How to (properly) charge vintage mobile phone using USB-C decoy/trigger board? - Page 1
Forum section: electronics / beginners

author: eventhorizon95
author_rank: Newbie
timestamp: February 09, 2025, 10:40:16 pm
content: Hello,I am attempting to use a so-called USB-C decoy/trigger board (https://www.aliexpress.com/item/1005006179191517.html) to charge a 1996 vintage PTT Telecom Pocketline Darwin+ (internally a Nokia 2110i) using the 100 W GaN charger (https://nl.pepper.com/aanbiedingen/re-load-wandlader-100-watt-met-usb-c-kabel-334597) I use for charging my smartphone. The phone contains a 6 V, 1.8 Ah Ni-MH battery (installed new 5× 4/5 A pack in December 2024) and originally uses an ACH-4E travel charger (specifications from service manual attached), which may be used in conjunction with the CHH-2 desktop dock/cradle. The battery takes ~2.5 hours to charge in a discharged state using these accessories.I have set the DIP switches on the decoy/trigger board for 12 V PD request voltage, verified output using my multimeter (Fluke 27/FM) = 12.2 V and connected both pairs of the ACH-4E's charging cable (2× black, 2× red, originally attached to an RJ10 connector) to the board's terminal block. The phone detects the 12 V input but unfortunately displays a 'NOT CHARGING' indication. So far, the only way I have managed to get the phone to charge the battery using the board is by connecting the multimeter in series with the 12 V output with the red test lead connected to the 320 mA input. The multimeter does display an overload warning, which is as expected since the original charger is rated for 800 mA output current. Using the 10 A input results in the same 'NOT CHARGING' indication. Shortening the cable and connecting only one black/red pair did not work either. (I assume the cable contains two pairs for current carrying capability? It uses a standard-looking DC barrel jack and the individual stranded aluminium wires are very thin.)Am I missing something obvious? Thanks in advance for your attention. I am happy to provide further information as required.Kind regards,Tobias

author: artag
author_rank: Super Contributor
timestamp: February 10, 2025, 02:35:22 am
content: Maybe the phone relies on the 800mA current limiting to avoid excess current in the battery, and expects the 12V to be reduced. When it sees the voltage at 12, it disables charging to avoid damage to the batteries. It might also be able to measure the current though that seems less likely. The 2.5 hour charging time does indicate that it's using 800mA - ie it's not just a safety limit but the actual current it uses.With your multimeter in 320mA mode, it will put a higher resistance in series which will limit the current, though perhaps not to the right level. A fluke 77 has about 5.5 ohm series resistance in 300mA mode which will limit the current to about 1A.If you have a working original charger, measure the voltage (and, if possible, the current) when charging. If it seems to be reducing the voltage to get 800mA current, put a similar limiter after your decoy board. Ideally that would be an 800mA constant-current circuit but you might get away with a 7.5 ohm 5W (or 10W) resistor. Higher would be better as that will run above 800mA when the battery is discharged. Note that in use the current limited will be dissipating about 5W and will get hot unless you use a switching circuit.

author: eventhorizon95
author_rank: Newbie
timestamp: February 10, 2025, 09:27:40 am
content: Thanks for the quick reply, artag. I was thinking about splicing in a terminal connector block (called 'kroonsteentje' in the Netherlands) to measure voltage and current while charging. I have a multitude of original chargers available for this phone, it was part of an auction of 20~25 pieces of the same model, probably originating from a company that upgraded their employees to a newer model in ~1998. Therefore, sacrificing one of the charger cables for this project was not an issue.

author: artag
author_rank: Super Contributor
timestamp: February 10, 2025, 09:48:50 pm
content: Better still, then, take one apart and find out how it works. I strongly suspect that it's not a 12V supply with 800mA burnout rating but an 800mA current source with 12V compliance. and that it will never reach 12V unless the batteries go open-circuit.