6502.org

All,

I'm new to 6502 programming. I bought a PAL-1 and I'm learning about 6502 programming and microprocessors. In the process, I found and read Don Lancaster's Micro Cookbooks... great books for folks like me, who never did any hardware or low level programming, btw. Anyway, one of the problems he presents (Discovery Module #4 - Audio Tone) requires the programmer to code up a 500us Low, 500us High square wave to fake an audio tone (it's not a sine wave, but it'll do). So I coded it up and ran it. Interestingly, it ran 500.25us low and 500.50us high. Here's my code and logic:

The flow was to:
Clear the decimal flag
Teach the port PB0 to be output
LOOP:
Output LO
Delay 491us
Output HI
Delay 491us

The Code:
There are 9 cycles of instructions - JMP (3), LDA (2), STA (4) between each delay loop (D1 and D2).

The desired cycle time for the square wave is 500us LO and 500us HI. So, the delay between states needs to be 500 - 9 cycles = 491us.

The delay loops use 5N - 1 cycles for the loop (5 cycles for each time through save the last, which only takes 4... DEX (2), BNE (3 for branching, 2 for not branching). Setup of the loop takes 2 cycles - LDX (2). So, each loop takes 5N+1 cycles.

Setting 5N+1 = 491 (the desired delay in us), N=98 (62H).

My question is - should I expect to get exactly 500us hi and 500us lo and if so, what kinds of things could be causing the discrepancy?

Have you checked the CPU clock is exactly 1MHz?

Also, I believe it should be 5n-1 = 491, or n=98.4. So setting it to 98 should actually make your half cycle time .25us too fast, not too slow. This is what makes me think the clock might not be exactly 1MHz.

Not sure why low does not match high though.

As for me, I think your results are good enough for a computer of this type.

This would throw off the timing off a bit. Even if you put in a NOP you would have a difference of 1us between high and low.

Welcome! Those I/O addresses seem familiar to me. By any chance are you running the code on a KIM-1?

I haven't counted up your clock cycles, but I did reformat your code slightly, and it's apparent that the HI loop and the LO loop have virtually identical instructions. (BDD, I'm sure the purpose of the "unneccessary" JMP GOHI is to waste 3 cycles and keep the two loops identical, timing-wise.)
Since the number of clock cycles in the HI loop is the same as that in the LO loop, I'm inclined to suspect that the apparent asymmetry is just a measurement error. By what means did you determine that it ran 500.25us low and 500.50us high?

BTW, if it is indeed a KIM-1 then the clock rate is 1 MHz. And a discrepancy of .25 us between the HI loop and the LO represents a fraction of one clock cycle, which seems highly dubious. One more reason to suspect some sort of issue with how the measurements were done.

-- Jeff

It looks like the regulars here do now know the PAL-1 system.

The PAL-1, a small SBC, is a clone of the MICRO-KIM by Vince Briel and as such a KIM-1 clone with 6532 RIOT instead of the 6530 RRIOT.

Can be expanded with motherboard, memory, second RIOT, KIM tape I/O etc.
Well documented, affordable and being a KIM-1, all KIM-1 software and documentation is applicable.

For software it is a KIM-1, original ROMs, same interfaces (bit banged keypad/LEDs and serial I/O).

Available as kit on Tindie, enthusiastic users on twitter and groups.io.

The clock is made with a 1 MHz can oscillator, the KIM-1 has a crystal of 1 MHz.

Yes, your analysis is correct. The JMP "eats" 3 cycles to keep the two identical and make them both 500us with a 1Mhz clock. It's a PAL-1, which is a clone of a KIM-1, with a can oscillator as Hans points out below, that supposedly clocks at 1Mhz. I'm inclined to think it might be measurement error. I used a 10 dollar 8 channel usb logic analyzer and just believed it cuz it seemed pretty accurate when I was looking at the microsecond level. I've got a Rigol DS1054Z. I'll hook that up and see how things look... and I'll read up on the logic analyzer specs (Saelae clone from China, I think), maybe fractions of microseconds at 1Mhz is beyond its limits somehow.

Will

And a beautiful thing it is, too! Hi Hans. Yours looks so much cooler than mine . I'm off to checking measurement error cuz I'm pretty sure the code's ok (not stylewize, but cyclewize). I will also do a quick rewrite with a subroutine to clarify what's going on, the repeated code and synchronizing JMP are sloppy and arcane at the same time. They do the job and illustrate necessity, but there's a clearer way.

Also check the oscillator on the PAL-1.

Your code should give you half cycle period of 498us and a frequency of ~1004 KHz if the oscillator is exactly 1.0000 +/- 0.00005 MHz

In any case 498us is within 0.4% of your goal. What precision are you after?

The code looks correct to me for 500us: Any error here is most likely a measurement error; sampling at 1MHz is not nearly fast enough.

Even using a scope can be problematic, but it seems the Rigol can be quite accurate if you use it correctly:
http://nerdralph.blogspot.com/2015/07/r ... uracy.html

What I find interesting is that the clock signal is dead on and seems to be measured accurately, whereas the hi/lo is off...

The main loops need to be 491us, each loop takes 5n-1 cycles:

Therefore
5n-1 = 491
5n=491+1
5n=492
n=98.4

He used n=98 so he will have 489us loops. He came up with 489 because he originally wrote 5n+1=491 instead of 5n-1=491.

Edit: Ahh, okay .. form what you did I see another error he made that exactly cancels out the other error. It turns out his loop do need to be 489us because it looks like he forgot to account the the LDX instruction such that there is actually 11us outside the loops, not 9us. It's nice when two wrongs make a right. LOL!. All's good.

The math again:

GOLO: LDA (2)
STA (4)
Delay (491)

' Delay Loop:
LDX (2)
DX: DEX (2)
BNE (3 or 2)
JMP (3)

GOHI: LDA (2)
STA (4)
Delay (491)

So, just focusing on the delay loop:
2 for LDX
5 for each iteration (except for the last, which is 4)

This is 2 + 5N-1, or 5N+1...

Between each delay, there are 9 cycles - JMP (3), LDA (2), and STA (4).
So, if 9 cycles are eaten by overhead, 500-9 = 491.

So, 5N+1=491, and N=98, or 62H.

Then:
Setup of output port happens once, then...
GOLO: LDA (2) STA (4) +
Setup delay counter: LDX (2) +
Delay: (5 * 98) - 1 => (489) +
Change State to HI - JMP (3)
GOHI: LDA (2) STA (4) +
Setup delay counter: LDX (2) +
Delay: (5 * 98) - 1 => (489) +
Change State to LO - JMP (3)
------------------
6 + 2 + 489 + 3 = 500
6 + 2 + 489 + 3 = 500

My mistake. I should read better. I was counting the LDX in the overhead as only the DEX and BNE are actually in the loop.

BTW, the two wrongs I mentioned in my previous post were mine.

Ha. I kept hoping it was the code! That’d be the easiest to fix. Meanwhile. I’m gonna keep digging!