The circuit has to be redesigned so it doesn't destroy the CPU. The mistake of the transistor orientation doesn't matter. If the transistor was an NPN, the voltage divider created by R916 and R240 still put 6V back onto the CPU. That means this CPU can't take any "injection current". The question about "injection current" has been asked before on this forum, and Freescale didn't answer the question: Some microcontrollers have a specification for "injection current" which means you're allowed to inject a few milliamps into the pins like this through a resistor. Your circuit provides way more than that and at quite a high current (10mA or so). The Data Sheet shows the MAXIMUM voltage allowed on any I/O pin is VDD+0.3V. With the PNP, the 12V will go directly through the transistor, emitter to base, through R239 and will likely destroy the CPU. You have a PNP transistor shown, not an NPN. I can't understand how that circuit works. If you don't want that you'll have to add a pulldown resistor to that pin to make it work properly at power-on. That shows that pin defaults to a 100k pullup on reset, and that is enough to turn Q6 on at power-on. Have you considered the power-on reset condition? Read "36.4.314 Pad Control Register The SS9013 has a minimum gain of 64, so that 10k base resistor is fine. You should use a 60V or 80V transistor instead. As well, the SS9013 is only rated to 20V. They will be needed for a "real product". It doesn't have any protection against any faults, and it doesn't have any EMI/EMC circuitry. Is that what you want? If it is, it is better to put a pullup to 3V3 rather than 12V. As well, during reset, that pin will default to ON. Why do you have R116 there anyway? The GPIO pin can turn the transistor on and off without it. So with that circuit the transistor will turn itself on at higher temperatures and voltages. Looking at the Data Sheets for a typical transistor like the following shows the transistor is on (drawing 1mA) with Vbe of 0.35V at 150C: Neither of those voltages will work at high temperatures. If this is in a car, the battery voltage tuns up to 15V or so, so it would be 0.15V + 15*220/10220 = 0.47V. The transistor might switch on and off (I'll work on that in a minute), but as the output CVBS_OUT_EN pin is connected directly to ground through two zero-ohm resistors, it isn't going to give much of a signal!Īre you within specs for the GPIO pin? The rated current into the GPIO is 1mA. The first circuit now protects the CPU pin as the voltage is limited by the BE voltage.
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