Installing battery charger into work radio.

I have two Milwaukee Job Site radios.

I just bought an 18V lithium ion tool kit. I have an extra charger with that kit that I required for free.
The new Milwaukee battery design does not fit into the old radios that I have.
And obviously, I want the battery to fit into it. And while I'm at it, I may as well install a charger into the radio as the radio itself will not charge batteries but only consume them.
The radio uses 12-28v batteries of the older design. But I don't own any of the older batteries/tools. So again. This is why I am converting this radio.

First things first, I made sure that the radio would run off the new 18v Li-ion battery, and it did! So I began modding. I even made enough room for the charger in the radio and installed it. I then modded the charger bay to holster the new battery size and terminals. I then decided to try the radio again to see if it worked. It didn't. After many frustrating minutes of trying to figure out why, I figured it out. I tested it with a different battery than I previously tested it with. A fully charged one. The one I used before was about 50% charged (going off the digital bat meter on the battery.) I even decided to try it with a 12V battery from a different tool brand, Hitachi. It works beautiful.

I just want to know why and what I can do to make the radio work with a fully charged battery. I'm assuming it's the amps. The amps on the new 18v battery must be higher than the old batteries were. I figured that I wouldn't have to do anything. As the radio is designed to accept 12-28v batteries. I was thinking of using some kind of resistor or even a voltage regulator. I just don't know what to do. This is where I need help.

What the radio does with the fully charged battery. It runs for a few seconds then it cuts off.

I'm just doing this for fun. Its an outside (water resistant design) radio. And I have two of them. If it blows up... It blows up I would just appreciate being able to work with a cordless radio once in awhile. Then to be able to charge my batteries without having to remove them from the radio.
I have everything else working except the problem of fully charged batteries not working with the existing radio's battery input lines. Help me if you can.
Thanks. Sorry for the long post.

Your lithium batteries have a circuit built into the battery itself. If you are drawing too much current, it will shut off.

If you are trying to operate the radio and charge the battery at the same time, it wont work.

This is why the chargers are not incorporated into the radios to begin with.

The circuit for charging is IN the battery. The holster touches the proper pins to communicate with the battery for the led indicator on the holster and to let the battery circuit know it is in a charger NOT a tool.(or radio)

When you turn it on, the battery works for a few seconds until the charger kicks on. At this point, the battery will go into a protection mode because the voltage to the charge circuit is wrong.

So when you use a fully charged battery, the charger circuit has not kicked in. As the battery drains, the charger is looking for a particular voltage on the battery terminals.

There is a difference in the charger holster and the radio battery holster pin outs.

The radio holstesr (And the all the tools) have only 2 pins where the charger has 3.

ive wanted to do the same for quite some time. if youre going all out and fabricating a new battery hole and everything in the radio, there is no reason you shouldnt be able to do this. youll have to add some circuitry that disconnects the battery from the radio electronically, that way it can charge while the radio gets power from the a/c input via an isolated dc converter.

the dewalt radios do it, all cellphones charge themselves while still running the phone... its certainly possible.

im actually making my own jobsite radio and it too charges a set of lithium batteries while plugged into a wall, i just used a set of charging diodes to isolate and switch the battery from charger to dc input.