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-ss c
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T A -
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200 G
'\mhtf'nl lrm1wro 1ur''
T A -
Fig . 2·5. A compadaon ot the variation• In operallonal
t h lrac tet lltlca due to c hanges In T A·
example from point A to point B.
the reverse current [lA] increases
from point A' to point B'.ltshould
be observed that in the reverse
biased state the reverse voltage
[VA] determines the current
through the diode [lA]. as oppoed
to IF controlling VF in forward
biased state. The battery voltage
can be increased to the limit Bv
(breakdown voltage) . Bv is stated
by the manufacturer and when
exceeded, the diode will enter the
avalanche region . Once a diode is
operated in the avalanche region
it will be destroyed .
Thermal Effects:
The resistance of the semi-
conductor material lies some-
where between that of an
insulator and that of a conductor.
The resistance of the semicon -
ductor material is determined by
the availability of current carriers
for a given volume . In conductors
(copper wire , etc .). current flow is
the result of free electrons
reacting to applied EMF (voltage) .
Current flow through semicon -
ductors results from electrons
and holes, which are the absence
of electrons from the internal
crystal structures. The amount of
electrons and holes in the semi-
conductor is the result of a careful
blend of impurities to the basic
material ; also, the resistance of
the block is inversely pro-
portionate to the number of
carriers (electrons and holes) .
PROGRAMMED TEST,
LESSON TWO
Instructions: The purpose of this test is to
guide you step-by-step through actual
circuit design problems. Also. the tests in
the Digital Circuit Design Course may
provi de you wi th addit ional design
technique. Most im portant, 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.
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Fig. 2-6 . A typical thermal denting curve •• g iven on manu -
lactwes data l hHIS. An lncreaae In T A rt1ul11 I n I dec:reaae
In Pd .
Ambu~ n l
te mpero turl"
Fit . 2· 7 . The 1tudent ahoul d eaUrMte Pd 1 M . . ) lor operation
at uo• c.
As the ambient temperature
[T A] is increased , more carriers
become available and the
resistance of the device lowers.
The increase in carriers is the
result of some of the electrons
absorbing energy from heat.
Inversely, should the ambient
temperature be lowered, elec-
trons would give off energy and
the result would be fewer
available carriers. The effects of
change in ambient temperature
are shown in figure 2-5. Note that
when the ambient temperature
decreases, making fewer carriers
available, VF increases .
Should the ambient tempera-
ture be higher than the manufac-
turers ' stated limit , the diode will
go into the condition of thermal
runaway , as was the case with
internal dissipation [Pd[maxJ] .
Most manufacturers provide a
derating curve or multiplier on the
data sheets which accounts for
both internal dissipation and
ambient temperature. Figure 2-6
is a typical derating curve
showing that when the ambient
temperature is increased, Pd[max 1
decreases. Quite often, however,
the manufacturer will state de-
rating information in the form of a
decrease in power dissipation per
degree centigrade increase in
ambient temperature above 25°
centigrade . For example , 150
milliwatts per degree centigrade
would be shown as (150 mw/° C.] .
When stated in this form , the
manufacturer is using a short-
hand method of describing figure
2-6. To use the stated informa-
tion , the designer need only to
subtract 25°C from the ambient
temperature and multiply the
difference by the power decrease
and then subtract the answer
from the stated Pd [max 1 at 25° C.
1
Refer to the text and return to BLOCK 11 .
The voltage drop across a diode when it is
forward biased is
a.
VF
GO TO BLOCK 20
b.
Bv
GO TO BLOCK 15
4 YOU ARE CORRECT!
2
YOU ARE INCORRECT!
EXAMPLE 1:
If the curve in figure 2-7 is to be
used to derate a particular diode,
determine the maximum dissipa-
tion at 150° C.
EXAMPLE 2:
A diode is rate d at 250mw at
25° C and must be derated
2mw/° C. We wish to operate this
d iode at 75° C . What is the
maximum allowable dissipation
at75°C?
First subtract to find the
difference in operating tempera-
tures:
75 ° C - 25°C = 50°C
Multiply the difference by 2mw .
2mw x 50
= 100 mw
Subtract the above from 250 mw .
250mw - 100 mw
><
150mw
ANSWERS TO EXAMPLES-
1. The diode can dissipate up to
65mv at 150°C.
2. The diode may dissipate a
maximum of 150mv at 75° C.
End of lesson two .
The reverse operating limit is
a.
Bv
GO TO BLOCK 11
b.
VR
TO BLOCK 23
5
YOU ARE INCORRECT!
Refer to the text and return to BLOCK 20
Refer to the text and return to BLOCK 25.
3
YOU ARE INCORRECT!
6
YOU ARE CORRECT!
9