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A means of closing a JFET switch with a postive voltage?

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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it's not a good idea to operate things above their rated voltage. Check the datasheet for the absolute maximum ratings.

What happens depends on the device. Some may apparently operate fine at higher voltages, others will rapidly self destruct.

and it's never easy to know whether the device you have is representative of the population or an outlier.
 

Old Steve

Jul 23, 2015
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AFex54, *steve*'s advice is good.
Just because a chip works with a higher-than-rated voltage, doesn't mean that it will keep working. It's bad practice.
It's life is likely to be shortened by the extra stress. Even running a chip right on the maximum is pretty iffy. The absolute maximum rating is usually only suitable for short durations.
If the manufacturer thought it was going to be reliable at 9V, they would rate it accordingly (because they'd sell more). And, after all, the manufacturer knows much more than you or I. ;)
Personally, I'd be looking for a device more suited to the application.
 

AFex54

Apr 10, 2015
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AFex54, *steve*'s advice is good.
Just because a chip works with a higher-than-rated voltage, doesn't mean that it will keep working. It's bad practice.
It's life is likely to be shortened by the extra stress. Even running a chip right on the maximum is pretty iffy. The absolute maximum rating is usually only suitable for short durations.
If the manufacturer thought it was going to be reliable at 9V, they would rate it accordingly (because they'd sell more). And, after all, the manufacturer knows much more than you or I. ;)
Personally, I'd be looking for a device more suited to the application.
thats a shame cos I already own 10 of these things and I cant see why I would ever be using a supply lower than 9V... infact from all the trouble 9V has caused me so far I'll be going higher!

I wonder could I simply lower the supply voltage to below 8V with a diode/resistor and run it with that?
 
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Old Steve

Jul 23, 2015
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As long as your inputs aren't going to go above the supply voltage, (I haven't read right back), a resistor and 8.2V Zener would bring you closer, at least. A resistor, then 6.8V Zener in series with a diode to the negative supply voltage would bring you even further down into the safe zone. (About 7.4V)
Either of these methods will result in a less stable voltage, though, which will vary more under load changes. Also, your battery won't last as long, due to the current through the Zener.
Alternatively, two series diodes in the supply rail, but that'll be even less stable.
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
May 8, 2012
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I'll go one level further than already stated. I won't operate any component over 90% of manufacturers maximums and for some even less.

As for what to do with these 10 chips..? Considering the depth of the work you're already involved in I find it highly unlikely that you're journey into electronics land is going to end here. You will undoubtedly have more projects. Some will be powered by 5V.

Chris
 

AFex54

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I have some free time so Im finishing this up today,
everything is working great now I just need to build on it a veroboard .
will let you guys know how it turns out

P.S would it be ok to just use 0.1μF bypass caps for all the ICs? this is at least a 'better than nothing' decision but maybe calculating the right value cap for each IC is necessary
 
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CDRIVE

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Are you referring to supply rail decoupling caps, which are synonymous?

Chris
 
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