Hi, I just bought the K4004B amp kit. I live in Japan with 100vac. I know close to nothing about electronics and don’t know which transformer I need to buy.
I was thinking it would be easier to use some kind of laptop AC adaptor than trying to find and install the proper transformer… is this possible? Maybe two AC 18v adaptors paralleled to the K4006 kit?
If I’m not making any sense, please spare the embarrassment and shoot me down ASAP LOL. But shoot softly
Hello again, i think I found the right AC adapter specs! For only 15 bucks a pair at Amazon!
Compatible with : PPP009L + All HP Pavilion DV1000,DV1300,DV1400,DV1300,DV1500,DV2000,DV3000,DV4000,DV5000,DV6000,DV7000,DV8000 Models.
AC Adapter Input: AC 100-240V Output: 18.5V 3.5APower: 65W
Connecter size: 1.7mm *4.8mm
Package Includes: 2 AC Adapter + 2 Power Cord
ClearMax AC adapters are made with the highest industry standards. Including safeguard features against incorrect voltage, short circuit, overheating. 100% compatibility with the original.
Yes, if the primary voltage is 100VAC.
Please make sure to check the K8060 aswell.
It exist assembled (vm100) and it is cheaper than the K4004 + APS200
You need two of them, as it is mono.
Ok great, guess work and a little internet search paid off, but man 80 bucks for a transformer is 80 bucks more than I expected to spend on this project…
but thanks for the tips VEL417!
Another question… I also want to make this K4004 amp Battery Operated using cordless power tool battery packs of 28 volts…would it be as simple as plugging the positive and negative terminals of one battery pack directly to the pos/neg of the K4006 capacitors, bypassing the bridge rectifier?
EDIT: BTW, yeah K8060 would be good for a future active mono subwoofer project, the K4004B has just enough power for my “active mono speaker/ portable stereo boombox project” and I got a real good deal from qkits.
Oh I just read about the Bridge Rectifier… I would still need to plug the battery pack to the Bridge Rectifier as its other job is to prevent accidental reversed polarity, correct?