ac to dc 100-240 v tell me if this is right.
1 rectify.. does this mean that 150 ac = 150 dc????
2 then switch thru center tap transformer to get desired dc v.... why does switching vary the voltage???
What you are doing here basically is getting say, 120V AC from a utility line voltage and converting it to a DC voltage.
No that does not mean that 150 V AC = 150 V DC.
To analyze the operation, first you need to know what the turns ratio is for the transformer being used. Let's say it is 10:1 (10 turns in the primary winding and 1 turn in the secondary winding). So we take V1 to be the input, which in this case is 120 V AC. 120 V is actually the rms value of the utility line voltage, so the rms value at the output of the secondary winding would be, 120 V (rms) / 10, 10 being the turns ratio, give us V2= 12 V (rms). Then, this voltage goes into a full-wave-center tapped rectifier, which implies that we are using a center-tapped transformer using 2-half-wave rectifiers.
Then we need to determine the peak voltage of V2. This is done by multiplying V2 by 1.414, the actual equation is Sqrt(2)*V2(rms), which gives us V2(peak) = 17V. What each of the half wave rectifers do is simple, it takes the sinousodial wave form, and clips the negatve portion of it, leaving us with only the positive half cycles. The output of the rectifier, lets call it Vo(peak) would then be 0.5*V2(peak) - 0.7V, which gives us 7.8V. The 0.7V is the voltage drop of a diode that is used within the rectifer circuit. This happens for both the negative and positive half cycles of the input AC waveform. After rectification, The output waveform resembles a DC waveform, except that it is a pulsating DC waveform. To calculate for Vo(DC) the equation is,
2*Vo(peak) / Pie or 0.6366*Vo(peak). Which would give us 4.96 V. Thats the DC voltage output. Then what can be done is pass this DC signal into a filtering capaciter to eliminate the pulsating characteristics of the waveform and make it look more like a smoother DC line.
I didnt clarify my question at all. forgive me. here it is again,,...
I know that most ambi-voltage ac to dc wall warts first rectify ac to dc and then filter and
then switch at a CERTain speed to get ,say 7 volts dc. How does this switching work to get a certain voltage just by turning it off and on quickly?????????????????
And why can these newer wall warts take any voltage in??? Is it because it doesnt matter what you have for a rectified dc voltage,, its the switching that matters..????
You can see how it seems that 120v and 220 v would give a different result.....but
theses new wall warts dont care ,they give out a certain dc voltage no matter.....???
Its a ratio thing. If you switch on for 1/10 of the cycle and off for the other 9/10 of the cycle (the "mark/space ratio") at a high frequency, then filter the result.... the filter will "Average out" that wave to something like DC that is only 1/10 of the original unswitched DC voltage. If you also use a circuit to compare the resultant output with a known reference voltage, then you can alter that "Mark Space" ratio from that circuit to keep the output at a specifed voltage..(within the limits of the circuitry).
In a "Wall Wart" (or adaptor as we here in the Antipodes call them) the switching is also done thru a small high frequency transformer to isolate the output from the mains voltage, and the resultant AC output then rectified before being filtered and referenced.
You can see, that by this method, quite a large variation of input voltage can still give a stable output.