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晚上好,我有一个很快的问题——我使用的是PIC16F628 A,在5伏威廉希尔官方网站
中供电,但是需要使用3.3伏处理器的MCLR线路。由于线路复位时低,AM我很好地连接MCLR,说一个10K拉电阻到5伏,然后运行MCLR到3.3伏I/O线树莓PI?然后,我可以通过拉动PI I/O线来重置PIC。我只是不确定我是否应该使用分压器威廉希尔官方网站
、二极管等,或者如果我能像刚才描述的那样直接。我使用BSS138晶体管作为两个芯片之间所有真正的数字逻辑通信的电平移位器,但我希望能用这条线做一些更直接的事情,因为它不必很快。任何帮助都将不胜感激!提前感谢!马歇尔
以上来自于百度翻译 以下为原文 Good evening, I have what I think will be a quick question - I am using a PIC16F628a powered in a 5 volt circuit, but need to use the MCLR line from a 3.3 volt processor. Since the line resets when low, am I good connecting MCLR to say a 10k pull up resistor to 5 volts, and then running MCLR to a 3.3 volt I/O line on a Raspberry Pi? Then, I can reset the PIC by pulling the Pi I/O line low. I'm just not sure if I should be using a voltage divider circuit, a diode, etc., or if I can just go direct as described. I am using BSS138 transistors as level shifters for all of the true digital logic communication between the two chips, but I was hoping to get away with doing something a bit more direct with this line, since it does not have to be fast. Any assistance would be greatly appreciated! Thanks in advance! Marshall |
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3个回答
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这里有一些因素需要考虑。在PIC16F628 A上,数据表告诉你thjat MCLR是一个“ST”输入。(表3-2:PIC16F62A/628 A/648 A引脚说明)“电规范”部分告诉您,ST引脚上的有效高输入是“0.8VDD”,因此如果VDD为5.0V,那么您需要至少4V的引脚被认为是一个安全的“高”。ES(在处理器内部),这将限制引脚电压到它自己的VDD加上大约0.3V,所以上拉电阻将不能将引脚拉到大于3.3±0.3=3.6V。您可以在PI和PIC之间插入串联电阻器,但是您必须确保当皮是拉低,它可以使MCLR引脚低于0.2VDD,这是1V.所以,从470R到1K可能工作。你必须做这些。一个更干净的解决方案是从NPI晶体管的基极从PI驱动,并使其拉低MCLR。(另一个考虑是,PIC16F628 A现在是一个14年的芯片。有PIN兼容PIC16F1XXX芯片,比PIC16F628 A更便宜、更快、更容易编程。
以上来自于百度翻译 以下为原文 There are a few factors to consider here. On a PIC16F628A, the datasheet tells you thjat MCLR is an "ST" input. ("TABLE 3-2: PIC16F627A/628A/648A PINOUT DESCRIPTION") The "Electrical Specifications" section tells you that a valid high input on an ST pin is "0.8VDD", so if VDD is 5.0V, then you need at least 4V on the pin to be considered a safe "high". The PIN ion the RaspBerry Pi will have ESD protection diodes on it (internal to the processor), which will limit the pin voltage to its own VDD plus about 0.3V, so a pullup resistor will not be able to pull the pin to more than 3.3 + 0.3 = 3.6V. You could insert a series resistor between the Pi and the PIC, but you'd have to make sure that when the Pi was pulling low, it could get the MCLR pin down below 0.2VDD which is 1V. so, something from 470R to 1k might work. You'd have to do the sums. A cleaner solution would be to drive the base of an NPN transistor from the PI, and get that to pull MCLR down. (Another consideration is, the PIC16F628A is a 14 year old chip now. There are pin compatible PIC16F1xxx chips that are cheaper, faster, and easier to program than the PIC16F628A. |
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Raspberry Pi的IO扩展器可以配置为开路漏输出:HTTPS://引脚。XYZ/引脚/ IoPiPixOn0*各种接口威廉希尔官方网站
:http://www. eluux.Org/rPIG-GPIOI接口威廉希尔官方网站
,单个小型NMOS器件将为任何IO引脚提供开路漏极能力。
以上来自于百度翻译 以下为原文 There is an IO expander for the Raspberry Pi that can be configured for open drain outputs: https://pinout.xyz/pinout/io_pi_zero# Various interface circuits: http://www.elinux.org/RPi_GPIO_Interface_Circuits A single small NMOS device would provide open drain capability to any IO pin. |
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非常感谢!事实上,QHB所提到的NPN晶体管方法和ELIUX.ORG POST无疑是最好的方法。我知道我有几个2N3904晶体管漂浮在楼下-完美!非常感谢!马歇尔
以上来自于百度翻译 以下为原文 Thanks so much! Actually, the NPN transistor method mentioned by qhb and in the elinux.org post are definitely the best way to go, given my current setup. I know I've got a few 2N3904 transistors floating around downstairs - perfect! Thanks so much! Marshall |
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