DATA SHEET
KH563
Case Temperature
As an example of calculating the maximum internal junc-
tion temperatures, consider the circuit of Figure 1 driving
2.5V, 50% duty cycle, square wave into a 50Ω load.
Tc
Case to Ambient
θ
ca
20°C/W
200°C/W
Termal Impedance
Tj(t)
Tj(q)
TA
Pt
Pq
Pcircuit
410Ω ⋅5
5 − 1
Ambient
Req = 50Ω
= 45.6Ω
Temperature
Figure 10:Thermal Model
= V /R total output current
Io = 2.5V/ 45.6Ω = 54.9mA
(
)
I
o
o
eq
2
2
2
1
IT
=
54.9mA + 54.9mA + .06
= 68.1mA
(
)
(
)
R A
f
L
with R = R
total load
eq
L
A − 1
L
PT = 68.1mA 15 − 2.5 − 0.7 − 15.3Ω ⋅ 68.1mA = 733mW
total power in both sides of the output stage
2
2
1
I =
I +
I
+ .06
(
)
t
o
o
2
Pq = 0.1 ⋅ 68.1mA 15 − 1.4 − 17.3Ω ⋅ 68.1mA = 84.5mW
total power in both sides of hottest junctions
prior to output stage
total internal output stage current
P = I ⋅ V − 1.4 − 17.3Ω ⋅ I output stage power
(
)
t
t
CC
t
Pcircuit = 1.3 ⋅ 15 ⋅ 2 ⋅ 68.1mA − 54.9mA + 19.2mA
(
)
− 733mW − 169mW = 1.058W
power in the remainder of circuit
P = 0.1 ⋅I ⋅ V − V − 0.7 − 15.3Ω ⋅I
(
)
q
t
CC
o
t
power in hottest internal junction
prior to output stage
With these powers and TA = 25°C and θca = 35°C/W
P
= 1.3 ⋅ V ⋅ 2 ⋅I − I + 19.2mA − P − P
(
)
T = 25°C + .733 + .169 + 1.058 ⋅ 35 = 94°C
circuit
CC
t
o
t
q
(
)
c
power in remainder of circuit [note V = |− V |]
case temperature
CC
CC
Note that the P and P equations are written for positive
From this, the hottest internal junctions may be found as
t
q
V . Absolute values of -V , V , and I , should be used
o
CC
o
o
1
for a negative going V . since we are only interested in
T t = 94°C +
.733 ⋅20 = 101°C output stage
j ( )
(
)
o
2
delta V’s. For bipolar swings, the two powers for each
output polarity are developed as shown above then
ratioed by the duty cycle. Having the total internal power,
as well as its component parts, the maximum junction
temperature may be computed as follows.
1
T q = 94°C +
j ( )
.0845 ⋅200 = 102°C
(
)
2
hottest internal junction
Note that 1/2 of the total P and P powers were used
T
a
here since the 50% duty cycle output splits the power
evenly between the two halves of the circuit whereas the
total powers were used to get case temperature.
T = T + (P + P + P
) • θ Case Temperature
ca
c
A
q
T
circult
θ
= 35°C/W for the KH563 with no heatsink in still air
ca
T
= T + P • 20°C/W
c t
output transistor junction temperature
j(t)
Even with the output current internally limited to 250mA,
the KH563’s short circuiting capability is principally a
thermal issue. Generally, the KH563 can survive short
duration shorts to ground without any special effort. For
protection against shorts to the 15 volt supply voltages,
it is very useful to reduce some of the voltage across the
output stage transistors by using some external output
T
= T + P • 200°C/W
j(q)
c
q
hottest internal junction temperature
The Limiting Factor for Output Power is Maximum
Junction Temperature
Reducing
airflow can greatly reduce the junction temperatures.
One effective means of heatsinking the KH563 is to use
a thermally conductive pad under the part from the pack-
age bottom to a top surface ground plane on the compo-
θ
through either heatsinking and/or
ca
resistance, R , as shown in Figure 9.
x
Evaluation Board
An evaluation board (part number 730019) for the KH563
nent side. Tests have shown a θ of 24°C in still air using
is available.
ca
a “Sil Pad” available from Bergquist (800-347-4572).
12
REV. 1A January 2008
CADEKA [ CADEKA MICROCIRCUITS LLC. ]
