Andy's explanation is correct but a little more explanation
may help.

A BSF and BCF instruction both implement their actions(a set
and a clear respectively)  at the same point in the cycle.
As BCF & BSF are 1 cycle instructions and the clock is 10MHz
and the clock is divided by 4, each cycle is 4 x 100nS =
400nS. In this case as the instructions are implemented one
after the other there is a 1 cycle delay between the rise
and the fall. The width of the high pulse is therefore 400
nanoseconds  as expected. If you implement a series of these
you get an overall cycle (one high pulse and one low pulse)
of 800 nS = 1.25 MHz.

The biggest confusion MAY come from the fact that the 10 MHz
clock is divided by 4 to produce one basic PIC cycle.

-----Original Message-----
From: Andrew Warren <fastfwd@IX.NETCOM.COM>
To: PICLIST@MITVMA.MIT.EDU <PICLIST@MITVMA.MIT.EDU>
Date: Thursday, 16 April 1998 20:16
Subject: Re: PIC 16X84


Andrejus Stavickis <PICLIST@MITVMA.MIT.EDU> wrote:

> i have made some hardware, but pic is not running fast
enought:
> ........
>         bsf     PORTB,5
>         bcf     PORTB,5
> ........
>
> in configuration i have used XT oscillator and in hardware
ceramic
> 10 MHz oscillator
> ....
> and in after this on 5 pin i have got sometnig like:
>
> +----+    +
> |    |    |
> |    |    |
> |    |____|
> :         :
> :         :
> freq something like 1.25 MHz. Is it normally or i have
missed
> somenthing ?

Andrejus:

This is exactly normal; each PIC instruction cycle takes 4
clock
cycles, so your "BSF" and "BCF" instructions each take 0.4
microseconds.

-Andy

=== Andrew Warren - fastfwd@ix.netcom.com
=== Fast Forward Engineering - Vista, California
=== http://www.geocities.com/SiliconValley/2499