Electronic Circuits and Projects Forum

Originally Posted by BOBKA

Hi, I am trying to get as much as possible from my serial connection. I am using 10Mhz xtal in HS+PLL mode (total of 40MHz).

at the top of my program I define

#define BAUD d'57600'
#define XTAL d'10'
#define X ((XTAL*d'1000000')/(d'64'*BAUD))-1

and then in the program I clear the BRGH bit (making the speed slow - and what is the limit here?).

Instead of the desired chars, I am getting some garbage. If I reduce the speed to 38400 b/s (or lower), everything works just fine.

I also tried
#define X ((XTAL*d'1000000')/(d'16'*BAUD))-1
and then set BRGH bit high, so I started to get some new garbage. I also tried to change the XTAL to 40, but no success.

Two questions:
1. what is the bitrates for BRGH=1 and BRGH=0?
2. is this a limitation of a pic/max/pc?

PS: I am using COM2 on pc which is set to 115200 in devices, and then set to appropriate BAUD in terra term. Also, fifo buffers are off.


Thank You.

Try the 10mhz high speed and x=10 or 40mhz, low speed, and x=10... i am surprised no one responded ... one has to wonder...

Dan

Originally Posted by Papabravo

Assuming integer division, and the parentheses the value of X is 1. I'm no expert but I think the error is kinda large. Using the value 1 the calculated baudrate is 78125. Sure enough this error is just too large. If you use BRGH = 1 to use a 16x baudrate clock then I compute a value of 9. This gives an actual baudrate of 62500. This is better but not quite what you want. If you need 57600 then I suggest that you consider changing crystals so that you get what you want. I think that 9.216 MHz. would work, and there may be other values

According to the data sheet:
--------------------------------
Solving for X:
X = ( (FOSC / Desired Baud Rate) / 64 ) – 1
X = ((16000000 / 9600) / 64) – 1
X = [25.042] = 25
--------------------------------
10000000/57600/16-1 (for BGRH=0=high speed, 64 is for low speed) = 10
Going the other way we get 56818 for an error of 1.3%
The moral of the story is: counters do not use integer division.

Dan

Thank you guys for you response. I have played with it a little bit more. It appears that the XTAL should be the actual clock at the clock pin and not the one after the PLL. I figured out that the clock division was the limiting factor, so I will be stack with it for a while.

I finally figured out how the BRGH works, to summarized:

There are two models for calculating SPBRG, one is based on low speed transmission (BRGH=0) and the other is based on high speed transmission (BRGH=1). Those high/low speed transmission settings can be derived from looking at the results of those two models.
Low speed transmission: SPBRG = (XTAL*d'1000000')/(d'64'*BAUD_RATE) - 1
High speed transmission: SPBRG = (XTAL*d'1000000')/(d'16'*BAUD_RATE) - 1
* the SPBRG should be rounded to the nearest integer
*BAUD_RATE is the desired baud rate (19200, 57600, etc)

Since the above SPBRG equations round the results, we want to calculate the actual transmission rate, and the percentage error due to this rounding.
low speed: ACTUAL_BAUD = XTAL*d'1000000' / (d'64'*(SPBRG+1))
high speed: ACTUAL_BAUD = XTAL*d'1000000' / (d'16'*(SPBRG+1))

The percentage error is calculated as (same eqn for both models):
% error = ((ACTUAL_BAUD-BAUD_RATE)/BAUD_RATE)*100%



Example: we want to get transmission rate of 19200 bits/s w/ crystal of 10M
=> XTAL=10, BAUD_RATE=19200

low speed: SPBRG = (10*1000000) / (64*19200) - 1 = 8.138 - 1 => 7
ACTUAL_BAUD = 10*1000000 / (64*(7+1)) = 19531.25
% error = ((19531.25-19200)/19200) * 100% = 1.73%

hish speed: SPBRG = (10*1000000) / (16*19200) -1 = 32.55 - 1 => 31
ACTUAL_BAUD = 10*1000000 / (16*(31+1)) = 19531.25
%error = 1.73% as for low speed, but if we were to round the SPBRG, we would get SPBRG= 32.55-1 = 32,
then ACTUAL_BAUD = 10*1000000 / (16*(32+1)) = 18939.4
and %error = ((18939.4-19200)/19200) * 100% = -1.14%

Results: using high speed model is always better cause it provides smaller % error. The only situation where it cannot be used, is when SPBRG for it is greater than 255. This will typically occur at high clock frequencies and very slow transmission rates. In this cases low speed transmission model should be used.

Hope this will help someone