On Tue, Jul 12, 2022 at 01:53:00PM +0200, Marek Behún wrote: > From: Pali Rohár <pali@xxxxxxxxxx> > > The baud rate generating divisor is a 17-bit wide (14 bits integer part > and 3 bits fractional part). > > Example: > base clock = 48 MHz > requested baud rate = 180 Baud > divisor = 48,000,000 / (16 * 180) = 0b100000100011010.101 > > In this case the MSB gets discarded because of the overflow, and the > actually used divisor will be 0b100011010.101 = 282.625, resulting > in baud rate 10615 Baud, instead of the requested 180 Baud. > > The best possible thing to do in this case is to generate lowest possible > baud rate (in the example it would be 183 Baud), by using maximum possible > divisor. Actually, the best way to handle this is to add a sanity check for the lowest supported check as you do in the next patch. That one makes this change superfluous. > In case of divisor overflow, use maximum possible divisor. Johan
