Hi @steve36485
Just verifying that you're connecting the 4.5V DC adapter to the appropriate battery connector terminals in the light's control box (assuming there's no provision to connect an external power supply in the control box e.g. DC power port) and not directly to the two wires connected to the LED lights?
If you are connecting directly to the two LED light wires that were connected to the control box then most probably what is happening is that the lights are alternately connected in reverse direction between the two wires so only half the lights will work depending on the polarity of the supply connection, (LEDs are diodes see image below)
The controller constantly reverses the polarity of the power supply to the lights so that only half the lights, then the other half, are lit at any one time. Doing it this way can mean that less batteries are required i.e. cheaper (3 x AA) rather than having a more expensive higher voltage battery e.g. 6LR61 9V, due to the higher voltage/current requirements if all were on at once.
The lights all appear to stay on due to the fast switching time and the perceived persistence of the light being emitted i.e. high frequency blinking of the lights not detectable by the human eye
If there's no provision to plug in/connect an external power supply, then if the battery holder looks like the one in the image below ( ignore the wires) connect the 4.5V DC adapter wires to the battery terminals as shown by the red and black arrows (or pink and green arrows if placement as per red and black doesn't work. The connections are dependent on the existing battery holder connections between each battery cell position).
Also ensure that the 4.5V DC adapter has an output current rating of 3A-5A. This is based on that the capacity of standard AA batteries is approx. 2.7A-3.0A so the adapter would have to be the same to make sure that the lights achieved their full designed brightness. The circuit will only use as much current to work as it needs so it is not a problem if the external supply has a higher output current. It would be a problem if the supply voltage was higher than specified though
(click on image)
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