Philips Semiconductors Product data
AU5790Single wire CAN transceiver
2001 May 18
16
Power Dissipation
Power dissipation of an IC is the major factor determining junction
temperature. AU5790 power dissipation in active and passive states
are different. The average power dissipation is:
P
tot
= P
INT
*Dy + P
PNINT
* (1-Dy)
where: P
tot
is total dissipation power;
P
INT
is dissipation power in an active state;
P
PNINT
is dissipation power in a passive state;
Dy is duty cycle, which is the percentage of time that TxD
is in an active state during any given time duration.
At passive state there is no current going into the load. So
all of the supply current is dissipated inside the IC.
P
PNINT
= V
BAT
* I
BATPN
where: V
BAT
is the battery voltage;
I
BATPN
is the passive state supply current in normal mode.
In an active state, part of the supply current goes to the
load, and only part of the supply current dissipates inside
the IC, causing an incremental increase in junction
temperature.
P
INT
= P
BATAN
– P
LOADN
where: P
BATAN
is active state battery supply power in normal
mode;
P
BATAN
= V
BAT
* I
BATAN
P
LOADN
is load power consumption in normal mode.
P
LOADN
= V
CANHN
* I
LOADN
where: I
BATAN
is active state supply current in normal mode;
V
CANHN
is bus output voltage in normal mode;
I
LOADN
is current going through load in normal mode.
I
LOAD
= V
CANHN
/R
LOAD
I
BATN
= I
LOAD
+ I
INT
where: I
INT
is an active state current dissipated within the IC in
normal mode.
I
INT
will decrease slightly when the node number
decreases. To simplify this analysis, we will assume I
INT
is
fixed.
I
INT
= I
BATN
(32 nodes) – I
LOAD
(32 nodes)
I
BATN
(32 nodes) may be found in the DC Characteristics
table.
A power dissipation example follows. The assumed values
are chosen from specification and typical applications.
Assumptions:
V
BAT
= 13.4 V
R
T
= 9.1 kΩ
32 nodes
I
BATPN
= 2 mA
I
BATN
(32 nodes) = 35 mA
V
CANHN
= 4.55 V
Duty cycle = 50%
Computations:
R
LOAD
= 9.1 kΩ / 32 = 284.4 Ω
P
PNINT
= 13.4 V × 2 mA = 26.8 mW
I
LOAD
= 4.55 V / 284.4 Ω = 16mA
P
LOADN
= 4.55 V × 16 mA = 72.8 mW
I
INT
= 35 mA - 16 mA = 19 mA
P
BATAN
= 13.4 V × 35 mA = 469 mW
P
INT
= 469 mW - 72.8 mW = 396.2 mW
P
tot
= 396.2 mW × 50% + 26.8 mW × (1-50%) = 211.5 mW
Additional examples with various node counts are shown in Table 4.
Table 4. Representative Power Dissipation Analyses
Nodes
R
LOAD
(Ω)
V
BAT
(V)
I
BATPN
(mA)
P
PNINT
(mW)
V
CANHN
(V)
I
LOAD
(mA)
I
BATN
(mA)
I
INT
(mA)
P
INT
(mW)
Dcycle
P
tot
(mW)
2 4550 13.4 2 26.8 4.55 1 20 19 263.5 0.5 145.1
10 910 13.4 2 26.8 4.55 5 24 19 298.9 0.5 162.8
20 455 13.4 2 26.8 4.55 10 29 19 343.1 0.5 184.9
32 284.4 13.4 2 26.8 4.55 16 35 19 396.2 0.5 211.5
2 4550 26.5 2 53 4.55 1 20 19 525.5 0.5 289.2
10 910 26.5 2 53 4.55 5 24 19 613.3 0.5 333.1
20 455 26.5 2 53 4.55 10 29 19 723 0.5 388
32 284.4 26.5 2 53 4.55 16 35 19 854.7 0.5 453.8
By knowing the maximum power dissipation, and the operation ambient temperature, the required thermal resistance without tripping the
thermal protection can be calculated, as shown in Figure 7. Then from Figure 5 or 6, a suitable PCB can be selected.
