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My pleasure Mamu.There's a lot of words in this post as I've tried to dot the i's and cross the t's but really there is not much to it. Feel free to ask if anything is not clear.This is the schematic of a typical switch mode power supply (smps). Yours may not be exactly like this and certainly the names and values of the components will be different but the basic principle is most likely the same.Firstly, beware of everything to the left of T1, the voltages there are dangerous. Even when power is removed that pesky capacitor C103 may be holding a nasty surprise.On to the good bit:The output of this supply is 5V, we can ignore the upper 12V part because it doesn't exist on your supply, the low voltage cable would connect to 5V and ground (0V).As you look down the right hand side of the schematic you can see +5V line connects to resistor R204 and then through R205 to ground, we'll call these resistors Rt for R204 the top one and Rb for R205 the bottom one. Two resistors in series like this are known as a 'voltage divider', the voltage that appears where Rt and Rb connect to each other, marked with 'fb' above, is 5V divided by the ratio of the two. The voltage at Vfb = (Vout / (Rt + Rb)) * Rb = (5V / (4k + 4k)) * 4k = 2.5VWhat we need to do is change either or both Rt and Rb such that whatever voltage we want out appears as 2.5V were the two resistors join.For your smps you need to locate this voltage divider, two resister near the output cable that join to each other and connect to the output +ve and 0v. There is not usually very much going on at the output of a simple smps so I think your chances are good.Measure the output voltage.Measure the voltage where the two resistors join.Unsolder one end of each of the two resistors you have identified, we need to unsolder at least one end so we can...Measure the resistance of the two resistors.Remembering the voltage divider equation above, do the maths and double check that the calculated Vfb voltage is the same as the measured Vfb voltage (it will probably be around 1V to 2.5V). If you are within about 0.2V you probably have the correct resistors.if Vfb = (( Vout / (Rt + Rb)) * Rb an you now know the value of Vfb you can calculate a new value of, say, Rt to give you the output voltage you require.Rt = ((Vout - Vfb) * Rb) / VfbIn this example, we want 4V out and changing 5V to 4V is not a big enough change to cause concern. Both Rt and Rb measure as 4k ohms and our measured and double checked Vfb = 2.5VRt = ((4V - 2.5V) * 4k) / 2.5VRt = 2.4k ohms2.4k is quite a large change from the existing 4k but this shouldn't cause any problems. You would pick a standard resistor with a value as close to the calculated value as possible. Higher values of Rt increase the output voltage, lower values of Rt decrease the output voltage.Now, if you are very unlucky and your smps does not use a voltage divider, looking something like this perhaps...then, unless you can hand draw a schematic for me, it's back to the diodes I'm afraid
Switch mode power supplies are very dangerous things to play with. High voltage in parts, the capacitors hold this voltage even when disconnected from the mains, the heatsinks are sometimes live etc. If anyone else tries this then please be aware of the risks, accept the dangers and do not blame anyone else if something, or someone, goes up in smoke.
never have to buy new batts for it, buy a batt charger, change out batts, or re-charge batts.
