Datasheet

LME49600
www.ti.com SNAS422E –JANUARY 2008–REVISED APRIL 2013
P
DMAX(DC)
= (I
S
x V
S
) + (V
S
)
2
/ R
L
(Watts)
where
• V
S
= |V
EE
| + V
CC
(V)
• I
S
=quiescent supply current (A) (3)
Equation (2) is for sinusoidal output voltages and Equation (3) is for DC output voltages.
2. Determine the maximum allowable die temperature rise,
T
RISE(MAX)
= T
J(MAX)
- T
A(MAX)
(°C) (4)
3. Using the calculated value of T
RISE(MAX)
and P
D(MAX)
, find the required value of junction to ambient thermal
resistance combining Equation (1) and Equation (5) to derive Equation (9):
θ
JA
= T
RISE(MAX)
/ P
D(MAX)
(5)
4. Finally, choose the minimum value of copper area from Figure 30 based on the value for θ
JA
.
Example
Assume the following conditions: V
S
= |V
EE
| + V
CC
= 30V, R
L
= 32Ω, I
S
= 15mA, sinusoidal output voltage, T
J(MAX)
= 125°C, T
A(MAX)
= 85°C.
Applying Equation (3):
P
DMAX
= (I
S
x V
S
) + (V
S
)
2
/ 2π
2
R
L
= (15mA)(30V) + 900V
2
/ 142Ω
= 1.86W (6)
Applying Equation (5):
T
RISE(MAX)
= 125°C – 85°C
= 40°C (7)
Applying Equation (9):
θ
JA
= 40°C/1.86W
= 21.5°C/W (8)
Examining the Copper Area vs. θ
JA
plot indicates that a thermal resistance of 50°C/W is possible with a 12in
2
plane of one layer of 1oz copper. Other solutions include using two layers of 1oz copper or the use of 2oz
copper. Higher dissipation may require forced air flow. As a safety margin, an extra 15% heat sinking capability is
recommended.
When amplifying AC signals, wave shapes and the nature of the load (reactive, non-reactive) also influence
dissipation. Peak dissipation can be several times the average with reactive loads. It is particularly important to
determine dissipation when driving large load capacitance.
The LME49600’s dissipation in DC circuit applications is easily computed using Equation (4). After the value of
dissipation is determined, the heat sink copper area calculation is the same as for AC signals.
SLEW RATE
A buffer’s voltage slew rate is its output signal’s rate of change with respect to an input signal’s step changes.
For resistive loads, slew rate is limited by internal circuit capacitance and operating current (in general, the higher
the operating current for a given internal capacitance, the faster the slew rate).
However, when driving capacitive loads, the slew rate may be limited by the available peak output current
according to the following expression.
dv/dt = I
PK
/ C
L
(9)
Output voltages with high slew rates will require large output load currents. For example if the part is required to
slew at 1000V/μs with a load capacitance of 1nF, the current demanded from the LME49600 is 1A. Therefore,
fast slew rate is incompatible with a capacitive load of this value. Also, if C
L
is in parallel with the load, the peak
current available to the load decreases as C
L
increases.
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