Replacing 6.5 nominal power supply with USB

[Ham]

I've got a Panasonic cordless phone handset that I would like to convert to USB supply. The rating on the phone is 6.5V rated 500ma, the power supply outputs 7.5V DC, and the handset has two AAA batteries, so 3V. I reckon I should be able to hack it to a USB output from a multi way anker supply. Any reason why not?

[Diver300]

You could just run a the handset from the 5 V supply. It might not work, but it certainly wouldn't do any harm.

If it doesn't work, you'll have to look at a way of increasing the voltage nearer to 6.5 V

[Kim]

Last time I had to solve this sort of problem (for 9V, IIRC), I discovered that cables with a USB plug on one end, a DC jack on the other and a boost regulator in the middle were available on eBay for less effort than I could be bothered to expend on a more elegant solution.

[Feanor]

Presumably the handset runs off the batteries when in use, so it's running off 3v.

The 6.5v supply can only be charging the batteries.
So the incoming supply must be regulated down internally.

So the question is: will the regulators come into regulation with an input of only 5v instead of 6 or 7v?
I think that's a suck-it-and-see kind of a thing.
I think there's a fair chance it will; 2v is usually a generous enough margin for regulation in this kind of scenario.  Especially if they have used Low Dropout regulators in the design.

Or is the PSU also powering Other Gubbins in a charging cradle?
In which case, all bets are off.

[Ham]

The cradle is simply charging, sucking and seeing has to be the way to go, I think. The Anker charger should eb able to look after itself.

[Ham]

OK, as a starter, I took the cradle apart and I discover there is a 12 ohm resistor in series. If I remember the calculation  rightly (? drop = IR) the voltage drop at 500mA would then be .5 x 12 = ....6 ?? dat has to be rong. OK, so the charging current must be lower, so, let's say 250-300mA? That would then mean the charging voltage would be c. 3V (this is 2 x AA series charging). So then, it suggests I should use a 6 ohm for a 1.5v drop? 

There's also something odd on the reverse, that looks a bit like an SMD, but it seems to go nowhere and do nothing, so I'm ignoring it

[Diver300]

The 500 mA rating would be worse case. There's no reason for the manufacturer not to put a generous number. It's not as though the power supply or wire would be any smaller or cheaper for 300 mA than 500 mA. If you make wire that will only take 300 mA, it will fall apart when you look at it. It would be a different story for bigger ratings where they would be more accurately calculated.

You are confusing the voltages. You have the power supply voltage, the battery voltage and the resistor voltage. In a simple circuit like that, power supply voltage = battery voltage + resistor voltage

With a 6.5 V supply, and 3 V of battery, there is 3.5 V across the resistor. A 12 Ohm resistor means that 3.5/12 = 292 mA charges the battery.

If you supply it with 5 V, still with a 3 V battery, that leaves 2 V across the resistor, and the charging current is 167 mA, so slower charging but still almost certain fast enough for most uses.

The surface mount resistor does connect. The light green tracks aren't visible under the white silk screen printing, but they are still there. I guess that the resistor is so that the handset can detect the charging base even if it's not plugged in, so that it will still answer a call when picked lifted off the base or hang up when replaced.

[grams]

The batteries are likely NiMH, so nominally 2.4V for two.

If the power supply is outputting 7.5 V, and assuming there's no further voltage drop in the handset then (7.5-2.4)/12 = 425 mA, which is a reasonable charge rate for NiMH AAs. You should probably check this with a meter and the original charger.

I'd guess it'll work as is with 5V, just charging slower (5-2.4)/12 = 217 mA. A 6 ohm resistor would double that.

The SMD think looks like it's in series, so could be an anti-backfeed diode. A bit small for carrying 400 mA though.

On preview: Pick which answer you want to believe.

[Ham]

Thanks for that - the chances of me getting it at least slightly wrong was, after all, close to 100%.

I was thinking of bypassing the input and connecting the 5v to the phone, but it looks like connecting to the DC input has to be the way to go.

[Feanor]

Yes, I wouldn't connect the 5v directly to the phone handset now that we know the cradle contains that current limiting resistor.
The current limiting resistor is required to set the charge current.

[Ham]

Success! I checked the voltage at the phone battery terminals beforehand - 3.3V, so there was obviously a regulating circuit in the phone. Hacked one of those dodgy-but-I'll-still-keep USB leads, to find that there were only 2 wires! good job I didn't need any more. 3.3V at the phone after.