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Hampton
Joined: 08 Jun 2006 Posts: 12
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Should I set analog pins low when not in use? |
Posted: Thu Jun 15, 2006 2:04 pm |
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In my program I take an ADC sample every so often. Using a PIC18LF2525. Is it advantageous to turn off the adc and set the pins low in between samples? Here's the pseudo code for what I'm trying to do:
Code: | while(1)
{
setup_adc_ports(AN0|VSS_VDD);
setup_adc(ADC_CLOCK_DIV_32|ADC_TAD_MUL_20);
set_adc_channel(0);
delay_us(4);
read_adc();
setup_adc_ports(ANALOG_OFF);
output_low(PIN_A0);
sleep();
} |
If I set pin A0 as an output while the PIC is sleeping do I need to make it an input when I enable the adc, or does the setup_adc_ports() function do that internally? |
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Ttelmah Guest
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Posted: Thu Jun 15, 2006 2:53 pm |
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The signal driving the ADC inputs, needs a low impedance if the value is to be 'right'. If you set the pin low, you will be shorting that signal to ground, possibly leading to overheating, and certainly leading to wasted power.
You would need to set the signal line _back_ to being an input, or it won't work at all, but you need to work out if you really want to do this at all...
This type of approach is used, on signals where (for example), a capacitor is charged by an external source, and discharged by the processor gate.
The approach would then be:
Code: |
while(1)
{
setup_adc_ports(AN0|VSS_VDD);
setup_adc(ADC_CLOCK_DIV_32|ADC_TAD_MUL_20);
output_float(PIN_A0);
set_adc_channel(0);
//You need more than 4uSc delay - typically 10uSec minimum,
//check the data sheet...
delay_us(4);
read_adc();
setup_adc_ports(ANALOG_OFF);
output_low(PIN_A0);
sleep();
}
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Notice the 'output_float', which turns off the pin drive, and the comment about the delay.
Best Wishes |
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