Pic 18f4455 multiplication and division

I need help in using multiplication and division operation using PIC18F4455 in assembly language.

So, my project is like this: I have a voltage divider circuit, with the input connected to a voltage reference of 5V and the output connected to the analog input of the PIC. I used the ADC feature of PIC to obtain the 10-bit voltage output. The project aims to determine the value of an unknown resistance given the other resistance is fixed at some value.

The problem begins with the actual manipulation of the ADC output. I can't figure out a way to perform operation on a 10-bit data, divided into two registers (ADRESH:ADRESL). My pseudocode is like this:

* Get the decimal equivalent of the ADC output: Vout=(Vin/1024)*analogvalue
* Solve for Runk = Rk[(Vout/Vin)-1] where Runk is the unknown resistance and Rk is the known resistance

However, I got stuck at the first step. How do I multiply analogvalue with (Vin/1024) considering the 2 most significant bits are in another register? Anyone got an idea on this? Or better yet, is there a better way to solve for the unknown resistance, without going through this process?

Thanks in advance.

Multiplying or dividing by a power of 2 (such as 1024) is quite easy, for clear reasons, if you are using assembler.
Multiplying or dividing by any value other than a power of 2 involves repeated additions, or 'tricks' for certain values.
This is why, whenever someone wishes to do more than trivial math, you either need to find a pre-existing math
library in assembler, or move to C. You'd have no problem dealing with numbers larger than 8 bits in C.
I suggest you move to C, and make life easy for yourself.

The following routine is a 10-bit X 10-bit assembly routine written by Martin Sturm for the PIC18F series:



On a side note a 10-bit ADC result represents 1024 unique values, ranging from 0 to 1023, therefore you need to divide the result by 1023, not 1024, to obtain the proper ratio.

The correct algorithm:

Vout=(Vin/1023)*analogvalue

Click to expand...

Instead of:

Vout=(Vin/1024)*analogvalue

Click to expand...

BigDog

sky_123 said:

This is why, whenever someone wishes to do more than trivial math, you either need to find a pre-existing math
library in assembler, or move to C. You'd have no problem dealing with numbers larger than 8 bits in C.
I suggest you move to C, and make life easy for yourself.

Click to expand...

I am actually thinking of repetitive subtraction and addition for division and multiplication, respectively. However, my problem is how to account for the 2MSBs located in another register (ADRESH). It's easy dividing 8bits by 8bits using repetitive subtraction (or so I think hehe) but I'm actually dividing 10bits by 10bits to maintain precision.

By the way, as much as I would like to use C, I can't because this course is restricted to assembly language only. (

Thanks!

- - - Updated - - -

bigdogguru said:

The following routine is a 10-bit X 10-bit assembly routine written by Martin Sturm for the PIC18F series:

On a side note a 10-bit ADC result represents 1024 unique values, ranging from 0 to 1023, therefore you need to divide the result by 1023, not 1024, to obtain the proper ratio.

The correct algorithm:



Instead of:




BigDog

Click to expand...

Thank you for the code! I will look into it, and try to insert it into my code. Oh, 1023, right! Thanks thanks.

Here is a 3-bit A/D.
There are 8 different voltage steps including zero and 1.
Its resolution is 8, not 7.

I believe Vref / 1024 is correct.

btbass said:

View attachment 78335

Here is a 3-bit A/D.
There are 8 different voltage steps including zero and 1.
Its resolution is 8, not 7.

I believe Vref / 1024 is correct.

Click to expand...

I get it now. I think 1024 is correct, since Vref/2^n gives the resolution of the ADC, not Vref/(2^n-1). Thanks!

btbass said:

Here is a 3-bit A/D.
There are 8 different voltage steps including zero and 1.
Its resolution is 8, not 7.

I believe Vref / 1024 is correct.

Click to expand...

If you examine my initial reply, it makes no counter assertions as the resolution of a 10-bit ADC and in fact I also point out there are 1024 unique values ranging from 0 to 1023.
And I should also add the choice of dividing by 1023 rather 1024 is less due to the devices resolution then the apparent choice of coding scheme and point which follows.

btbass said:

View attachment 78335

Click to expand...

Nice graph, unfortunately in this context, it incorrectly suggests the ADC output of 0b111 represents Full Scale or Vref, which is clearly NOT the case.



The output of an ADC can be coded using one of several widely used ADC coding schemes, some are more applicable to bipolar, than unipolar ADCs.

They all have their advantages and disadvantages and the choice largely depends on the application requirements, quantization characteristics, output range, minimization of various sources of error, etc.

In this case the chosen coding scheme, which is the simplest and commonly referred to as natural or straight binary coding.

For a unipolar ADC utilizing straight binary coding, the maximum return value is not Full Scale (FS) or Vref, it is instead Vref minus the Least Significant bit (LSb).


In the case of your 3-bit ADC with a Vref = 5v, the maximum value analog value the output from the ADC can represent is 4.375v, not the 5v Full Scale of Vref.

The reason for is, of course, the largest value returned by the ADC is 0b0111 or 7, which represents a quantization uncertainty or error of 1 LSb which equates to 0.625v in this case.

The following formula/algorithm, where ADo represents the output of the ADC, has a range of 0v to Vref - LSb, or if Vref =5v, a range of 0v to 4.375v.

Ain = (Vref / 2^n) * ADo

Click to expand...

Or in the case of 3-bit ADC:

Ain = (Vref / 8) * ADo

Click to expand...

The following are more accurate graphics depicting the output limitations of straight binary coding:






Another example of a 4-bit ADC and the output limitations of straight binary coding in table format:





If the full scale output range of 0v to Vref is required, which is often the case in may application such as a ohmmeter then some addition software or hardware techniques must be utilized to achieve this output range. These techniques can be observed in real devices as a Tare/Clear or Calibration features. An ADC coding scheme which produces results over the full range should be beneficial in this application as it would allow a zero ohm reading to be obtained when the probes or test points are shorted, as similar results would be expected from an ohmmeter.

One of the simplest techniques to achieve full scale output is the following formula/algorithm:

Ain = (Vref / (2^n -1)) * ADo

Click to expand...

Or in the case of 3-bit ADC:

Ain = (Vref / 7) * ADo

Click to expand...

Or in the case of 10-bit ADC:

Ain = (Vref / 1023) * ADo

Click to expand...

Admittedly the use of the phrase, "correct value" in my initial reply, was a poor choice as the implementation of the technique is not appropriate in all cases.


The technique does not effect the ADC device resolution, however it does have the effect of scaling the output over the entire range of 0v to Vref. The technique does slightly increase quantization uncertainty/error and introduce some small amount of scaling error, however these errors are far less than the 1% accuracy of a precision resistor.

Additional information concerning ADC/DAC coding schemes is available in the following document:

FUNDAMENTALS OF SAMPLED DATA SYSTEMS - ANALOG-DIGITAL CONVERSION

Of course there are other techniques which can be implement in either hardware or software which provides coding output over Full Scale, however most are considerably more elaborate.



On a side note, for ADC or DSP intensive applications, typically the more useful coding scheme of fractional binary coding is utilized, which is always normalized to Full Scale (FS). However, this coding scheme does require considerably more arithmetic operations of a more elaborate ADC hardware module.


I hope I've cleared any confusion as to the motive of my suggestion.

BigDog

Wow! Very detailed explanation. Got it! Super thanks!

Just a question about macros.. This will be my first time using this. In order to use your 10x10bit multiplication code, I just have to specify the destinations of the data right? So for the macro, MULT_10x10 MACRO aH,aL, bH,bL, rH,rM,rL, I have to specify 7 registers, the first four registers are my operands, and the last three will store the result. Am I on the right track? And what is the purpose of the "mmac" macro? Thanks!

Do you happen to have a code for 10x10bit division also?

Thank you so much for replying. I am finally getting a grasp on this. And I'm sorry for asking too much. :|

Hi guys, I got this 16x8 multiplication macro in piclist and it's just what I needed. However, I am having trouble using it since it's my first time using macros. Here's the code:


Heeeelp! Everytime I build it, an error occurs at the line where I call the macro. Is there something wrong in the "call" line? Or did I misplace the macro? Do macros need to be defined first? Heeeeelp!

Right now, I want to know if the code actually works so I can incorporate it in my code. Help, anyone?