I have this Philips One Blade trimmer (disassembly video) and I have probably by mistake tried to charge it with charger from the older Phillips shaver which had almost identical connector but 15V instead of 4.3V used by this one - the result is it can't charge. I opened it and there is one burned component on the PCB, but I can't identify what is it (I am not much of electronics expert).

I would be glad for help with identifying that component or a replacement alternative to repair the trimmer.

Solution summary: The part was diod in SOD323 package. I replaced it with 1N4148WS-SOD323 and it seems to work. Thanks for all help.

PCB Photo


3 Answers 3


The 'A' and 'K' markings suggest a diode, probably installed for reverse poalrity protection.

  • \$\begingroup\$ Thanks. So the diode would have anode on Ch+ (which is + from the charging power supply) and cathode on the B+ which is the + of the battery (given the power supply is 4.3V and battery is 1.2V NiMh). Would that make sense? \$\endgroup\$
    – Michal
    Commented Mar 21, 2020 at 15:08
  • \$\begingroup\$ Yes. @user245763 suggestion is a good one. Remove the diode, check for shorts with a multimeter. If there are any, something else is wrong. \$\endgroup\$
    – polwel
    Commented Mar 21, 2020 at 16:10
  • \$\begingroup\$ I sourced several SMD diodes with packages which could roughly match. It the SOD232 package seemed to fit the best. With diode 1N4148WS-SOD323 the trimmer seems to both charge and work. \$\endgroup\$
    – Michal
    Commented Mar 24, 2020 at 18:48

@polwel i agree most likely a diode for RPP incase connection is reversed on entry, check for continuity or very low resistance to ground on both sides of the pad with a multimeter

if its not shorted to ground then any or even no (but bad idea) diode will do

  • \$\begingroup\$ I'm not sure if "any" diode will do, you want one with a low forward voltage drop. Using a 3.3V battery and a 1v Vf diode won't leave much left for the circuit to operate. Less important in this circuit is the current handling capability, but you don't want to have a 100mA diode when the circuit pulls 500mA during operation. \$\endgroup\$
    – Ron Beyer
    Commented Mar 21, 2020 at 17:11
  • \$\begingroup\$ @user245763 It was not shorted to the ground. So I tried to replace the diode with a wire, which resulted in battery charging with 1.4V (per DC multimeter). However, the motor did not start. So I removed the wire and now it works. I now have time till next charging to determine and source the right diode. \$\endgroup\$
    – Michal
    Commented Mar 21, 2020 at 21:00
  • \$\begingroup\$ @RonBeyer The battery is 1.2 NiMh. The charger is rated 4.3V/70mA - and the diode is between charger+ and battery+. So you would suggest a diode with low voltage drop and current higher than 70m (so let's say 200 to be safe)? \$\endgroup\$
    – Michal
    Commented Mar 21, 2020 at 21:05

Old post but I wanted to share my thoughts since there is not much info about the oneblade PCB. It must be a zener diode, to drop the charger voltage and also act as isolation, so voltage is not exposed on the charging pins. Since battery is a 1.2 battery and charger is up to 4.3V there is something that must be dropping the voltage down. So 4.3V - 1.2V = 3.1V zener diode.

And to prove my theory I found a YouTube video where a guy repaired a switch on it and indeed the component had "3V" written on it.

Fun fact, my charger is 6.7V unloaded, 5.6V with 90mA load and cuts off at 100mA :O

  • \$\begingroup\$ Are you saying that you think the charging circuit takes in 4.3 V, but drops about 3 V using a Zener diode in order to obtain about 1.3 V to charge the battery with? That seems like an extremely strange design; why would they design it so that more than two thirds of the power turns into waste heat in the diode? That wouldn't really provide isolation, either, since it would only drop up to about 0.7 V while discharging, meaning the charging pins would still have about 0.5 V on them at minimum. \$\endgroup\$ Commented Oct 8, 2023 at 0:15
  • \$\begingroup\$ @CassieSwett The charging circuit is external. It is built into the power plug. But else yes. It is a strange design but I guess they are just reusing the standard 4.3V power plug. If they was unsure if they would go for lithium or nimh then they could just put that diode in and drop it to 1.2-1.3v Good point about the diode drop voltage. There is also a mosfet on the GND input so I guess it must have a sense signal that turns on the mosfet. I measured mine earlier and there was no voltage on the contacts. \$\endgroup\$
    – Stefanhg
    Commented Oct 9, 2023 at 6:25

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