Rolfiboy 
Hi i am new in c programming, and now i have a problem/question.
My ADC will only show 4092mV althoug i measure on the 5V VCC point, and i came down to that it has something to do with this line of code:
ADC_result = (ADC_result * (5000 / 1023));
Can it realy be right that i need to use a float/double as sub calculations to get the right result?
I am curretly running the code on a atmega16 and i would like to keep it simple and low on ressources
5000/1023 = 4 using integer math. Using 50000/1023 and then dividing by the final result by 10 will get you closer. There is probably another ratio approach that will get you even closer without using float.
Or do the multiplication first
ADC_result = (ADC_result * 5000UL) / 1023;
(the parenthesis isn't really needed).
Rolfiboy Thanks but think i use a double, its more precise.
clawson 
Thanks but think i use a double, its more precise.
To be honest I don't know why you'd use float/double for this anyway. Surely ints are far tighter/faster and if you want to hold more accuracy scale them.
Rolfiboy Or do the multiplication first
ADC_result = (ADC_result * 5000UL) / 1023;(the parenthesis isn't really needed).
This metod work even better than with a double:)
But how can it work, if i do the math 1023 * 5000 is 5115000, way more than ADC_result can represent as it is a unsigned int?
Notice the UL in 5000UL, it means that it's an unsigned long (32 bits in avr-gcc), so the calculation is done in 32 bit math, and when ADC_result * 5000UL is divided by 1023 the result fits in 16 bits.
Koshchi Surely ints are far tighter/faster and if you want to hold more accuracy scale them.
Rolfiboy Thanks, every new day is a day to learn.
theusch 
Thanks, every new day is a day to learn.
In this case, let me put it this way: You said "Thanks but think i use a double, its more precise." But that ignores the fact that you are starting with a 10-bit A/D value. In all likelihood you are losing a few bits of accuracy unless you have a squeaky-clean analog subsystem in your app. At most, no matter how many decimal places you print out from your "double" you still only have about three useful significant digits. You can print out 1.23456789 but that really represents about 1.23 +/-0.01.
jwatte Thanks but think i use a double, its more precise.
I hope you are aware that the ADC only has 10 bits of resolution, and unless you pay close attention to your AREF and your AVCC and your timing and de-coupling, you may actually get less than that.
So, an unsigned long has all the precision you will ever need for this application, and spending time with the floating point emulation does nothing but waste perfectly good CPU cycles.
indianajones11 Also, you should be dividing by 1024, NOT 1023.
Why? (ok it's cheaper). Isn't 1023 max?
octave:8> 5000*1023/1024 ans = 4995.1 octave:9> 5000*1023/1023 ans = 5000
indianajones11 Because it's the correct equation. An ADC can't read full-scale Vcc value, which the wrong eq'n implies ( A 3 bit system would read to 7/8*Vcc, NOT 7/7*Vcc = Vcc for example ). 1023 is max COUNT, but there are 1024 STATES. It's always ( 2^N ) -1 / 2^N.
theusch
Also, you should be dividing by 1024, NOT 1023.
Why? (ok it's cheaper). Isn't 1023 max?
... and others. In practice it >>usually<< doesn't matter much even when trying to get "good" ADC results: except in squeaky-clean analog subsystems there is almost always a few LSB of uncertainty in the ADC result--ripple anywhere for example. So in some apps I've even done 5000/1025 for fast and small units conversion in a tight Mega48 app. That's 5000mV/1025 "approximate" counts. Why? 5000*ADC/1025 => 200*ADC/41 and everything fits nicely into 16-bit operations.
In practice it >>usually<< doesn't matter much even when trying to get "good" ADC results:
NOT
Koshchi Because it's the correct equation. An ADC can't read full-scale Vcc value, which the wrong eq'n implies ( A 3 bit system would read to 7/8*Vcc, NOT 7/7*Vcc = Vcc for example ). 1023 is max COUNT, but there are 1024 STATES. It's always ( 2^N ) -1 / 2^N.
indianajones11 indianajones11 wrote:No, you are wrong. Think of it as a pie. If the pie is cut into 8 pieces, and you have all 8 pieces, then you have 8/8's of a pie. But the pie has 9 "states" ( 0-8 ). With your logic, if you had all 8 pieces you would have only 8/9's of a pie.Because it's the correct equation. An ADC can't read full-scale Vcc value, which the wrong eq'n implies ( A 3 bit system would read to 7/8*Vcc, NOT 7/7*Vcc = Vcc for example ). 1023 is max COUNT, but there are 1024 STATES. It's always ( 2^N ) -1 / 2^N.
Edit: Steve, your pie only has 8 states, 0-7 with a max. cnt of 7 ! :wink: Now I need to go to Marie Callender's !
