Play Meter

Issue: 1981 May 01 - Vol 7 Num 8

Digital Circuit Design Course
LESSON 5:
Emitter Follower Design
Editor 's Note : The material below is a serialization of the Kurz Kasch correspondence course for electronics, designed
spec ifically for the coin -operated amusemen~ industry. Th is course is copyrighted and. owned_ by Kur~ K_asch of f?ayto n,
Oh io and its reprinting is being sponsored jotr_Jtly by Kurz Kasch and Play Meter magaztne. Th1s matenal 1s authonzed fo r
publication exclusively by Play Meter magaztne.
The operation and design of
emitter followers is the topic of
this lesson . The circuit is
analyzed· both mathematically
and graphically.
Definition: An emitter follower. or
common collector circuit, is a
circuit configuration in which the
output is from the emitter to
ground. In the case of the
inverter, or common emitter
circuit of Lesson Four, the output
was taken from the collector to
ground . The emitter follower is
used primarily as a current ampli-
fier. It is used in digital circuits to
drive high current devices, such
as lamps and relays , and also to
drive low impedance trans-
mission lines.
Emitter Follower Operation:
Figure 5-1 is a schematic diagram
of an emitter follower. Notice that
the collector is connected
directly to Vee and the resistor
connecting the emitter to ground .
The input to an emitter follower is
from base to ground, and the out-
put is from emitter to ground. The
circuit operates as follows:
During time, t less than 0, the
input is at a low level (OV). the
base is then zero biased, and the
only current flowing is the
leakage current. The leakage
current is too small to cause a
voltage drop across RE worth
accounting for; therefore, all of
the voltage is across the transis-
tor, collector to emitter. This
operating condition is graphically
shown at point A on the emitter-
follower load load line of fig. 5-2.
At point A, VeE is approximately
equal to Vee and le is ap-
proximately zero. Because all of
the voltage is across the transis-
tor, the voltage across RE must
then be zero .
12
The base-emi tter diode will begin to
forward bias at the knee in the curve
of fig . 5-2, and this point is termed
VB E(cu tin). As a general rule-for
silicon transistors VBE(cutin ) = 0.5 V
and for germanium transistors :
= 0 .1
VBE (cut in)
V.
- \ c;c
IIIIHII
II\ ~ J
I • II
Our pol
~v"
ov 1· 0 1 -
F ig . 5· 1. An em iUtr fo llower chc~o~l t . The Input and output
w avef orms are In phase .
.
~
~
\,._,lJ
"•
At t = 0 the voltage [Vin] rises from
OV to a level high enough to
forward-bias the base emitter
diode, a current begins to flow in
the emitter circuit [IE]. IE causes
a voltage drop across RE which
subtracts from VeE. The voltage
dropped across RE will continue
to rise with the input voltage and
continue to subtract from
VeF. Let us assume that Vin has
risen to Vin3 . This operating
condition is represented at point
B. Projecting the voltage inter-
cept, we find that VeE has de-
creased and that VE has in-
creased proportionately . The
current le can be found by pro-
jecting the current intercept.
Also , notice that le has increased .
When the input rises to the steady
voltage V, the emitter voltage will
also reach a steady state voltage.
This operating cond ition is
shown at point C. Notice that VE
has risen to almost Vee and that
VeE is almost zero. The steady
state voltage across the em itter
resistor [VE] will be equal to the
input voltage minus the voltage
base to emitter, or
VE
\ f' .. 0
\ Ct-: • 0
F ig . 5·2. The family of Input -out put t tates fOf an emi tter ·
foll ower . The output voltage W i ng the Int ercept of the load
lint wllh a partlc u ler V 111 cwrve .
VE
O urpul
Fie. 54. Aneqw lvel"t c lreu lt of tM Input circu it of en tm l lttr·
follower . The output volt ... le Ye t: Ieee t"-n the Input volta .. .
= Vin -VBE
An equivalent circuit of the
base-emitter relationship is
shown in fig . 5-3. The diode
represents the base-emitter
diode. This illustrates clearly why
the o·utput voltage is VBE less than
the input voltage.
Also, note from fig. 5-1 that the
input and output voltages are in
phase.
Collector Dissipation:
Because the transistor is
basically a resistive device, as
current flows there will be dissi-
pation and , therefore, heat. The
effects of heat regarding VF and lA
fora diode and regarding leo fora
PLAY METER, May 1, 1981
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 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)
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