我假设你知道RA4是一个开路漏极输出,所以ULN2003的所有驱动器都必须来自4.7K电阻。在ULN2003输入上至少需要2V来打开它,所以在电阻上只有3V,所以只有0.63Ma。这可能起作用,但是它正好在不工作的阈值上。G.我建议减少上拉电阻到1K。此外,如果引脚仅在2V,那么当按钮不预设时,您将难以读取它。RA4是ST输入,所以对于高输入,至少需要0.8VDD,当VCC=5V时是4V。
以上来自于百度翻译
以下为原文
I assume you are aware that RA4 is an open drain output, so all the drive for the ULN2003 has to come from the 4.7k resistor.
You need at least 2V on the ULN2003 input to turn it on, so there's only 3V across the resistor, so only 0.63mA.
This might work, but it's right on the threshold of not working.
I'd suggest reducing the pullup resistor to 1k.
Also, if the pin was only at 2V, you'd have trouble reading it as a high when the button is not pressed.
RA4 is an ST input, so needs at least 0.8Vdd for a high input, which is 4V when Vcc=5V.
我假设你知道RA4是一个开路漏极输出,所以ULN2003的所有驱动器都必须来自4.7K电阻。在ULN2003输入上至少需要2V来打开它,所以在电阻上只有3V,所以只有0.63Ma。这可能起作用,但是它正好在不工作的阈值上。G.我建议减少上拉电阻到1K。此外,如果引脚仅在2V,那么当按钮不预设时,您将难以读取它。RA4是ST输入,所以对于高输入,至少需要0.8VDD,当VCC=5V时是4V。
以上来自于百度翻译
以下为原文
I assume you are aware that RA4 is an open drain output, so all the drive for the ULN2003 has to come from the 4.7k resistor.
You need at least 2V on the ULN2003 input to turn it on, so there's only 3V across the resistor, so only 0.63mA.
This might work, but it's right on the threshold of not working.
I'd suggest reducing the pullup resistor to 1k.
Also, if the pin was only at 2V, you'd have trouble reading it as a high when the button is not pressed.
RA4 is an ST input, so needs at least 0.8Vdd for a high input, which is 4V when Vcc=5V.
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