thank you for your replies, is this achievable with minimum number of components? an how simple could this circuit be? I found reference to a similar project with 4 LED's using the following code:
// attiny85
// reset -+---+- power
// (on while touching) pb3 -+* +- pb2 (toggled by touch)
// (touch input) pb4 -+ +- pb1 (fading while touching)
// ground -+---+- pb0 (fading always)
int fadepin1 = 0; // the led that fades on and off
int fadepin2 = 1; // the led that fades on and off while you're touching the input pin
int togglepin = 2; // the led that's toggled when you touch the input pin
int steadypin = 3; // the led that's on while you're touching the input pin
int calibration = 0;
int previous;
int randomval = 0;
int fadeval = 0, fadestep = 1;
int togglestate = LOW;
void setup()
{
pinMode(fadepin1, OUTPUT);
pinMode(fadepin2, OUTPUT);
pinMode(togglepin, OUTPUT);
pinMode(steadypin, OUTPUT);
delay(100);
for (int i = 0; i < 8; i++) {
calibration += chargeTime(PB4);
delay(20);
}
calibration = (calibration + 4) / 8;
}
void loop()
{
int n = chargeTime(PB4);
if (n > calibration) digitalWrite(steadypin, HIGH);
else digitalWrite(steadypin, LOW);
analogWrite(fadepin1, fadeval);
if (n > calibration) analogWrite(fadepin2, fadeval);
else analogWrite(fadepin2, 0);
fadeval = fadeval + fadestep;
if (fadeval == 255) fadestep = -1;
if (fadeval == 0) fadestep = 1;
if (previous <= calibration && n > calibration) {
if (togglestate == LOW) togglestate = HIGH;
else togglestate = LOW;
digitalWrite(togglepin, togglestate);
}
previous = n;
delayMicroseconds(500);
}
byte chargeTime(byte pin)
{
byte mask = (1 << pin);
byte i;
DDRB &= ~mask; // input
PORTB |= mask; // pull-up on
for (i = 0; i < 16; i++) {
if (PINB & mask) break;
}
PORTB &= ~mask; // pull-up off
DDRB |= mask; // discharge
return i;
}
I cant seem to work out how to trim the code down to a single LED, any help much appreciated