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Christophe
Joined: 10 May 2005 Posts: 323 Location: Belgium
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What is faster? ADC() conversion and multiplication + divis. |
Posted: Tue May 23, 2006 2:59 am |
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Hi,
here is my routine to measure analog Battery Voltage:
Code: | //***********************Meten Batterijspanning*******************************//
//Vbat = (150+120)/120*waarde*3/1023
//digitale waarde = 10 bits
void Meet_Vbat()
{
setup_adc_ports(AN0_AN1_AN3);
setup_adc(ADC_CLOCK_DIV_32);
set_adc_channel( 0 );
delay_us(20);
Vbat = read_adc() * (270.0 / 120.0) * (3.0 / 1023.0);
setup_adc(ADC_OFF);
setup_adc_ports(NO_ANALOGS);
} |
Now my Q is about the line:
Code: | Vbat = read_adc() * (270.0 / 120.0) * (3.0 / 1023.0); |
wouldn't it be faster to do the arithmetic at first and stating this:
Code: | Vbat = read_adc() * 0.006598240469; |
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Ttelmah Guest
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Posted: Tue May 23, 2006 4:01 am |
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If you have 'constants' like this, the compiler works them out at compile time, _provided_ the arrangement is such that this part of the equation can obviously be solved first.
So:
Vbat = read_adc() * ((270.0 / 120.0) * (3.0 / 1023.0));
(notice the extra pair of brackets), will result in exactly the same code as using your 'presolved' version.:-)
As written, the compiler is not quite smart enough to realise that the whole block can be solved.
As a seperate comment, consider using integers instead. If you use integer 1/100th volts, then you can use integer arithmetic. Much faster and smaller, and also quicker for tests/printouts too. The compiler does have the ability to print an integer as a 'scaled' value. Might be worth considering.
As a final comment, the value at the end of the equation, should be 1024, not 1023. A ten bit ADC, gives 1024 'steps', with the full scale voltage (3v), never actually being reached. If it was, it'd be a count of 1024, not 1023, with the transition exactly 'at' the full scale voltage.
Best Wishes |
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