For a god quality PC power supply;
120mV on the 12V rail
50mV on the 5V and 3.3V rails
a linear power supply the noise or ripple can be reduced to mv. However it is bulky heavy and inefficient A switching power supply it is easier to filter its output but the hi frequency noise and spikes are not that ease to get rid off. The weight and size can be greatly reduced and the efficiency greatly improved
Obviously, you don't want the voltage to sag. For computer power supplies, you want the power to be even, pure, and clean without any ripple.
Power supplies are rated at the maximum wattage they can put out. Of course, a computers requirements vary depending on what the computer is doing, so a computer power supply should be of the same, or greater, wattage requirement of the computer when under it's greatest load.
Remove and replace the power supply.
You first have to know the type of power supply your are testing. Then you could test whether the power supply is receiving anything from its source. Next check if it is giving any output.
In an ideal DC power supply, there is no ripple.
When checking for ripple on a power supply, you should set the meter to the AC voltage setting. Connect the meter probes across the output terminals of the power supply while it is under load, if possible. This allows you to measure any AC voltage fluctuations (ripple) superimposed on the DC output. Ensure that the range is appropriate for the expected ripple voltage level.
Ripple, in DC power supplies, is technically unitless. Ripple voltage is specified in Volts/Volt, or a percentage. For example, a 12VDC power supply with 120mV (pk-pk) of ripple voltage is (0.12/12) = 1% ripple voltage.
I think the cause of ripple voltage would be from a bad ground or capacitve voltage.
The maximum power of a class 2 power can be between 40-50v. You should always make sure you are aware what is the maximum output for your electronics.
THe Filter capacitor value depnds on the maximum current I of the Power supply , Switching frequency and the permissible ripple C= (I * (1/2f ))/ ( V * %Ripple) - for a full wave rectifier C= (I * (1/f ))/ ( V * %Ripple) - for a Half wave rectifier Where C= Capcitance in Farads I = Current in Amps f = Switching Frequency V = Nominal voltage in this case 12 V Reji J Thoppil
A multimeter draws next to no current, so will not effectively test the power supply under load. The voltage on the meter is averaged out and so will not show ripple current, or spurious dips in the supply. You will need an oscilloscope to check for ripple current (poor smoothing) and the supply needs to be under a load.
A: Ripple is a residual voltage evident as voltage following the AC input frequency. The ripple magnitude is a function of not enough of both filtering capacitance or overloading the output. Increasing capacitance will reduce the ripple or reducing the loading
If the 2 amps is the output amperage of the power supply, the maximum that should be drawn from the unit is 2 amps. The load amperage that is connected to the power supply should govern the amperage of the fuse used. There is not much range there, the fusing could go from .25 to 2 amps. If the input amperage is 2 amps then the input and output voltage of the power supply should be stated.
Ripple.
A capacitor helps improve the ripple factor in power supply circuits by smoothing out the fluctuations in voltage that occur after rectification. When connected in parallel with the load, the capacitor charges during the peaks of the rectified voltage and discharges during the troughs, effectively reducing the voltage ripple. This results in a more stable DC output, which is particularly important for sensitive electronic devices. By minimizing the ripple, the capacitor enhances the overall performance and reliability of the power supply.
the maximum power it will supply