Difference between revisions of "Embedded System Tutorial UART"
(Created page with "<font color="red"> This article is under construction</font> == Introduction == '''UART''' stands for '''U'''niversal '''A'''synchronous '''R'''eceiver '''T'''ransmitter. Th...")
Revision as of 20:48, 31 July 2013
This article is under construction
UART stands for Universal Asynchronous Receiver Transmitter. There is one wire for transmitting data, an done wire to receive data. A common parameter is the baud rate known as "bps" which stands for bits per second. If a transmitter is configured with 9600bps, then the receiver must be listening on the other end at the same speed.
UART is a serial communication, so bits must travel on a single wire. If you wish to send a char over UART, the char is enclosed within a start and a stop bit, so to send 8-bits of char data, it would require 2-bit overhead; this 10-bit of information is called a UART frame. Let's take a look at how the character 'A' is sent over UART. In ASCII table, the character 'A' has the value of 65, which in binary is: 0100.0101 If you inform your UART hardware that you wish to send this data at 9600bps, here is how the frame would appear on an oscilloscope : <Picture needed>
A micrcontroller can have multiple UARTs in its hardware, and usually UART0 is interfaced to a "USB to serial" converter chip which is then connected to your computer. In this exercise, you will write a driver for UART-2 and attempt to communicate between two boards.
Clock System & Timing
A crystal drives a processor clock, and it is usually less than 20Mhz. A processor usually uses a "PLL" or "phased-lock-loop" to generate a faster clock than the crystal. So, you could have a 4Mhz clock, and the PLL can be used to internally multiply the clock to provide 48Mhz to the processor. The same 48Mhz is then fed to processor peripherals, and sometimes you have a register that can divide this higher clock to slower peripherals that may not require a high clock rate. Remember that lower clock speed means lower power consumption.
9600bps means that one bit takes 1/9600 = 104uS (micro-seconds) per bit. The idea is that we want to divide the peripheral clock to UART hardware by a number that yields roughly 104uS per bit. The Software Driver section goes over how to configure your UART driver to divide the clock to yield the desired baud rate.
There is not much hardware design other than to locate UART-2 pins on your processor board and connecting these wires to the second board. Each pin on a microcontroller may be designed to provide specific feature. So the first thing to do is identify which physical pins can provide UART-2 signals.
After you identify the physical pins, you would connect these pins to the second board. Remember that your TX pin should be connected to second board's RX pin and vice versa. Connecting two TX pins together will damage your processor. After you connect the Rx/Tx pairs together, you also need to connect the ground wire of two boards together. Not connecting the ground reference together is essentially like asking the other board "How far is my hand from the ground" when the ground reference is missing.
The UART chapter on LPC17xx has a really good summary page on how to write a UART driver : <Picture needed>
What you've done so far is wrote a polling UART driver. If you used 9600bps, and sent 1000 characters, your processor would basically enter a "busy-wait" loop and spend 1040ms to send 1000 bytes of data. You can enhance this behavior by allowing your
uart_putchar() function to enter data to a queue, and return immediately, and you can use the "THRE" or "Transmitter Holding Register Empty" interrupt indicator to remove your busy-wait loop while you wait for a character to be sent.