January 1965 Electronics World
Table
of Contents
Wax nostalgic about and learn from the history of early electronics. See articles
from
Electronics World, published May 1959
 December 1971. All copyrights hereby acknowledged.

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 custommade heat sink. This article from a 1965 issue
of Electronics World magazine 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. At the time TO−8 and TO−3 metal cans were a couple of the most
common sizes for which a large variety of heatsinks were available.
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 heatproducing 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 pnpn
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 turnon power pulses is very low. The heat produced by the SCR due
to forward drop loss is given by P_{loss} = V_{f} X I_{load}
where V_{f} 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 I_{load} and the forward
loss is assumed to be 1.15 I_{load}. This is strictly a worstcase 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 worstcase assumption.
The physics book says that Q = H_{c} X A X ΔT where Q is watts
of heat transferred by convection, H_{c} is convection transfer constant,
A is surface area, and ΔT is temperature difference between ambient and heat
sink.
Varieties of TO3 package style heatsinks.
H_{c} is a constant of heat transfer which is given by H_{c}=
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 powertool 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 reliability.
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 fullon operation.
The geometry assumed for the nomogram is a vertical, square, 1/8inch 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 halfwave circuit (or a fullwave 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 onehalf 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 anodizedaluminum 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 twosided heat sink is required for a 1kw., 115v.a.c. light
dimmer using an SCR in a halfwave circuit?
Answer: The 1000va. line is followed horizontally across the nomogram
till it intersects the halfwave 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 onesided
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 1/8inch 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 2.2amp.
electric drill as a powertool control?
Answer: The drill voltampere rating is given by 115 * I or 115 * 2.2
= 253 va. Following the value horizontally to the fullwave curve and reading downward
gives 3 1/2 inches square.
3. Two SCR's are used as a contactor for a 115v.a.c., 1h.p. induction motor,
both mounted on the same heat sink. What size is required?
Answer: Two halfwave SCR's are the same as one fullwave one, so the
fullwave 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 8inch heat sink is required.
4. A germanium transistor is used in a 400cps static inverter that draws five
amperes from a 28volt d.c. line. As the circuit is pushpull, 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
switchingpower 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 fullwave 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 turnon losses and starting transients. Three inches would be a good
choice.
Posted November 4, 2020 (updated from original post on 2/6/2015)
