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Strange Use For a Varistor ?

CHARGER.png

In this circuit a 26.5 volt mains adapter charges a 22volt 1300ma/hr lithium cell.
I found the Varistor had failed to a short circuit and cannot identify the 7mm disc device marked
xxx x40
ZTjB (I marked it wrong in the schematic)
I find the circuit, extracted from a Hoover Charger, rather strange in that the break point of the varistor is probable above the 26.5 working voltage and the rating of the Mains Adapter is only 300ma. It maybe that the charge current uses the non linear varistor impedance below the breadown point to linearise. However a zinc oxide device probably provides a very high resistance below the breakdown point whilst a silicon carbide device might better sustain the charge current?

I would be interested to hear from anyone who indentifies the varistor - to enable me to replace it.
Whilst an unusual use of a varistor the circuit could not be described as an ideal charger.
I would also be interested in comments on the circuit. If a battery is connected the wrong way round D2 would probably explode and if not the battery might explode instead. At full charge the charge voltage remains although a LED circuit monitors the voltage across R2 and the \led is extinguished if this falls below about 1 volt.
Very many thanks,
Docwat.
 

bertus

Moderator
Hello,

It is likely not a varistor, but a PPTC.
it is the X040 PPTC, wich has a hold current of 0.4 Amp and a diameter of 7.6 mm.
See the attached datasheet for more info.

Bertus
 

Attachments

  • X-FUSE(XX Series)-2.pdf
    221.8 KB · Views: 4
Very many thanks bertus. I will study the datasheet. I certainly could not understand use of a varistor here. The PCB marked the device with a varistor symbol.

Grateful for your info,
Thanks
Docwat
 
Hi bertus,
Your identification of my varistor as a presettable fuse solves all my confusions ! That squares with the 300ma power source and handles excessively discharged Li Batts. The charge current is only controlled by R2 and the battery voltage but that 100% displays the function of the circuit. Didn't cross my mind that this was a fuse.

Thanks,
Docwat
 

bertus

Moderator
Hello,

Imagine what happens when the battery connections are shorted.
Then the current would be 26 Volts / 12 Ohms = 2.17 Amp.
The selfresettable fuse will peotect the power supply in that case.

Bertus
 
Yup and even an excessively discharged battery will break the fuse and not charge at all.
This circuit is a commerial masterpiece in cost engineering and self protection !

Regards,
Docwat
 
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