why you use particularly 9600 in your project

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bagavathi

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Hi friends,

I attend robert bosch interview , i said i used 9600 baud rate in uart they ask question like , why used very low baud rate in your project ? . the answer i dont know . then in another project i said 10 bit adc i used n my project . they asked why 10 bit adc you used , why not 12 bit ,8 bit adc you use. i didn't say .

for this question how i have to answer . pls help.
 

Hello there,

9600 baud rate means 9600 bits per seconds f = 9600 Hz or t = 1/9600. when you send data ,one bit will be sent in 1/9600 seconds.That bit can be 1 or 0. To measure a signal with an osiloscope ,that signal must be periodic.

Some uart chips (8250 UART,Intel 8251)Obsolete with 1-byte buffers. These UARTs' maximmum standard serial port speed is 9600 bits per second if the operating system has a 1 millisecond interrupt latency. which meant that communications software performed poorly at speeds above 9600 bits/second, especially if operating under a multitasking system or if handling interrupts from disk controllers. High-speed modems used UARTs that were compatible with the original chip but which included additional FIFO buffers, giving software additional time to respond to incoming data.

Universal asynchronous receiver/transmitter:- it is a universal standard serial communication so 9600bps is default value. According to your requirement you can set you baudrates. (Based on your lenght of communication ,normal IBM PC interface,based on speed of communication you required ..etc).

project i said 10 bit adc i used n my project . they asked why 10 bit adc you used , why not 12 bit ,8 bit adc you use

it quite easy.

Answer is how much resolution do you need?? meansThe minimum milli voltage you can measure through adc??

As you know adc can measure upto your vreference voltage.

For example your vref=5V=5000mV
so the minimum voltage you can measure through adc is

minimum_voltage=Vref/2^n(here n is your adc bit resolution)
like
8-Bit
minimum_voltage_8bit=Vref/256=5000/256=19.53125mV
b'coz(2^8=256).

10-Bit
minimum_voltage_10bit=Vref/1024=5000/1024=4.8828125mV
b'coz(2^10=1024).

12-Bit
minimum_voltage_12bit=Vref/4096=5000/4096=1.220703mV
b'coz(2^12=4096).

So well did you pass the interview??

best regards,
 
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