Hello

My prof required us to make a full bridge AC to DC power supply. The input voltage is 230Vrms.

Design Requirements:

Ripple - should not exceed 1vpk-pk
Regulated output should be 3V and 12V
And, Efficiency = 85% <<< Oh, also I dont get this one. Books said that the maximum efficiency for a full bridge rectifier is 81.2%...

Components I currently have:

Step-down transformer (from 220Vrms to 12Vrms)
4 Diodes (1N4001)
Resistor (1k ohms)
Capacitor (1000uF)

Vout = [12(sqrt2) - 2(.7)] x 0.707 = 11.01 (Is my equation correct?). If so, is it means that the values of capacitor and resistor doesn't vary the output voltage?

My main problem here is that I'm having difficulties to achieve the output voltages and the required efficiency. Pls. help me, tnx a lot! ^_^

 

Your equation is wrong.
Here is the solution for 120 VAC


Vp =(V pri rms x 1.414) - 1.4 V
Vp=(120x1.414)-1.4V
Vp = 168.28 V

Do the same for the secondary to find the peak voltage there. (should be approx 16.8 volts for a 12 volt secondary)

 

Efficiency can be improved with schottky diodes or synchronous rectification:
http://en.wikipedia.org/wiki/Synchronous_rectification

 

Your specs state 12vdc and 3vdc REGULATED outputs. 12vdc is a typical requirement, 3 volts is curious - is this correct? Is it possibly 3.3vdc or 5 vdc?
As kermit2 states The circuit shown will only provide about 16.8vdc, but this is UN-REGULATED DC.
To provide regulated voltage will require a voltage regulator IC. An easy way is to use a 3-pin type such as 7812 for 12 volt DC regulated.
The load on the supply is also important.
Is the 1k resistor specified by the prof? - this is the load which determines the current drawn from the circuit shown and affects the ripple voltage.

 

How is your prof defining efficiency? I mean, a transformer alone may not exceed 85% efficiency. The only way to get end-to-end efficiency of at least 85% is SMPS.

Diode voltage drop, 1.2V for a normal rectifier, is a big efficiency loss when the voltage is just 12V and of course a huge factor at 3V. So the schottky suggestion is a good one.

I wonder if instead of using a linear regulator (and the large dropout loss), you could rectify to a peak of 12V and use a large filter capacitor to hold the voltage at that peak. By "large" I mean large enough to minimize ripple into that small load.

Not sure what to do for 3V while maintaining efficiency.

 

Super tnx guys for the replies!

@kermit2
Vout = 12(1.414) - 1.4 = 15.57V, this one sir?

@tubeguy
Nope, not specified. Maybe sir let's disregard the output voltage of 3V for awhile, I'll ask my prof about it. I think sir we can't use the 3-pin that u said becoz we are restricted to these basic components.

@wayneh
He defined it as Pout/Pin. First, he wanted us to have a maximum output power of 36W but he changed it to efficiency = 85%. Also sir, we can't use the suggestion of yours about schottky becoz we haven't tackled it yet in our lecture.

If Vout is 16.8 (approx 15.5V in multisim), then it is too far from 12V. What should I do to have a Vout of 12V? Should I change the step-down transformer?

Tnx again!

 

Super tnx guys for the replies!

@kermit2
Vout = 12(1.414) - 1.4 = 15.57V, this one sir?
...
...
lozadarod:
If Vout is 16.8 (approx 15.5V in multisim), then it is too far from 12V. What should I do to have a Vout of 12V? Should I change the step-down transformer?

Tnx again!

Between the input and output of 7812, there is a drop down voltage about 2.5V, but you should be count to 3V is more safely, so the input voltage on input side of 7812 is ≧15V and the output is 12V.

7812 datasheet, page 3 : 14.5 ≦ VIN ≦ 30
http://www.datasheetcatalog.org/datasheets/150/44435_DS.pdf

 

As has already been stated, you will not be able to achieve 85% efficiency unless you use an SMPS (switch-mode power supply) design. Also, you won't be able to achieve 85% efficiency with a simple SMPS design, like a flyback.

You don't have a current or wattage output requirement. You said that previously, the professor had 36 Watts as a requirement; well that's 3A if the output is 12V. You'll be hard-pressed to get a linear regulator to output that much without acting as a room heater.

You need better specifications from your professor.

 

@wayneh
He defined it as Pout/Pin.

Well yes, but power in WHERE? If he means power in from the wall and OUT is regulated power into a load, you can stop now. It's not possible. You could get closer to a perpetual motion machine than you can to 85% end-end efficiency with that technology.

Now, if he means, power drawn from the bridge as "IN", so that you need to lose no more than 15% in the regulation step only, you have a small chance.

 

@ScottWang
Tnx!

@SgtWookie
Ok, i'll better ask my prof regarding this one. Tnx

@Wayneh
Oh sry, I don't know where.

Tnx again guys, I know now what questions should I ask to my prof regarding to my project. I'll just notify you guys if I gathered enough info that you asked.