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Can't figure out how to design a circuit for EE class

S

sdaccord01

Senior member
Hi,

I'm having trouble with making an AC 110V to DC 6V @ 20A circuit. I have been looking online and found something that converts 240VAC to 5VDC but I don't understand how it works except for the fact that you need a combination of capacitors and transformers to convert an AC signal to DC. Any help is much appreciated.
 
Link to the 5VDC circuit please. There are many circuits that will do this for you. A majority of them well have these two circuits in it.

1) Rectifier
2) Low pass filter
 
1. transformer +
2. bridge rectifier +
3. filter capacitor +

= quick n' dirty power supply.

Now, for 20A, you'll need a decent transformer, mind you.
 
yeah 6V at 20A will need a real beefy transformer with a regular diode bridge rectifier + capacitor
probably wont be cheap

why not a switchmode power supply, all you need is a high voltage diodebridge rectifier, transformer, cap, oscillator/mosfet/power diode/inductor, error amp.. i forgot what else its been a while

or just use a old pc power supply - they should be able to crank out 5VDC at ~30 amps or something
 
And remember, silicon diodes drop a certain amount of voltage.



What about a few transformers in parallel to get to 20A?
 
Well.. the schematic you're showing basically breaks down into these stages.

1st Stage: 240:9 voltage transformer

2nd Stage: Diode Rectifier Circuit. Since the input to this is still an AC signal (meaning it will swing back and forth between the two rails), it will attempt to draw out only the positive portion of it. Imagine the two extremes:

The first being where the top is positive and the bottom is negative. The diode in the lower left will clamp the bottom to being 0V (GND) - Vdiode. The diode on the top right corner will draw the far right side up to Vtop - Vdiode and store it on the capacitor.

The second is the opposite where the top is negative and the bottom is positive. The diode in the upper left will clamp the top to at lowest 0V-Vdiode and the diode in the lower right will pull the right side to be Vbottom-Vdiode.

3rd Stage: So up to this stage, it merely takes the input AC signal and does a rough 'absolute function' on it. Having the capacitor will actually take only the peaks and store it since the only way it can really lose its charge is going backwards through a diode or into the 7805 device. So whatever it stores, it can only store more hence taking the maximum. So on the 470uF capacitor, it will hold a little less than 9V.

4th Stage (7805): 5V Voltage Regulator. It can take any DC signal and control its output to maintain a 5V output.

Edit: I'm not too sure how the 7805 Voltage Regulator works until I get a more detailed schematic.
 
uh do you really need 20A@6VDC ->120W ?

7805 is a linear regulator - it uses a voltage divider to sample the output and compare that with a internal reference, the result is used to bias an output element ( it will try to bias the element to keep the output voltage constant )

i think the power dissipated ( if i remember from the datasheet )in the part would be heheh (Vin-Vout)*Iout, so we're lookin at (9-5)*20 = 80W!

i dont think you could do this even with a beefy pass transistor
 
I've been looking around digikey, and havent found a lot of what you need, and if its not there, its a real pain to find anywhere else. 20 amps is a whole lotta power. You sure thats the output right? And not what the input is (a standard 20 amp 120V outlet)? It's still possible, but not with the design you have. Voltage regulators generally arn't rated for more then 1.5 Amps, max, and I don't think you want 14 of them in parallel to do the job.
 
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