需要多少电流电压来引起闩锁

一般来说，你不应该超过VDD，或者你可能会闩锁或其他奇怪的行为。超过“绝对最大值”将造成永久性损坏。



      以下为原文

    Generally speaking, you should never exceed VDD, or you may get latch-up or other strange behaviour. Exceeding the "absolute maximum" will cause permanent damage.

输入/输出钳位电流是ABS最大20mA的每一个PIC数据表我见过。我不认为这是一个闩锁问题，这么多的烧毁ESD二极管问题。我偶尔使用ESD二极管，我已经看到了应用程序笔记，他们利用这种电压电平转换，但你必须小心不要超过IO绝对最小/最大值为0.3至VDD + 0.3，因此4MA可能太多。这些不是肖特基，二极管V/I特性没有我知道的。我会把钳位电流保持在几百微安下，所以在这种情况下10K系列电阻最小。如果信号不快，越大越好。在内部二极管箝位的情况下，有一个3.3V多通道ADC（非微芯片）。一个LogAMP，我们以前没有使用，它不应该输出& gt；2.5V在正常操作意外驱动到5V时启用引脚变低。有一个4.7K电阻在线，100-200 UA钳位电流导致所有其他ADC通道读数明显改变。



      以下为原文

    The input/output clamp current is abs max 20mA on every PIC datasheet I've ever seen. I don't think it's a latch-up issue so much of a burn out the ESD diode issue. I've occasionally used the ESD diodes and I've seen app notes that take advantage of them for this kind of voltage level translation, but you have to be careful not to exceed the IO absolute min/max of -0.3 to vdd+0.3, so 4mA is probably too much. These are not schottky's, and the diode V/I characteristic isn't published that I know of.
 
I would keep the clamp current under a couple hundred microamps, so 10K series resistor minimum in this case. If the signal isn't fast, bigger is better.
 
The one case where I did have issues with internal diode clamping was with a 3.3V multichannel ADC (not Microchip). A logamp that we had not used before and which should never output >2.5V in normal operation unexpectedly drove to 5V when the enable pin went low. There was a 4.7K resistor in line and that 100-200uA clamp current caused all other ADC channel readings to change noticeably.