參數資料
型號: TJA1054
廠商: NXP Semiconductors N.V.
元件分類: CAN
英文描述: Fault-tolerant CAN transceiver(容錯CAN收發(fā)器)
中文描述: 容錯CAN收發(fā)器(容錯的CAN收發(fā)器)
文件頁數: 7/16頁
文件大?。?/td> 75K
代理商: TJA1054
Philips Semiconductors
APPLICATION HINTS
Systems Laboratory Hamburg, Germany
Fault-tolerant CAN transceiver
V2.0
Page 7 of 16
3.3. Average Supply Current at Absence of Bus Short-Circuit Conditions
In recessive state the Transceivers are consuming a Vcc supply current as listed in the corresponding
data sheets. In dominant state the Vcc supply current is calculated by the addition of the IC-internal
supply current (Data sheet, no load condition) and the output current at pins CANH and RTL.
3.3.1. Maximum dominant supply current (without bus wiring faults)
I
cc_dom
= I
cc0_dom
+ I
CANH_dom
+ I
RTL_dom
(1)
I
RTL_dom
= (Vcc - V
CANL_dom
) / R
T
(2)
3.3.2. Example calculation
Maximum dominant supply current without bus wiring faults:
Item from Data Sheet / Assumptions
Max. Vcc supply current dominant, no load
CANH dominant current
Assumed termination resistor
Assumed CANL dominant voltage
Symbol
I
cc0_dom
I
CANH_dom
R
T
V
CANL_dom
PCA82C252
35 mA
40 mA
1 k
1 V
TJA1053
35 mA
40 mA
1 k
1 V
TJA1054
27 mA
40 mA
1 k
1 V
PCA82C252
:
I
cc_dom 252
= 35mA + 40 mA + (5V - 1V) / 1k = 79 mA max.
(Ex 1.1)
TJA1053
:
I
cc_dom 1053
= 35mA + 40 mA + (5V - 1V) / 1k = 79 mA max.
(Ex 1.2)
TJA1054
:
I
cc_dom 1054
= 27mA + 40 mA + (5V - 1V) / 1k = 71 mA max.
(Ex 1.3)
3.3.3. Thermal considerations (without bus wiring faults)
For thermal considerations the average supply current at pin Vcc is relevant considering the transmit
duty cycle. In the following example a continuously transmitting node is assumed. This might happen
e.g. if a node starts a transmission while the rest of the network does not respond with an
acknowledge for some reason. Typically a much lower duty cycle is relevant since a node transmits
messages within certain time slots only, depending on the applications network management.
With an assumed transmit duty cycle of 50% on pin TxD, the maximum average supply current is
I
cc_norm_avg
= 0.5 * (I
cc_rec
+ I
cc_dom
)
(3)
3.3.4. Example calculation
Thermal considerations without bus wiring faults:
Item
Symbol
PCA82C252
TJA1053
TJA1054
Vcc supply current recessive, max.
I
cc_rec
10 mA
10 mA
11 mA
PCA82C252
:
I
cc_norm_avg 252
= 0.5 * (10mA + 79mA) =
44.5 mA max.
(Ex 3.1)
TJA1053
:
I
cc_norm_avg 1053
= 0.5 * (10mA + 79mA) =
44.5 mA max.
(Ex 3.2)
TJA1054
:
I
cc_norm_avg 1054
= 0.5 * (11mA + 71mA) =
41 mA max
.
(Ex 3.3)
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