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11 years, 3 months ago.
How many ADC pins does the FRDM-K64F have?
I'm a bit confused about the FRDM-K64F platform. In the microcontroller datasheet (MK64FN1M0VLL12) it says it has 2 ADC. But on the FRDM-K64F mbed page it shows I have 6 AnalogIn pins.
https://mbed.org/platforms/FRDM-K64F/
How this is possible?
Are those 6 pins muxed against those two internal ADC?
I need to adquire 3 analog signals on 3 different pins. Although I don't need them to be adquired simultaneously. I can adquire them sequentially (first one pin, then the second, finally the third...). So, I don't need 3 ADC, I just need to be able to mux the two ADCs against 3 pins...
Is this possible?
Apologise for my english.
1 Answer
11 years, 3 months ago.
Yes, as you say they are muxed. This is the complete list:
static const PinMap PinMap_ADC[] = {
{PTC2, ADC0_SE4b, 0},
{PTC8, ADC1_SE4b, 0},
{PTC9, ADC1_SE5b, 0},
{PTD1, ADC0_SE5b, 0},
{PTC10, ADC1_SE6b, 0},
{PTD5, ADC0_SE6b, 0},
{PTC11, ADC1_SE7b, 0},
{PTD6, ADC0_SE7b, 0},
{PTB0 , ADC0_SE8 , 0},
{PTB1 , ADC0_SE9 , 0},
{PTB2 , ADC0_SE12, 0},
{PTB3 , ADC0_SE13, 0},
{PTC0 , ADC0_SE14, 0},
{PTB10, ADC1_SE14, 0},
{PTB11, ADC1_SE15, 0},
{PTC1 , ADC0_SE15, 0},
{PTA17, ADC1_SE17, 0},
//{PTE24, ADC0_SE17, 0}, //I2C pull up
//{PTE25, ADC0_SE18, 0}, //I2C pull up
{NC , NC , 0}
};
I don't know why the last two are commented, since I would assume that it is only the case for that specific board.