Thank you for your reply
The original security light comes with a 3AAA battery pack internal. When the pir is triggered, it comes on and goes off about 10 seconds after the motion that triggered it leaves. There is about 30 minutes of light in the original design. It is solar powered so the batteries recharge each day. The light was originally 15 white LEDs. I remove the white LEDs and replace them with 7 red and 8 green LEDs. I put a 4.7ohm resistor between the red and green LEDs so that the greens will come on. I wanted to make it last longer so I developed the 6volt external battery pack. It has a 6 volt 4.5ah Lead acid battery with a voltage regulator in a weatherproof box. I plug this into the solar panel jack on the light. The voltage regulator is set to 4.0 volts so it charges the internal batteries.
I thought that if I could use the 6 volt battery and remove the regulator, and the internal battery, that I could put a fader type circuit in the space where the original batteries were inside the light. I also purchased a voltage booster so that I could bring 12 volts from the battery box to the light. I tried to make a fader from some of the designs I saw online. I used 12 volts with a 2300uf capacitor and a 10K resistor connected to the base of a PN2222a transistor, but when connected to the light board, the maximum voltage I can get is 2.8 volts. the green LEDs do not brighten up. The transistor gets really hot too so I think I am overloading the transistor. I didn't know if I put 2 transistors in parallel if that would divide the current and allow more to get to the LEDs. When I try to use the 6 volt battery, I only get 2.4 volts and the green LEDs do not light up.
When the original light triggers, with its 3.6 volt battery pack, the voltage across the light board is 3.2 volts. That is what I am trying to achieve. Is it possible without too many complex components? Thanks again