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'[EE]: +-15V Square Resistor value?'
2001\04\04@071740 by

Hi all,

Thanks for the help before on this topic. I have just one more question.

I am going for the "high value resistor, and let the internal diodes do
their thing" approach. but....

How do I spec the "high value" resistor? What is a safe value of resistor
for a +- 15V signal? Is this a gut feeling type of thing, or is their some
calculation that can be done? (I hope so!).

I do not want to smoke my pic!

Thanks

Graham
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> Hi all,
>
> Thanks for the help before on this topic. I have just one more question.
>
> I am going for the "high value resistor, and let the internal diodes do
> their thing" approach. but....
>
> How do I spec the "high value" resistor? What is a safe value of resistor
> for a +- 15V signal? Is this a gut feeling type of thing, or is their some
> calculation that can be done? (I hope so!).
>
> I do not want to smoke my pic!

Step 1: Imagine the equivalent circuit (I don't do ASCII art :-) :

Source signal (assumed low impedence) thru Rx to junction of two protection
diodes.

Leakage current input/out of PIC pin (very small number, check spec sheet).

Step 2: Compute lower bound of resistor value:

Compute an Rx so that the maximum possible current is well below the
specifications in the 'Absolute Maximum Ratings' section of the datasheet.
I'd look for a current 100x-1000x smaller, at least.

Step 3: Compute an upper bound of resistor value:

Assuming worst case leakage current, compute the largest resistor that won't
drop enough voltage to have the input voltage out of spec as  a valid
digital signal (ie: near or at the power rails).

Step 4: Throw a dart to pick a value between the Step 2 and Step 3
computation. I'd tend toward the larger size if possible.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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Graham,

As you may ( or may not) know, resistors limit current.  So to find
the value you want, just use ohms law.   Lets say for instance you
want to connect a PIC pin to 100 volts.  And lets further say that
the spec sheet for the PIC (I don't have one in front of me, so
these values are illustrative only) says that the input current
(IIL and IIH) to any pin is 100 uA.  So, to protect ourselves, we'll
limit the input current to 50 uA.   So now we divide. 100 volts
divided by 50 uA is 2,000,000 ohms or 2 Megohms.  Get the idea?

Regards,

Jim

On Wed, 04 April 2001, Graham North wrote:

{Quote hidden}

jimjpes.com

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At 07:49 AM 4/4/01 -0400, you wrote:

>Step 4: Throw a dart to pick a value between the Step 2 and Step 3
>computation. I'd tend toward the larger size if possible.

Assuming, of course, that it turns out that Upper bound >= Lower bound. ;-)

Best regards,
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