Has anyone tried if 7135 die when the polarity is wrong?
I couldn’t find something useful/hopeful in the datasheet, the only thing which gives a hint is that the voltage is –0.3V min.
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I am asking because I build a little led lamp on top of a standard e10 bulb socket and I simply soldered two 7135 in it. The only thing I fear is that sometimes the polarity is switched than what I used now for my specific application…
Of course I could have used a diode before the enable pin…but I thought about that after I epoxied everything together…
I don’t know for sure but I would think that as the D in LED is for diode :bigsmile: that the voltage could not flow in the wrong direction therefore protecting the 7135s.
Cheers David
ps If I am wrong, be gentle with me, I am a bit sensitive and just might rip your arm off and hit you with the soggy end :evil: Edit, and I am wrong, see the post by Rufusbduck, who was gentle so he keeps his arm
Led would protect led- side of 7135 but pwm pin(Vdd) would still be exposed to full reverse polarity voltage. There are two current loops, one going through the led and one just through the 7135.
I just tried a 2X7135 board with LED (from an old Ikea reading lamp): put a CR123 3V battery on it reverse for 5 seconds. The chips got quite hot real fast but still work. I’m betting they will go poof if left longer or higher voltage.
That –0.3V in the spec is likely for a good reason.
Hey, pommie! I wasn’t describing a driver, for the 7135 to pass current, B+ Has to be applied to the Vdd pin and if the cell is in backwards Vdd isn’t protected by the led the way the output pin is. I don’t know whether the -.3V limit is on Vdd, the output pin, or both.
I greatly enjoy having arms.
I wasn’t describing a driver, for the 7135 to pass current, B+ Has to be applied to the Vdd pin and if the cell is in backwards Vdd isn’t protected by the led the way the output pin is. I don’t know whether the -.3V limit is on Vdd, the output pin, or both.