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deepakomanna
Joined: 06 Mar 2007 Posts: 92 Location: Pune,India
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12f629...voltage? |
Posted: Mon Aug 20, 2007 3:16 am |
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Dear sir,
i am using 12f629 ic,I want to know, if +5v is given to pin number 3 and pin no. 1 (vdd)kept open.Then voltage showing at pin no 1 is approx. +4.5v
And if +5v is given to pin no. 4 then voltage at pin no. 1 is 0v.
Then what changes i have to make so that i get 0v at pin no 1 if +5v is given to pin no 3.
I am not understanding why is it so. _________________ Thank You,
With Best Regards,
Deepak. |
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Ttelmah Guest
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Posted: Mon Aug 20, 2007 4:34 am |
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You can't.
This is because of the internal protection diodes on the output FETs. Think of it as if each pin is connected to the Vdd, via a diode, that is normally reverse biased (when the pin's voltage is below Vdd). However if the voltage rises above Vdd, this diode starts to conduct, preventing damaging voltages on the pins. The exception, is pin4, which has no output FET (and hence no diode of this sort), and connects internally to a charge pump, used when the chip is being programmed. This will draw current, when the voltage gets to about 12v. At this point, you risk destroying the contents of the PIC's ROM....
If you look at the data sheet, section 12. It gives the voltages on the various pins. The key ones are:
'voltage on all other pins with respect to Vss', which is Vdd+0.3v max. With Vdd at 0v, this allows just 0.3v, before the diodes will start to actually draw Vdd upwards.
The MCLR pin (pin4), instead has 13.5v max.
If you have signals that you want to connect to the PIC, and leave the PIC unpowered, then you need to add something like a 4066 gate, and disable this when the PIC is turned off.
Best Wishes |
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