Add a couple of diodes in series with the 5V out.
2 x 1N5004 diodes will work ok. Each diode will drop around 0.7V so the result will be pretty close.

The one thing that can cause a disaster is overheating the cells and making them explode. I have seen that happen a few times, never done it myself.
One amp seems like a very high charging current, that would deliver the 1200 Ma Hours in just 1.2 hours, not 16. Could the desired current actually be 0.10 amps? THAT would seem a lot more reasonable. But feel the batteries every few minutes to see if they are getting hot.

 

The one thing that can cause a disaster is overheating the cells and making them explode. I have seen that happen a few times, never done it myself.
One amp seems like a very high charging current, that would deliver the 1200 Ma Hours in just 1.2 hours, not 16. Could the desired current actually be 0.10 amps? THAT would seem a lot more reasonable. But feel the batteries every few minutes to see if they are getting hot.

on 30% now, charging at 0.12A, maybe the extra amperage is for the display and or to be able to use the device while charging, maybe the device has a further rate limiter inside? Check the pics attached, maybe it will answer some questions.

 

I managed to get 6 diodes from an old set-top box and put them in series, it dropped the voltage to 3.5v from the 5v, its been about an hour and the device hasn't charged 1% even.

6 diodes are to many. 2 is what you need. Without a load on it, the voltage drop across each diode may not be apparent as the meter draws very little current. Try measuring the output voltage with 2 diodes with a 1Kohm resistor across the output.

After putting all these resistors in series I get 42.3 Ohms, voltage hasnt dropped at all, still 5v.

You will not drop the voltage across the resistors without current flowing, so just measuring the output without it connected to the load is not the way to do it.
The voltage dropped accross the resistors will change according to the current flow. That is why diodes are the preferred way to go as the voltage drop across them is "almost" constant when the current is flowing.

very crude charger using a 1R 3W resistor and two 1N5401 diodes in series could be made, powered from thr 5V power supply.

And the series resistor is a good idea. The resistor is used to limit the max current to the battery, and to protect the charger. Not really control the voltage. But maybe use a larger value as @Dodgydave says, but like a 23R or 10R to allow a faster charge.

 

on 30% now, charging at 0.12A, maybe the extra amperage is for the display and or to be able to use the device while charging, maybe the device has a further rate limiter inside? Check the pics attached, maybe it will answer some questions.View attachment 204911View attachment 204912

This is a whole lot more complex than my battery powered clippers. They have a red LED to show that they are charging..And the battery pack certainly has a bit more to it.

 

I have the following diodes available:

IN4004 x4
RL207 X 1
SHK75-11 X 1
SR506 X 1
2 Diodes without markings measuring 670 on the diode mode on my multi meter

I have the following resistors available (Measured using my multi meter, not so good with the color coding, some colors are difficult to make out especially grey, gold, silver, violet):

2.6 KOhm
10 Ohm
23 Ohm
11 Ohm
47 KOhm
2 KOhm
182 KOhm
3 KOhm
.465 KOhm when measured on the 2K setting on my multi meter.

 

And the series resistor is a good idea. The resistor is used to limit the max current to the battery, and to protect the charger. Not really control the voltage. But maybe use a larger value as @Dodgydave says, but like a 23R or 10R to allow a faster charge.

Sorry, can you please tell me your terminology here: 23R, 10R, 1R 3W . Does the R stand for Ohms?

 

One amp seems like a very high charging current, that would deliver the 1200 Ma Hours in just 1.2 hours, not 16. Could the desired current actually be 0.10 amps? THAT would seem a lot more reasonable. But feel the batteries every few minutes to see if they are getting hot.

How do you calculate the charging time?

 

Sorry, can you please tell me your terminology here: 23R, 10R, 1R 3W . Does the R stand for Ohms?

Yes. "R" is ohms, as "K" is kilo or 1000 ohms.
Sometimes you see a value written as 2K7 instead of 2.7K and it is a good way to write it so as not to miss a small "." in the writing.
And 3W is 3 Watts, the power rating of the resistor.

 

How do you calculate the charging time?

Charging time is rated at 10% of the aH rating of the cells, so your cells are 1200mA, that's a charging current of 120mA for 10Hours or 60mA for 20 Hours.

 

Charging time is rated at 10% of the aH rating of the cells, so your cells are 1200mA, that's a charging current of 120mA for 10Hours or 60mA for 20 Hours.

THAT was the basis for my comment about charging current and charging time. My statement was not a random guess. And the one amp did not seem reasonable.

 

Do you have a multimeter that can measure current?

Bob

 

This is a bit long. With the diodes it didnt charge at all, maybe too much diodes taking away all the charging current?

More likely they were installed backwards. Which was lucky for you, because it stopped you from destroying the batteries.

Bob

 

Imhave recharged batteries from a lab power supply that had both a regulated voltage mode and a regulated current mode. Also from an adjustable voltage regulated supply using a resistor to limit the current. All of the charging times went a bit long.