DATA SHEET
KH561
Case Temperature
4572).
Tc
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.
Case to Ambient
θ
ca
20°C/W
200°C/W
Termal Impedance
Tj(t)
Pt
Tj(q)
TA
Pq
Pcircuit
410Ω 5
Ambient
Temperature
Req = 50Ω
= 45.6Ω
5 −1
Figure 10: Thermal Model
I = V / R total output current
I = 2.5V / 45.6Ω = 54.9mA
(
)
o
o
o
eq
2
2
1
IT =
54.9mA + 54.9mA + .06
= 68.1mA
(
)
(
)
2
R A
f
L
withR = R
total load
eq
L
P = 68.1mA 15 − 2.5 − 0.7 −15.3Ω 68.1mA = 733mW
A −1
[
]
T
L
total power in both sides of the output stage
2
2
1
I =
I +
I + .06
o
(
)
t
o
P = 0.2 68.1mA 15 −1.4 −17.3Ω 68.1mA = 169mW
2
[
]
q
total power in both sides of hottest junctions
prior to output stage
total internal output stage current
P = I −1.4 −17.3Ω I output stage power
V
(
)
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 = 0.2 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
T = 25°C + .733 +.169 +1.058 35 = 94°C
(
)
P
= 1.3 V
2 I −I +19.2mA −P −P
q
c
(
)
circuit
CC
t
o
t
case temperature
power in remainder of circuit [note V = | −V |]
CC
CC
From this, the hottest internal junctions may be found as
Note that the P and P equations are written for positive
t
q
T t = 94°C + 1 .733 20 = 101°C output stage
V . Absolute values of -V , V , and I , should be used
j( )
(
)
o
CC
o
o
2
for a negative going V . since we are only interested in
o
T q = 94°C + 1 .169 200 = 111°C
j( )
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.
(
)
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 KH561 with no heatsink in still air
ca
Even with the output current internally limited to 250mA,
the KH561’s short circuiting capability is principally a
thermal issue. Generally, the KH561 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 • 20°C/W
c t
j(t)
output transistor junction temperature
= T + P • 200°C/W
T
j(q)
c
q
hottest internal junction temperature
The Limiting Factor for Output Power is Maximum
Junction Temperature
Reducing
θ
through either heatsinking and/or
resistance, R , as shown in Figure 9.
ca
x
airflow can greatly reduce the junction temperatures.
One effective means of heatsinking the KH561 is to use
a thermally conductive pad under the part from the pack-
age bottom to a top surface ground plane on the compo-
Evaluation Board
An evaluation board (part number 730019) for the KH561
is available.
nent side. Tests have shown a θ of 24°C in still air
ca
using a “Sil Pad” available from Bergquist (800-347-
12
REV. 1A January 2004
CADEKA [ CADEKA MICROCIRCUITS LLC. ]
