January 1965 Electronics World
People old and young
enjoy waxing nostalgic about and learning some of the history of early electronics. Electronics World
was published from May 1959 through December 1971. See all
Electronics World articles.
Not everybody with a high temperature semiconductor application
in need of heat dissipation has access to a thermal management
program with a database of available commercial heat sinks and/or
an ability to analyze a custom-made heat sink. This article
contains simple equations, a handy chart, and instructions on
how to use them to figure out what kind of heat sink you need
for your project.
Semiconductor Heat Sink Design Chart
By Frank D. Gross
Simple method of determining how large a heat sink area is
required for SCR's and other heat-producing semiconductors.
Heat sink information seems to be rarely, if ever, in a usable
form, particularly for SCR and other switching circuits. This
nomogram directly relates the load an SCR is controlling to
the required heat sink area. The nomogram may be extended to
apply to any semiconductor.
These curves will give a 55° F (13° C) temperature
rise of the heat sink above room temperature when using a square,
vertical 1½ -inch aluminum plate with a minimum of 1/8
inches of side clearance and unobstructed top and bottom access
to ambient air. The heat sink may be any color as long as any
coatings used on the aluminum are as thin as possible.
The heat produced in an SCR is caused by two factors; namely,
a brief power pulse during turn on, and the continuous power
loss due to the forward drop of the p-n-p-n junction. In all
power frequency circuits (1 kc. or less), the turn on of the
SCR is so fast that only the forward drop need be considered.
Put another way, the duty cycle of the turn-on power pulses
is very low. The heat produced by the SCR due to forward drop
loss is given by Ploss = Vf X Iload
where Vf is the forward drop which varies with the
load current but never exceeds 1.1 volts when the SCR is run
within its continuous power rating. Let us make the conservative
assumption that the forward drop is always exactly 1.15 volts
and that the SCR is on all the time (or half the time in a halfwave
circuit). This means we can assume that the power loss in an
SCR is equal to 1% of the maximum load power since the load
power is given by 115 Iload and the forward loss
is assumed to be 1.15 Iload. This is strictly a worst-case
assumption as the power loss will be considerably less when
lower conduction angles (less load power due to speed or brightness
setting) are chosen.
Typical examples of commercial heat sinks where fins are
used to increase the effective cooling surface.
Heat transfer is accomplished in two ways by the heat sink:
convection and radiation. Convection is almost always the stronger
of the two transfer mechanisms. Radiation is very much a function
of the color and roughness of the heat sink surface and can
approach zero for a smooth, highly polished surface. Convection
is independent of these parameters. Let us make a second assumption
that all of the heat transfer is provided by convection. Again,
this is a worst-case assumption.
The physics book says that Q = Hc X A X ΔT
where Q is watts of heat transferred by convection, Hc
is convection transfer constant, A is surface area, and ΔT
is temperature difference between ambient and heat sink.
Hc is a constant of heat transfer which is given
by Hc= 0.0022(L/ΔT)1/4 where L is
the vertical length of a square metal plate in inches and T
is the temperature difference in degrees centigrade between
the plate and the ambient air.
An SCR is capable of safely operating at case temperatures
that can cause serious burns to humans. In any SCR control,
consideration should be given to what the operator or user of
the equipment can stand and not to the ultimate temperature
damaging to the SCR. This is especially true in small dimmers
and power-tool controls where the case doubles as a heat sink.
Operation at heat sink temperatures safe to humans allows the
SCR to run well within its ratings, enhancing circuit life and
A metal plate at 140°F (60°C) may be described as
alarmingly hot. No burn damage will occur, but substantial will
power is required to hold onto a metal plate at this temperature.
Above this temperature, the probability of a burn rapidly increases.
A choice of 55°F (13°C) of allowable temperature rise
permits the heat sink to stay below the critical temperature
for any ambient temperature below 85°F. This is quite reasonable
for most SCR applications. The heat sink is normally well below
this design temperature except during full-on operation.
The geometry assumed for the nomogram is a vertical, square,
1/8-inch thick piece of aluminum with both sides exposed to
the cooling air. A minimum clearance of 1 1/2 inches on either
side is assumed. It is also assumed that there is no obstruction
to ambient air either above or below the heat sink. A bit of
thought will allow this geometry to be distorted into any heat
sink geometry required for a specific application.
Generally, the nomogram will give quite conservative results,
e.g., the heat sink temperatures will be less than predicted.
If the SCR is used in a half-wave circuit (or a full-wave
circuit in which the alternate half cycles are conducted by
a diode or other SCR), only half the normal SCR power is produced
since the SCR is only on half the time. Because of this, a heat
sink of one-half the area (or a square of 0.707 L) is required.
If only one side of the heat sink is available to the cooling
air, then twice the required area (or 1.41 times each side)
must be used.
Actually, the nomogram is simply a plot of how many watts
a heat sink can transfer and is by no means limited to SCR's.
Any semiconductor or, for that matter, any heat producer will
provide the same results.
If higher heat sink temperatures are permitted, the reduction
in area is proportional to the allowable temperature rise. For
instance, if a 110°F rise is permitted, only half the required
area for the 55°F case is needed. If higher temperature
operation is used, the SCR must have a very low thermal resistance
to the heat sink. This means that at most a thin mica or anodized-aluminum
insulating washer may be placed between SCR and heat sink. The
use of silicone grease is mandatory in this case.
Here are some examples that show nomogram use.
1. How large a two-sided heat sink is required for a 1-kw.,
115-v.a.c. light dimmer using an SCR in a half-wave circuit?
Answer: The 1000-va. line is followed horizontally
across the nomogram till it intersects the half-wave curve.
The required size is read vertically downward. The answer is
five inches square.
If only one side of the heat sink has access to cooling air,
the same area is still required. The area is 2 * 5 * 5 = 50
in.2 This is equal to a one-sided square slightly
over seven inches on a side. The area need not be calculated
if you are using a square geometry. Simply multiply the original
side by 1.41, or in this case 5 * 1.41 = 7 inches.
These curves will give a 55° F. (13° C) temperature
rise of the heat sink above room temperature when using a square,
vertical 1/8-inch aluminum plate with a minimum of 1 1/2 inches
of side clearance and unobstructed top and bottom access to
ambient air. The heat sink may be any color as long as any coatings
used on the aluminum are as thin as possible.
2. How large a heat sink is required for a bilateral SCR
operating a 2.2-amp. electric drill as a power-tool control?
Answer: The drill voltampere rating is given by
115 * I or 115 * 2.2 = 253 va. Following the value horizontally
to the full-wave curve and reading downward gives 3 1/2 inches
3. Two SCR's are used as a contactor for a 115-v.a.c., 1-h.p.
induction motor, both mounted on the same heat sink. What size
Answer: Two half-wave SCR's are the same as one
full-wave one, so the full-wave curve must be used. The voltampere
load of the motor must be found. One horsepower equals 746 watts.
The efficiency of the motor at full load is probably above 90%,
and the power factor is most likely to be 0.8 or better. The
voltamperes drawn are then equal to 746/(0.8 * 0.9) = 1036 va.
An 8-inch heat sink is required.
4. A germanium transistor is used in a 400-cps static inverter
that draws five amperes from a 28-volt d.c. line. As the circuit
is push-pull, the transistor is only on half the time. What
size heat sink is required?
Answer: The low frequency of the inverter allows
us to assume that most of the heat produced is during the conduction
or on time and that we may neglect switching-power losses during
on time. The saturation voltage may be found on the transistor
data sheet, or it may be assumed to be less than 0.3 volt. The
heat produced is then equal to 5 amps X 3 volt = 1.5 watts.
Since the transistor operates on a 50% duty cycle, .75 watt
of heat must be continuously removed by the heat sink. The right
ordinate of the nomogram and the full-wave curve are used to
give an answer of one inch. As this is quite small, it might
be better to consider a larger value to account for turn-on
losses and starting transients. Three inches would be a good
Posted February 6, 2015