PIC18F4580T-I/ML Microchip Technology, PIC18F4580T-I/ML Datasheet - Page 341

• Manufacturer Part Number: PIC18F4580T-I/ML
• Description: IC PIC MCU FLASH 16KX16 44QFN
• Manufacturer: Microchip Technology
• Series: PIC® 18Fr

Specifications of PIC18F4580T-I/ML

• Core Processor: PIC
• Core Size: 8-Bit
• Speed: 40MHz
• Connectivity: CAN, I²C, SPI, UART/USART
• Peripherals: Brown-out Detect/Reset, HLVD, POR, PWM, WDT
• Number Of I /o: 36
• Program Memory Size: 32KB (16K x 16)
• Program Memory Type: FLASH
• Eeprom Size: 256 x 8
• Ram Size: 1.5K x 8
• Voltage - Supply (vcc/vdd): 4.2 V ~ 5.5 V
• Data Converters: A/D 11x10b
• Oscillator Type: Internal
• Operating Temperature: -40°C ~ 85°C
• Package / Case: 44-QFN
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant

24.9.2
As already mentioned, the Time Quanta is a fixed unit
derived from the oscillator period and baud rate
prescaler. Its relationship to T
Rate is shown in Example 24-6.
EXAMPLE 24-6:
The frequencies of the oscillators in the different nodes
must be coordinated in order to provide a system wide
specified nominal bit time. This means that all oscilla-
tors must have a T
It should also be noted that although the number of T
is programmable from 4 to 25, the usable minimum is
8 T
8 T
© 2009 Microchip Technology Inc.
T
T
Nominal Bit Rate (bits/s) = 1/T
This frequency (F
frequency used. If, for example, a 10 MHz external
signal is used along with a PLL, then the effective
frequency will be 4 x 10 MHz which equals 40 MHz.
CASE 1:
For F
Nominal Bit Time = 8 T
T
T
Nominal Bit Rate = 1/10
CASE 2:
For F
Nominal Bit Time = 8 T
T
T
Nominal Bit Rate = 1/1.6 * 10
CASE 3:
For F
Nominal Bit Time = 25 T
T
T
Nominal Bit Rate = 1/1.28 * 10
Q
BIT
Q
BIT
Q
BIT
Q
BIT
Q
Q
= (2 * 1)/16 = 0.125 μs (125 ns)
= (2 * 2)/20 = 0.2 μs (200 ns)
= (2 * 64)/25 = 5.12 μs
. There is no assurance that a bit time of less than
(μs) = (2 * (BRP+1))/F
in length will operate correctly.
(μs) = T
= 8 * 0.125 = 1 μs (10
= 8 * 0.2 = 1.6 μs (1.6 * 10
= 25 * 5.12 = 128 μs (1.28 * 10
OSC
OSC
OSC
= 16 MHz, BRP<5:0> = 00h and
= 20 MHz, BRP<5:0> = 01h and
= 25 MHz, BRP<5:0> = 3Fh and
TIME QUANTA
Q
(μs) * number of T
OSC
OSC
CALCULATING T
NOMINAL BIT RATE AND
NOMINAL BIT TIME
that is an integral divisor of T
-6
Q
Q
) refers to the effective
OSC
= 10
Q
:
:
-6
:
s)
-6
BIT
(MHz)
6
-4
BIT
s = 625,000 bits/s
-6
bits/s (1 Mb/s)
= 7813 bits/s
s)
Q
and the Nominal Bit
per bit interval
-4
(625 Kb/s)
(7.8 Kb/s)
s)
Q
,
PIC18F2480/2580/4480/4580
Q
Q
.
24.9.3
This part of the bit time is used to synchronize the
various CAN nodes on the bus. The edge of the input
signal is expected to occur during the sync segment.
The duration is 1 T
24.9.4
This part of the bit time is used to compensate for phys-
ical delay times within the network. These delay times
consist of the signal propagation time on the bus line
and the internal delay time of the nodes. The length of
the propagation segment can be programmed from
1 T
24.9.5
The phase buffer segments are used to optimally
locate the sampling point of the received bit within the
nominal bit time. The sampling point occurs between
Phase Segment 1 and Phase Segment 2. These
segments can be lengthened or shortened by the
resynchronization
Segment 1 determines the sampling point within a bit
time. Phase Segment 1 is programmable from 1 T
8 T
before the next transmitted data transition and is also
programmable from 1 T
due to IPT requirements, the actual minimum length of
Phase Segment 2 is 2 T
equal to the greater of Phase Segment 1 or the
Information Processing Time (IPT). The sampling point
should be as late as possible or approximately 80% of
the bit time.
24.9.6
The sample point is the point of time at which the bus
level is read and the value of the received bit is deter-
mined. The sampling point occurs at the end of Phase
Segment 1. If the bit timing is slow and contains many
T
line at the sample point. The value of the received bit is
determined to be the value of the majority decision of
three values. The three samples are taken at the sam-
ple point and twice before, with a time of T
each sample.
24.9.7
The Information Processing Time (IPT) is the time
segment starting at the sample point that is reserved
for calculation of the subsequent bit level. The CAN
specification defines this time to be less than or equal
to 2 T
define this time to be 2 T
must be at least 2 T
Q
, it is possible to specify multiple sampling of the bus
Q
Q
to 8 T
in duration. Phase Segment 2 provides a delay
Q
. The PIC18F2480/2580/4480/4580 devices
Q
SYNCHRONIZATION SEGMENT
PROPAGATION SEGMENT
PHASE BUFFER SEGMENTS
SAMPLE POINT
INFORMATION PROCESSING TIME
by setting the PRSEG<2:0> bits.
Q
Q
.
process.
long.
Q
Q
to 8 T
Q
, or it may be defined to be
. Thus, Phase Segment 2
The
Q
in duration. However,
DS39637D-page 341
end
Q
/2 between
of
Phase
Q
to