Suraj143

I have an AD result placing problem.

When I right justify the LSB is in ADRESL & the MSB is in 1st bit of ADRESH. That’s ok

But when I left justify where is the LSB & the MSB in a 10 bit result?

Thanks

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gualala

the bits are counted from right to left, so that for a byte:
bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0
therefore, for left justify, the MSB is the bit7 of ADRESH. LSB is bit6 of ADRESL.
for right justify, MSB went to bit1 of ADRESH where LSB is bit0 of ADRESL.

Suraj143

Hi gualala excellent reply.Thats what I want to know.

The left justify makes hard to read from the software because the 2 LSB is in the ADRESL.

But right justify easy to read.

Thats fine.

gualala

well, if you need 8-bit resolution, you can left-justify it and use ADRESH as the results.

futz

The reason for left justify is not for ease or difficulty of reading. It is so that you can easily use the high 8 bits of the result in ADRESH as an 8-bit value, dropping (ignoring) the two least significant bits in ADRESL. In many applications they're probably just noise anyway.

So if you only need an 8-bit result, use left justify and read ADRESH.

If you need all 10 bits, right justify and read both ADRESH and ADRESL.

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Suraj143

How did you read my mind? thats what i was going to ask from you.
its very nice.

I need a preset connect to RA0 (AN0) when the preset turns from zero to maximum the result value must vary between 0 to 255.

The problem is when I turn due to 10 bit it will roll over 4 times.
I dont need to rolls over 4 times I need to vary between 0-255 one time.

Do you get what I mean?

Suraj143

Oh thats the point ignoring the LSB is not a good idea.

I need a 8 bit result format with 2 LSB also.It must vary between 0-255.

Let say the AD result value is decimal 9 = B'00001001'

reading just ADRESH will show the value = d'2' that is wrong.

Whats your idea?

futz

If you need precision and your measurement won't exceed 8-bits, right justify it and ignore ADRESH, which should never exceed zero anyway if you've set up your circuit and VREF+ correctly.

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Suraj143

This is correct but need bank switching as well.ref are Vdd & Vss.

I need to connect a preset to analog input & when I turn the preset from min to max it must vary the result from 0 to 255.

The PIC doesn't have 8 bit conversion its doing from 10 bit.So when I read the result registers its from 10 bit format.

When I turn it its rolls over 4 times thats the problem.256 X 4 = 1024

futz

OH!!! Finally I understand what you were saying earlier! Gotcha!

Hmm... I never had to worry about that in anything I've done that needed 8-bit A/D. I just left justified and read ADRESH. Good enough.

Anyway, if your circuit is built correctly (and if necessary, you provide a proper VREF+ to get your readings in the proper range) either way should work fine.

You may need to spend some time with a calculator figuring out what voltage range you need to feed to the A/D pin to get the readings you want, and modify the circuit to get the voltage you're measuring into that range. You may find that you need to give the chip an external VREF+ to get it to read the kind of range you require. I had to do this for my accelerometer.

To stay in the low 8-bits right-justified with 5V VREF+, you need to be measuring a voltage of 0 - 1.25V.

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Suraj143

Oh no again you are telling to left justify then it will still missing 2 LSB

It must right justify.

My supply is 5V.Preset is connect to 5V rails & the middle pin to PIC pin.

The maximum voltage is 5V.

futz

They're insignificant! It works. Try it. An 8-bit measurement can't be that accurate anyway, or it would be a 10-bit measurement.

What are you measuring?

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gualala

hmmm... not really quite understand what you want but...

LSB means least significant bit. They can be discarded if you don't need that much resolution.
For a 10-bit ADC, it outputs 10-bit result from 0 to 1023. If we drop the last 2 bits, (i.e. set them to 00), you get 0 to 1020 with step size of 4. divide it by 4 (i.e. right shift 2 bits) results in 0 to 255 with step size of 1, which is the LEFT 8 bits of the result.

Suraj143

Ok I‘ll tell what I’m doing.

There is a preset (variable resister) connected to PIC analog channel & the presets other two pins connected to 5V supply.

I need 10 values which correspond to AD result.

Means

If the AD result is between 0-25 value must show 1
If the AD result is between 25-50 value must show 2
If the AD result is between 50-100 value must show 3
---
---
If the AD result is between 225-250 value must show 10

When I turn the preset it must give these outputs to a register.

Now both of you got it?

futz

Yup. Left justify and read ADRESH. Gives a nice 8-bit result. It works fine. Those two least-significant bits are irrelevant - that's why they're called "least significant". If you need that kind of precision then do a 10-bit read.

For an 8-bit result, bits 1 and 0 of ADRESH are your two LSB's. ADRESH will vary between 0-255. It really does work that way. Try it.

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