164335413 发表于 2015-12-1 16:50:11

关于Arduino readCapacitivePin()函数,读取电容值。

最近在搞一点小玩意,利用Arduino做电容按键的处理,按键用铜箔(2cm*2cm)贴在不透明的亚克力背面,焊接上导线连接到Arduino。
但出现的问题是,直接接触铜箔,数值可以大于2,但是经过亚克力以后,手触摸亚克力,数值只能2。增加到3个按键后,其他两个按键的值一直是2,第一个按键触摸后变为2,偶尔还会受到干扰,自己跳变。我知道直接触摸导线或铜箔数值会增大到十几,但还是想隔着亚克力进行操作。
希望高手们指教一下!
先附上参考代码:
int ledPin = 13;
int capval0,capval1,capval2;
void setup()
{
pinMode(ledPin, OUTPUT);
Serial.begin(9600);
Serial.println("Touch senser");
}
void loop ()
{
digitalWrite(ledPin,LOW);
capval0 = readCapacitivePin(8);
capval1 = readCapacitivePin(9);
capval2 = readCapacitivePin(10);
Serial.println("Touch senser");
Serial.println(capval0, DEC);
Serial.println(capval1, DEC);
Serial.println(capval2, DEC);
if (capval0 > 1) {
    // turn LED on:
    digitalWrite(ledPin, HIGH);
    delay(10);
}
delay(300);
}

uint8_t readCapacitivePin(int pinToMeasure) {
// Variables used to translate from Arduino to AVR pin naming
volatile uint8_t* port;
volatile uint8_t* ddr;
volatile uint8_t* pin;
// Here we translate the input pin number from
// Arduino pin number to the AVR PORT, PIN, DDR,
// and which bit of those registers we care about.
byte bitmask;
port = portOutputRegister(digitalPinToPort(pinToMeasure));
ddr = portModeRegister(digitalPinToPort(pinToMeasure));
bitmask = digitalPinToBitMask(pinToMeasure);
pin = portInputRegister(digitalPinToPort(pinToMeasure));
// Discharge the pin first by setting it low and output
*port &= ~(bitmask);
*ddr |= bitmask;
//delay(1);
// Make the pin an input with the internal pull-up on
*ddr &= ~(bitmask);
*port |= bitmask;

// Now see how long the pin to get pulled up. This manual unrolling of the loop
// decreases the number of hardware cycles between each read of the pin,
// thus increasing sensitivity.
uint8_t cycles = 17;
if (*pin & bitmask) { cycles = 0;}
else if (*pin & bitmask) { cycles = 1;}
else if (*pin & bitmask) { cycles = 2;}
else if (*pin & bitmask) { cycles = 3;}
else if (*pin & bitmask) { cycles = 4;}
else if (*pin & bitmask) { cycles = 5;}
else if (*pin & bitmask) { cycles = 6;}
else if (*pin & bitmask) { cycles = 7;}
else if (*pin & bitmask) { cycles = 8;}
else if (*pin & bitmask) { cycles = 9;}
else if (*pin & bitmask) { cycles = 10;}
else if (*pin & bitmask) { cycles = 11;}
else if (*pin & bitmask) { cycles = 12;}
else if (*pin & bitmask) { cycles = 13;}
else if (*pin & bitmask) { cycles = 14;}
else if (*pin & bitmask) { cycles = 15;}
else if (*pin & bitmask) { cycles = 16;}

// Discharge the pin again by setting it low and output
// It's important to leave the pins low if you want to
// be able to touch more than 1 sensor at a time - if
// the sensor is left pulled high, when you touch
// two sensors, your body will transfer the charge between
// sensors.
*port &= ~(bitmask);
*ddr |= bitmask;

return cycles;
}


164335413 发表于 2015-12-1 16:55:59

铜箔是那种屏蔽用的铜箔,另一面有双面胶。

164335413 发表于 2015-12-2 13:04:05

要沉贴了?@弘毅,帮忙诶~~~

zoologist 发表于 2015-12-2 15:56:41

涉及到不同材质的,看起来很难办

要不你试试玻璃?

164335413 发表于 2015-12-2 16:28:27

zoologist 发表于 2015-12-2 15:56 static/image/common/back.gif
涉及到不同材质的,看起来很难办

要不你试试玻璃?

谢谢回复,用ad转换结果是电容充放电的曲线值,用帖子中的方法就是直接触摸到铜箔可以,隔着一层东西就不行了。我再找找其他的办法!

zoologist 发表于 2015-12-2 17:04:48

164335413 发表于 2015-12-2 16:28 static/image/common/back.gif
谢谢回复,用ad转换结果是电容充放电的曲线值,用帖子中的方法就是直接触摸到铜箔可以,隔着一层东西就不 ...

哦 这样的话绝缘的东西都不行

164335413 发表于 2015-12-8 09:19:01

昨天尝试用Pf级的电容和一些简单的电路做了一下,抗干扰能力有提升,我觉得铜箔的大小也会有关系,有效面积越大。看来还是模拟电路没搞好啊!

风宇声 发表于 2020-12-11 17:12:37

好东西,点个赞!
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