12V DC, 4 Ampere, 48 watt is load
I have to give AC power as input. How much AC power is required for 48w DC?

 

48W + loss at rectifier which may be 5-10%.

 

diode loss = forward voltage drop * current.
In general forward voltage loss in a diode is around 0.7V.

48W, 12 V ==> current = 4A.
rectifier loss = 4*0.7=2.8W.

AC power = 48+2.8 + other conduction losses (very low may be neglected) = 51W approx.

 

NKT... the technical reply, of course, has a number of "It Depends On" answers. So, here is a very elementary approach, depending on how the 12Vdc is derived:

a) Half-wave rectification (1 diode) but a lot of ripple effect.

b) Full-wave rectification (2 diodes) but less ripple effect.

c) Full-wave diode-bridge (4 diodes) least ripple effect.

Neglecting power supply losses, then only diode losses have to be added. Typically, diode-loss is,

Wd =(Idc)^2 x Rd x T, where,

o Wd = loss (Watts).

o Idc = load current (Amperes).

o Rd = forward conduction resistance of diode (Ohms)

o Td = Diode conduction time for a full AC-cycle (decimal).

The values for Td are: 0.5 for a); 1.0 for b); and 2.0 for case c).

I repeat, this approach is very elementary!

Regards,
Phil Corso

 

NKT... correction:

Multiplier for case c) is 4.

Phil