From: gherold@teachspin.com   
      
   On Tuesday, April 9, 2019 at 11:50:13 AM UTC-4, default wrote:   
   > On Tue, 9 Apr 2019 05:57:21 -0700 (PDT), George Herold   
   >  wrote:   
   >   
   > >On Tuesday, April 9, 2019 at 7:32:10 AM UTC-4, default wrote:   
   > >> I'm using some cmos IC gates in a project.  All seem to have an   
   > >> internal protection network to insure that the inputs can't exceed   
   > >> supply or go more negative than ground with internal diodes.   
   > >>   
   > >> My question has to do with those protection diode ratings.  How much   
   > >> current can they clamp to prevent exceeding the input voltage rating?   
   > >>   
   > >> For instance, I'm running one part of the circuit at 8 volts and   
   > >> another at 12.  It is necessary to input a 12volt output to a 8 volt   
   > >> input.  (that's a no-no according to absolute maximum ratings)   
   > >> HOWEVER, the 12V side is an open collector output with a 10K pull-up   
   > >> resistor.  It works... so apparently the internals diodes don't have a   
   > >> problem clamping ~400 microamps, but I'd feel better knowing what the   
   > >> spec. is and haven't found one.   
   > >>   
   > >> The datasheet shows two sets of protection diodes for every input with   
   > >> a current limiting resistor to further protect the inputs.   
   > >   
   > >How about a link to data sheet, or some part numbers.   
   > >Did you measure the 400 uA?  As a WAG you can probably dissipate   
   > >~100mW in the diode/series R.  Try blowing some up maybe?   
   > >(How about a resistor divider to drop the 12 volts to 8?)   
   > >   
   > >George H.   
   >   
   > Can you answer the question? I already know how to return the pull-up   
   > to the 8 volt side - but want to keep the 12V swing for other circuits   
   > using that same signal line.  I'd rather not add anything since the   
   > build is such that I'd prefer to stay out of it...  Likewise, I could   
   > add a zener clamp as the easiest bulletproof solution, but if the   
   > internal diodes are happy with it why bother?   
   >   
   > Texas Instruments CD4520B   
   > Dual up counter   
   >   
   > Datasheet acquired from Harris Semiconductor   
   > SCHS067D - Revised Mrch 2004   
   >   
   > http://www.ti.com/lit/ds/symlink/cd4520b.pdf   
   >   
   > Page 4 shows the protection network   
      
   Huh, what does the dashed line between the diodes going to the positive   
   supply mean?  Do you have some series resistance between the 12V output   
   and 8 V input?  Otherwise you've either fried the first diode, or maybe   
   it's current limited by the driver.  What's the voltage at the input to   
   the 8 V part?   Typical currents are listed as ~6mA... I'm guessing that's   
   not enough to blow the first diode, but I really don't know.   
      
   George H.   
      
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