Wattage question about N64...
Moderator: Moderators
Wattage question about N64...
I noticed that many people are able to run the N64's 12 volt input at many varing voltages. With my basic knowledge of electricity (being that the N64 will always need the same amount of power or watts), does the N64 draw more amps to make up of lower voltages in order to get the same amount of power? BTW, don't hesitate to lecture on how electricity or electric circuits really works if I dont have a clue what im talking about. The only way I'll learn is if you tell me.
no, it has a linear regulator which (i think) draws the same amps as it outputs. if it had a dc-dc convertor (switching regulator), then it would draw more amps at lower voltage and less amps at high voltage. that's why most people use switching regulators for the 3.3v, the battery is much higher voltage than 3.3v so it draws less amps and has longer battery life
yeah, thats basically it.
a linear regulator will drop voltage while keeping the current the same. so. lets go through a hypothetical scenario:
the 12v line needs maybe .5A. the linear regulator outputs 5v, so 5v * .5A = 2.5 watts. now if you supply the 12v line with 12v, the nintendo 64 still sees 5v coming off the regulator, and draws .5A through it. so its using 12v * .5A = 6 watts. where did the other 3.5 watts go? they are lost as heat in the regulator.
for this reason regulators usually get pretty hot. and as you can see, more than 50% of the power you deliver to the regulator at 12v is lost. which is why dc-dc converters are preferred (although no one uses them for the 12v line, its current draw is pretty insignifigant compared to the 3.3v line)
so a dc-dc will do a conversion of voltages, instead of 'burning' the extra voltage off as heat:
if it needs to output, say, 3v at 1A (3 watts) then it might take 12v at .33A (4 watts) in. ideally it would take in the same amount of power it puts out, but it needs some power to operate, and there are losses. usually you acheive about 85% efficiency.
a linear regulator will drop voltage while keeping the current the same. so. lets go through a hypothetical scenario:
the 12v line needs maybe .5A. the linear regulator outputs 5v, so 5v * .5A = 2.5 watts. now if you supply the 12v line with 12v, the nintendo 64 still sees 5v coming off the regulator, and draws .5A through it. so its using 12v * .5A = 6 watts. where did the other 3.5 watts go? they are lost as heat in the regulator.
for this reason regulators usually get pretty hot. and as you can see, more than 50% of the power you deliver to the regulator at 12v is lost. which is why dc-dc converters are preferred (although no one uses them for the 12v line, its current draw is pretty insignifigant compared to the 3.3v line)
so a dc-dc will do a conversion of voltages, instead of 'burning' the extra voltage off as heat:
if it needs to output, say, 3v at 1A (3 watts) then it might take 12v at .33A (4 watts) in. ideally it would take in the same amount of power it puts out, but it needs some power to operate, and there are losses. usually you acheive about 85% efficiency.
"Linux is only free if your time is worthless"