Calculating UART DLL and DLM contents from Baud Rate and Clock Frequency

[Stu80]

Homework Statement

A PC16550D UART has a clock running at18.432 MHz and its
baud rate is set to 2000. Determine the HEX contents of its DLM and
DLL registers

Relevant Equations

Baud rate = Clock frequency/(16×N)

I understand that,

Baud Rate = Clock Frequency / (16 x N)

and that N = DLL + DLM

when I substitute the figures to get an answer for N to be 576, how do I then convert this into the HEX contents of DLL and DLM, especially considering the maximum value for DLL or DLM is 255?

I have uploaded the PC16550D datasheet for reference if needed.

 

[berkeman]

Fun problem, and very practical. I do this sort of thing all the time at my work.

I didn't look at the datasheet, but look for a built-in divider or shift by a couple of bits. For example, in the USB standard, the requested Icc by a device being plugged in is a factor of 2 higher (in mA) than the hex value that is reported (requested current = 2x the configuration value), and the USB host knows that.

 

[Tom.G]

See paragraph 8.5.1 on pg. 17 of the datasheet, and immediately following, the last two columns of Table 1.

A clue is this sentence in paragraph 8.5.1:
Two 8-bit latches store the divisor in a 16-bit binary format.
These Divisor Latches must be loaded during initialization ...

There is more information a few pages later.

In fact, I recommend you read the whole datasheet, even the confusing parts. You will remember a few of the ideas the next time you use this or similar devices, and will gradually be able to put together a mental picture of their general operational approaches.

Cheers,
Tom

 

[Stu80]

Thank you Tom, I had already read that specific paragraph but it hadn't clicked regarding the 16-bit divisor - I was wrongly reading them as 2 separate 8-bit latches.

Question answered!

Cheers