16 Sep
2013
16 Sep
'13
12:47 p.m.
I think we
need a little more info as to why you have a
source of 1.52volts?
If I create a sine waveform (or even a square wave centered
around 0V)
using 5VDC, chopping a 5V signal using a transistor would yield 4.3V PtP
(.7v drop using the transistor). If I consider the AC voltage as
centered around the 2.15V mark, that's 2.15VAC(max). If the waveform created is a sine (or approximation of a sine wave),
I'd multiple
2.15V(max) by .707 = 1.52VAC(rms). Now, please correct me if I am
wrong, but that's how I got the measurement.
5V is not a 'natural voltage' -- it doesn't sound like you're tryign to
run the C64 from a car battery, for ecxample. Where is that 5V coming
from? Can you not find a more suitable voltage further back in the chain?
-tony
Why not just start with 110VAC?
Because
the power source at hand is 5VDC @ 10A. The 64 need 4.3A of
that, leaving a bit more than 5A for this step-up solution.