PIC18F-LF1XK50 MICROCHIP [Microchip Technology], PIC18F-LF1XK50 Datasheet - Page 273

• Manufacturer Part Number: PIC18F-LF1XK50
• Description: 20-Pin USB Flash Microcontrollers
• Manufacturer: MICROCHIP [Microchip Technology]

EQUATION 22-1:
EXAMPLE 22-2:
 2010 Microchip Technology Inc.
For this example, the following assumptions are made about the application:
• 3.3V will be applied to V
• This is a full-speed application that uses one interrupt IN endpoint that can send one packet of 64 bytes every
• A regular USB “B” or “mini-B” connector will be used on the application circuit board.
In this case, P
the IN endpoint. All 64 kBps of data could potentially be bytes of value, 00h. Since ‘0’ bits cause toggling of the
output state of the transceiver, they cause the USB transceiver to consume extra current charging/discharging the
cable. In this case, 100% of the data bits sent can be of value ‘0’. This should be considered the “max” value, as
normal data will consist of a fair mix of ones and zeros.
This application uses 64 kBps for IN traffic out of the total bus bandwidth of 1.5 MBps (12 Mbps), therefore:
Since a regular “B” or “mini-B” connector is used in this application, the end user may plug in any type of cable up
to the maximum allowed 5 m length. Therefore, we use the worst-case length:
Assume I
USB bandwidth is shared between all the devices which are plugged into the root port (via hubs). If the application
is plugged into a USB 1.1 hub that has other devices plugged into it, your device may see host to device traffic on
the bus, even if it is not addressed to your device. Since any traffic, regardless of source, can increase the I
current above the base 218 A, it is safest to allow for the worst-case of 2.2 mA.
Therefore:
The calculated value should be considered an approximation and additional guardband or application-specific prod-
uct testing is recommended. The transceiver current is “in addition to” the rest of the current consumed by the
PIC18F1XK50/PIC18LF1XK50 device that is needed to run the core, drive the other I/O lines, power the various
modules, etc.
Legend:
1 ms, with no restrictions on the values of the bytes being sent. The application may or may not have addi-
tional traffic on OUT endpoints.
L
CABLE
PULLUP
V
P
P
L
I
PULLUP
CABLE
USB
ZERO
IN
= 5 meters
:
ZERO
:
= 2.2 mA. The actual value of I
:
:
:
= 100% = 1, because there should be no restriction on the value of the data moving through
ESTIMATING USB TRANSCEIVER CURRENT CONSUMPTION
CALCULATING USB TRANSCEIVER CURRENT
I
is fully utilized (either IN or OUT traffic) for data that drives the lines to the “K” state most of the time.
use cables no longer than 5m.
the host or hub end of the USB cable, 15 k nominal resistors (14.25 k to 24.8 k) are present which
pull both the D+ and D- lines to ground. During bus Idle conditions (such as between packets or during
USB Suspend mode), this results in up to 218 A of quiescent current drawn at 3.3V.
Voltage applied to the V
Percentage (in decimal) of the IN traffic bits sent by the PIC
Percentage (in decimal) of total bus bandwidth that is used for IN traffic.
Length (in meters) of the USB cable. The USB 2.0 specification requires that full-speed applications
Current which the nominal, 1.5 k pull-up resistor (when enabled) must supply to the USB cable. On
PULLUP
I
XCVR
USB
I
XCVR
is also dependant on bus traffic conditions and can be as high as 2.2 mA when the USB bandwidth
and V
=
=
(60 mA • V
DD
(60 mA • 3.3V • 1 • 0.043 • 5m)
, with the core voltage regulator enabled.
Pin =
USB
USB
(3.3V • 5m)
1.5 MBps
PULLUP
pin in volts. (Should be 3.0V to 3.6V.)
64 kBps
Preliminary
(3.3V • 5m)
• P
ZERO
will likely be closer to 218 A, but allow for the worst-case.
= 4.3% = 0.043
• P
IN
• L
CABLE
+ 2.2 mA = 4.8 mA
PIC18F/LF1XK50
)
®
†
+ I
device that are a value of ‘0’.
PULLUP
DS41350E-page 273
PULLUP