In phase
GO TO BLOCK 11
STEP 2: Find Is.
The formula for Ia is taken from
Lesson One, formula [3) .
=
2 YOU ARE CORRECT!
a.
Formula 2A from Lesson One is used
to find lc. There is no need for
concern as to which value of alpha to
use, since the range is from 0.98 to
lc
1
The input to an emitter-follower is from
a. Emitter to ground
GO TO BLOCK 13
b.
Base to ground
GO TO BLOCK 20
I·
I Vcc l - I VoE
STEP 2: Find lB.
lo
Instructions: The purpose of this test is to
guide you step-by-step through actual
circuit design problems. Most important,
these tests will provide you with a gauge to
establish your degree of understanding of
the material covered in the lesson
text. The test is programmed: start at
block 1 and follow the numbered
instruction associated with your answer.
IE =----------------------
GO TO BLOCK 3
12YOU ARE INCORRECT!
Referto the text and return to BLOCK 20.
13 YOU ARE INCORRECT!
Refer to the text and return to BLOCK 1.
[Test continues on next page)
13