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Squibcakes
- 31st July 2006, 04:59
Hi,

I have a -15V signal (steady state through 100k resistor) that I want to pulse to 0V every 2 secs.

Originally I wanted to use a P-MOSFET but, I need a negative gate voltage to turn it on etc. Problem is, the PIC outputs 0 to 5volts.

Instead:
Can I do away with the MOSFET all together and use an I/O pin to toggle the -15V signal.

I was planning to do this by changing the state of an I/O pin from High empedance (input pin) to low (output pin) to ground the -15V signal through the pic.

The -15V signal comes through a 100k resistor so current should be around .15ma which is well within the 25ma spec for current sinking. But is the 15V too high?

Or will this Deep fry the Chip?

Cheers
Squib

keithdoxey
- 31st July 2006, 12:08
All PICs have protection diodes on the pins which prevent the pins going outside the range Vdd+0.3V and Vss-0.3V.

Because your feed is via a high impedance resistor you wont damage the PIC but when you are expecting the output to be -15V it will only be -0.3V

I am guessing that because you have -15V in your circuit you have dual supply rails so could you not use an Opamp to invert the pic output to be in the range 0V to -5V. You would then have -ve voltage to drive your PIC.

Another possibility is to use a hardware PWM in the PIC to generate a negative voltage. Search for "negative voltage" and "LCD".

Hope that helps

Dave
- 31st July 2006, 12:12
Squibcakes, All of the pins on the Pic's that I have seen have reversed biased protection diodes from VSS to VDD. If you were to limit the current of the -15 volts signal and connect it to one of the processor pins it would place that pin at around -.7 volts in reference to ground. This is not a good practice and a great way of destroying a good Pic processor. Have a look at using an optoIsolator to do the interface.

Dave Purola,
N8NTA