AC regulation and active rectification circuit

[ACharnley]

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

 

[ACharnley]

With a bit more thought, the 18V zenier is too high? I forgot the Darlington has a 0.7v turn on.

 

[MikeMl]

I forgot the Darlington has a 0.7v turn on.

More like 1.3V

About the easiest is to use a FW bridge made from Schottky rectifiers.

Dont understand what you are expecting to do with the PFETs?

 

[ACharnley]

The MOSFET's are so called 'active rectification', no voltage drop. Schottkies still give about 0.8v (and the silicon bridge chip I used prior to that was 1.1v).

Previously the load was connected to one darlington, so power loss would only ever occur if there was a high load and high speed. Not I'm sacrificing a bit of that for low voltage performance.

 

[AnalogKid]

First, there *always* is voltage drop, although it can be less than with a standard diode if you oversize the FETs.

The description in post #1 is not clear at all. Among other things, I don't think you want to feed AC from a bike generator *into* a USB port. Also, why do you think FETs can't tolerate circuit voltages above 20 V? If you are talking about Vgs, then maybe, but some MOSFETs have VDS ratings above 1000 V. You need to be more clear about the voltages you are describing.

Overall, it sounds like you want to use synchronous rectification to improve the efficiency of converting a variable-frequency, variable-amplitude AC power source to 5 Vdc. Is that it?

ak

 

[ACharnley]

Synchronous, right! Unfortunately I'm quite new to all this. I only today learned what a MOSFET was. Bare with me.

The idea with the FET's is to replace the diode's, which will give virtually no voltage drop.

The FET Vgs can't go over 20V, hence my choice of zener diode in the darlington shunt arrangement.

Originally I was going to use 2N7000 FET's but from what I read they are 250mA. I'd potentially have 1A going through them (dynamo is 6W/6V rated but voltage can go up).

 

[ACharnley]

From what I now read, synchronous rectification is another term for swich mode voltage regulation, which is already sorted (not part of this circuit). This circuit must a) step the AC down to AC below 20V and b) rectify it to DC as efficiently as possible.

 

[Nigel Goodwin]

From what I now read, synchronous rectification is another term for swich mode voltage regulation.

Not at all, synchronous rectification is nothing to do with switch-mode, in fact it's not necessarily anything to do with 'electronics' either

A classic example would be synchronous vibrators used to power valve radios in cars (long ago) - extra contacts were used inside the vibrator (basically an oscillating relay) to perform the rectification, a purely electro-mechanical system.

 

[ACharnley]

Assuming there's nothing wrong with the circuit, I'm looking at a 2A FET, something along the lines of a 2N60, but does anyone know the P-channel equivalent, or know of a matched pair?

 

[AnalogKid]

On sites such as Digi-Key you can do a parametric search. As you narrow the parameters it updates a list of everything they have that meets them.

Synchronous rectification is used in very low level low frequency amplifiers and radio detectors.

And you seem to be hopping back and forth between the terms FET and darlington. These are not the same, and have radically different biasing requirements. For Fets, a standard way to prevent a Vgs overvoltage is to place a 12 V zener between the gate and source, and drive the gate with a resistor.

Also, if you preregulate the input AC so it is low enough not to damage the switching post regulator, that is two regulator stages in series. Sometimes that's the only way to get the job done, but consider a synchronous rectifier that delivers 60 Vdc followed by s switching buck regulator that can handle a 60 V input.

ak

 

[Les Jones]

Hi ak,
I think you meant to say place a 12 V zener between the gate and source. (Rather than place a 12 V zener between the gate and drain.)

Les.

 

[ACharnley]

Sorry, just to explain a bit more, before the FET rectification there is a Darlington/Zener shunt, which will drop anything over 18V (that's the zener I was referring to) as heat. This is to protect the FET's with their typically 20v max gate voltage.

I wasn't able to find a switch mode chipset that could handle 60V, and the higher voltages they go to tend to mean an increased minimum voltage. With the dynamo voltage being all over the place, this is part of the problem.

I've spent about two hours searching for a good N & P Mosfet which can handle 2A with a 20v gate rating. I'm unsure how important getting a good N & P match is. The frequency from the dynamo should be pretty low.

I've also been looking for dual and quad MOSFET packages but haven't found any with 2x P and 2x N within them. This article mentioned one but it looks like it never made production.

https://digitaldiy.io/articles/elec...no-voltage-drop-bridge-rectifier#.VnclynUS-V4

Thanks!

Andrew

 

[spec]

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

Hi Andrew,

Intresting project. Another approach is to simplify and also use a bit more of the energy from the cycle alternator. Here is an outline schematic to give an indication of what I am thinking about. The rectifier diode loss is only around 500mv.

​

 

[ACharnley]

Brilliant! - though I don't really understand what's going on. Would it work down to 6v? Also, I see power having to go through two diodes; D2, D6, which would have those losses?

I'd very much like not to dump the extra voltage as heat, though cyclists will rarely go so fast that this will occur.

I finally found a dual MOSFET package as well, the AOP605. **broken link removed**

 

[spec]

The arcitecture suggested achieves the opposite to what you think. With six volts output from the alternator the voltage across the two capacitors would be around double the alternator output voltage ie 12V. At 100V output the voltage across the capacitors would be 200V. Current flows through one diode for one half of the alternator output and through the other diode for the other half cycle. Thus there is only one diode drop in the circuit at any one time. The MOSFET is a switcher which stores energy at one voltage and then dumps it in the lower capacitor at the 5V output for the USB power. The energy is stored in the inductor. The control box takes care of all that.

The priciple of the circuit is known as constant energy, so there is no energy loss, unlike a linear approach. In this application all the components would run as cool as a cucumber. So there is no dumping heat however fast you go; that is the principle of a constant energy approach.

 

[ACharnley]

I see now where the 12v comes from, clever!, I'll have to follow the rest through tomorrow morning (as I'm off to bed!). Many thanks for taking the time to draw it out.

So the maximum drop here would be that of a Schkotty, so let's say 0.4v, and it would work up to the max voltage, whereas my idea would drop less, perhaps 0.1v but only work efficiently up to 18v.

I guess the control circuit would be quite complicated, not something available off the shelf?

 

[AnalogKid]

Hi ak,
I think you meant to say place a 12 V zener between the gate and source. (Rather than place a 12 V zener between the gate and drain.)

Les.

oops. fixed.

 

[spec]

I see now where the 12v comes from, clever!, I'll have to follow the rest through tomorrow morning (as I'm off to bed!). Many thanks for taking the time to draw it out.

So the maximum drop here would be that of a Schkotty, so let's say 0.4v, and it would work up to the max voltage, whereas my idea would drop less, perhaps 0.1v but only work efficiently up to 18v.

I guess the control circuit would be quite complicated, not something available off the shelf?

That is corect about the s diode.

The control function is standard stuff and would be done by a simple, freely available and cheap chip.

I'm off to bed too.

 

[ronsimpson]

Have some questions? What current on the 5 volt output?
How much current is available from the generator at (full speed) and (at low speed)?
Is this a cell phone charger?

Using Mr SPEC's idea:
Using a voltage doubler to get 2x higher voltage.
Using a 4 to 140 volt PWM. at 400mA
Will need a change so the PWM will output less current at low input voltage. (easy)

 

[spec]

Have some questions? What current on the 5 volt output?
How much current is available from the generator at (full speed) and (at low speed)?
Is this a cell phone charger?

Using Mr SPEC's idea:
Using a voltage doubler to get 2x higher voltage.
Using a 4 to 140 volt PWM. at 400mA
Will need a change so the PWM will output less current at low input voltage. (easy)
View attachment 96266

Mr RON,

You are a genius