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AC regulation and active rectification circuit

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Have you tried this?

No, can't do physical moment.

Pretty certain approach is good though- well proven principle (famous last words). A big junction area BJT is the key. It gives a very low Vsat at decent IC (see graph).

Just to clarify, the circuit will only start functioning around 1.2V in theory, but 3V or so say, in practice, but the voltage drop (equivalent to rec diode VF) will only be 10s of mV as previously stated.

So, for example, with 3V input there will be no output, but with 4V input there will be 4V output, as near as damn it that is.

Don't worry about the gross error on circuit- that is standard with me and is due to laziness.

Feel free to analyse the circuit and make suggestions for improvements- I would appreciate that. Also point out any goofs.

No doubt, the pair of us can turn it into a nice practical circuit. Then it can be optimised and job done.
 
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Feel free to analyse the circuit
comment 1:
There is a wire labeled 0 to +100V in one place and labeled 0 to -100V in another place.
Have you thought what happens when the wire goes from -100V to +100V?
 
Hi All,

I'm improving a circuit for a dynamo, which is to feed the AC into a USB 5v output. The dynamo has a wide range, theoretically anything up to 100v is possible, but at lower speeds it'll generate about 6v. As bicycle speeds are generally all over the place, it can differ quite widely. I know that I've had over 50v downhill at speed.

Why this matters, is at lower speeds the voltage is crucial to getting enough wattage out to the USB. Previously I've used a standard bridge, with a 1.1v drop, a 24v darlington regulator and then the switch mode buck.

I'm trying to improve lower speed efficiency, at the expensive of higher speed by replacing the 1.1v bridge with mosfets. This means I'm limited to 20v unless I add in more resistors and zeners, but the switch mode can only handle a maximum of 24v anyway. So, my thinking now is a 18v zener/darlington on protection and then the fets should be safe. Does the x1.41 factor still apply here?

Please take a look at the circuit and see if I've made any errors. I'm completely new to PCB design (you'll notice I haven't figured out how to switch the P-FETS around yet).

Many thanks!

Andrew
What does this alternator look like?
Does it ride on the hub, part of the wheel or???
PS:
Google has pictures. :D
 
comment 1:
There is a wire labeled 0 to +100V in one place and labeled 0 to -100V in another place.
Have you thought what happens when the wire goes from -100V to +100V?
Yes, the the labels are a hangover and are nonsense. That common line should read sine wave from +-0V to +-100V peak. Yes, I had considered what would hapen when the voltage reverses. that is the essnce of the circuit. At the moment the capacitor would discharge through the CB juction of the power transistor and then through the protection diode- that is the gross error that I mentioned.

I was in a real rush when I did that circuit- it's the principle that counts though
 
I ended up going back to four sckhotties and two large zenier diodes to dump anything over 18V. When the dynamo has a load on it the voltage will go back down to 6V so really the protection is just for spikes.

Next up, a digital relay. Time for a new thread!
 
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