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Converting 6vac to 5vdc, A little help please.

Mysteryname

Sep 6, 2010
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Hey everyone.
This looks like a great site full of information. (I admit that I only just registered but still poking around)

The idea is to create a circuit that can input a AC voltage and output 5Volts DC (USB standard) to charge any USB device.

I've hit a bit of a problem with a circuit that I am designing.

I have a hub generator on my bicycle that is rated at 6v 3w power. While I spin the wheel by hand the amount of voltage it produces varies from 6.2v to 7v. So I can expect the voltage to be higher while I ride the bike. (This make simulation a little tricky.)

My idea is using a full wave rectifier a capacitor and a zener diode to 'regulate' the voltage to the output. I want to know if this is correct.

When I simulate a Voltage of 7vac.
The output is at 5.2v

But if I simulate a voltage of 8v the output is at around 5.7v Which is way too high for my liking

Is there are better or simpler way to create this circuit with out using linear regulators?

-Mystery
 

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Open circuit, those hub generators can easily generate voltages of 60 (yes sixty) VAC.

Simply rectifying the output and connecting a voltage regulator is going to kill the voltage regulator in very short order.

The issue is that as the load falls, the voltage rises. you need to either devise some sort of pre-regulator for the 3-terminal regulator that can protect it from high voltages, or maintain a load on it so that its output voltage cannot rise too high.

The issue with the latter solution is that it will impose drag on you as you ride.

Also, your circuit diagram is upside-down, and I'd be very careful about that ground symbol as I'm pretty sure you'll find that the bicycle body is connected to one AC rail.
 

Mysteryname

Sep 6, 2010
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I'll have to keep an eye on the voltage of the hub.

or maintain a load on it so that its output voltage cannot rise too high.

Could I use a zener diode, where the zener voltage is at 5v and some sort of power draining device. Some lights to warn me that the voltage is too high and a linear regulator to burn all the extra power.

Of if thats a bad idea, how could I design a pre-regulator? (higher capacitance at C2?) Sorry I'm new to this as this is the first circuit I have developed from what I have been learning at uni. (my last circuit was a mash of 2 put together with no real understanding of what was going on)

Also I'll only have the circuit connected when I need to charge the device. Otherwise it will be disconnected via a switch.


Also, your circuit diagram is upside-down, and I'd be very careful about that ground symbol as I'm pretty sure you'll find that the bicycle body is connected to one AC rail.

The hub generator is completely isolated from the chassis of the bicycle. But I have no -real- ground. I used a ground so I can simulate the circuit.

And the Diagram is upside down because I created the Diode bridge wrong so I adapted with the ground being on top. (bad practice I know)


Thanks for the reply.

-Mystery
 
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1) Make sure your bridge rectifier is rated at 200V or more.

2) set up a simple 20V (ish) regulator using an NPN transistor, a resistor and a zener (about 20V). The transistor needs to have a Vcebo of 200V or more and an Ic of 1A

3) use this to charge a capacitor (25V, 1000uF) and then regulate this using your regulator of choice.

The resistor/zener is going to be a little tricky as it will need to operate from about 8V up to (say) 60V. It's probably not the best option.

The alternative is to use an "amplified zener" as a shunt regulator (set to something a little less than the max voltage for your linear regulator). If set to a high enough voltage, the drag you experience may be quite small (or not -- you'd have to test it).

It is tempting to suggest using a switchmode regulator as your main regulator too...
 

Mysteryname

Sep 6, 2010
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Ok, I'll admit that most of what was said blew over my head. I'm obviously way out of my depth with designing and creating this circuit by-myself. But one thing did ring a bell. Switch mode regulator, after extensively reading about these regulators I found that they are used in high quality car chargers and easily drop the voltage down to 5v @ 1A from 7-40 volts.

I read a couple of datasheets and example circuitry of these regulators. I came across a component called "LM2575"

Its rated for 7-40v of unregulated input and provides 5v @ 1A output which is 5W of power.(the maximum power rating for the USB device)

This solves producing a regular voltage to the USB device rather than the non-regulated output of the first prototype.

Second I looked up "amplified zener" and found a few things out. But I'm not entirely sure if this is correct layout for the circuit. The NPN transistor that I have selected has a maximum voltage of 60v should I select a NPN transistor that is capable of higher voltages?

Here are the datasheets that I have read and used the example circuit from. I cited them here to double check that I understand correctly.
Zener Diode: http://www.ee.latrobe.edu.au/internal/workshop/store/pdf/1N957-973.pdf
NPN Transistor http://www.ee.latrobe.edu.au/internal/workshop/store/pdf/MJE3055.pdf
LM2575: http://www.micrel.com/_PDF/lm2575.pdf

Also Steve you have helped so much :D
I thank you for that.
 

Mysteryname

Sep 6, 2010
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Sorry Edit wouldn't let me add the Image.
 

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Firstly, note that a switchmode regulator needs an inductor. You have drawn a circuit that doesn't have one.

Secondly, a switchmode regulator needs somewhat more careful layout than a normal linear regulator. As usual, I wouldn't advise one as a first project.

I imagine you got the amplified zener circuit from here. the additional diodes (D11 in your case) are just to trim the voltage -- you won't need them.

A 60V transistor is fine as the shunt regulator will limit the voltage. I would recommend a 36 volt zener as it will clamp the input voltage well below the maximum input voltage for the regulator and will waste very little power.

An MJE3055 is overkill for this application. Something with a max current rating of 1A would be fine, and the power dissipation would be 3W in the absolute worst case (and that assumes an almost fanciful situation where the generator can supply 3W at a higher -- almost open circuit -- voltage).

If you edit, then click "go advanced" you can add images -- at least I'm pretty sure you can.
 

NickS

Apr 6, 2010
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The first DC-DC switcher I designed was with the LM2575. I worked out nice and easy. I attached a pic of the example circuit(circled the inductor). I think you will be ok trying this part under two conditions.

(1) Read the datasheet first(even if most of it is foreign to you you will still pick up some of the information and you will better understand what you are working with)

(2) Use the recommended parts and practices from the Test circuit section. This is essentially mandatory for first time designer.

I agree with Steve on using an amplified zener on the input to present a load in high voltage situations. However I think the 60V peak voltage was just thrown out as a precautionary measure. I could be wrong does anybody know for sure what the peak voltage of his generator is?

Anyway my point is that I suspect it would not take much of a load to pull the voltage back into range for the switcher. To test this theory I would connect my multimeter to the unloaded output with the max hold feature and see just how much you can get it up too. Then I would try again with a 10k 1 watt resistor in parallel with the meter. If that pulls it back into range for your LM2575 then you could simply add a 10mA LED and 10k resistor in series with the BJT in your amplified zener circuit and you would probably never have a problem. But I want to reiterate that that is just a guess too. You should definitely test before you run with it.
 

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However I think the 60V peak voltage was just thrown out as a precautionary measure. I could be wrong does anybody know for sure what the peak voltage of his generator is?

For this model, who knows?

I did a test on one of mine after it kept blowing tail-lamps whilst riding at high speed. I recall the open circuit voltage was well over 100V, but naturally fell significantly (to about 8V) under load. I mentioned 60V because I simply can't remember the peak voltage I measured, and I know this was well within its reach.

I no longer have such a generator to perform the tests on.

A shunt regulator will provide enough load at the worst of times to reduce the peak voltages.

Incidentally, by solution was to replace the rear lamp with flashing LEDs and a regulator, and to use the headlamp as a load to keep the voltage down. A failure of the headlamp would have blown the tail-light sky-high again (I recall I had a zener across the DC rail, but it would not have helped much I expect)

Have these generators been redesigned to improve their load regulation? No Idea -- but I have my suspicions.
 

NickS

Apr 6, 2010
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For this model, who knows?

I did a test on one of mine after it kept blowing tail-lamps whilst riding at high speed. I recall the open circuit voltage was well over 100V, but naturally fell significantly (to about 8V) under load. I mentioned 60V because I simply can't remember the peak voltage I measured, and I know this was well within its reach.

I no longer have such a generator to perform the tests on.

A shunt regulator will provide enough load at the worst of times to reduce the peak voltages.

Incidentally, by solution was to replace the rear lamp with flashing LEDs and a regulator, and to use the headlamp as a load to keep the voltage down. A failure of the headlamp would have blown the tail-light sky-high again (I recall I had a zener across the DC rail, but it would not have helped much I expect)

Have these generators been redesigned to improve their load regulation? No Idea -- but I have my suspicions.

Sorry, I did not realize you actually had one. And I absolutely agree that the unloaded V could be disastrously high. I also agree totally that a precautionary measure should be taken. I was just thinking if he knew exactly how bad the spike could be it might help verify the scale of the solution.
 

Mysteryname

Sep 6, 2010
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Currently I'm at uni so I won't elaborate on the circuit for now.
I should not of forgotten the inductor.

I did some digging for the spec sheet of the Dyno, Sadly it says in the front page.
specs said:
The hub dynamo generates an extremely high voltage. Never touch the connection
terminal of the hub dynamo directly while riding the bicycle or while the wheel is
spinning. Touching the dynamo terminal may cause an electric shock.

So I must watch this element of the dynamo as it does not seem that there is a voltage regulator...

I have a breadboard to test the circuit before I solder it together, the Uni has kindly allowed me to use there equipment to test loads and inputs. So I can test most circumstances in a controlled manner.

-Mystery
 

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That's not sad, that's just a fact.

If you have access to the appropriate equipment, then you should look at the waveform produced open circuit, or into a very small load (say a 10k resistor) at various speeds.

Then have a look at the effect of a shunt regulator. It would be most instructive to be able to determine the power dissipated in a 35V (say) shunt regulator. This represents power lost (and more importantly, additional drag on the bike).
 

Mysteryname

Sep 6, 2010
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I just run some tests with the HUB
No load: the max I managed the get the hub was 17VDC
LEDs as the Load: the highest I managed to get the potential was 10VDC at 12mA
Riding slowly the average voltage was about 7VDC and the current was <12mA

With these results I think I'll reduce the Voltage limiter to a much smaller transistor, this transistor can handle 45v at 100mA which I think is still overkill for the circuit but gives plenty of head room.

With C3 Should I go for a larger capacity or lower capacity? or should I just weigh up the options myself?
High: More drag but may allow the switching reg to stay on for longer
Low: Less drag as the cap will fill up quicker but will it effect the switching reg too much?
If not I think it will be easier to use the lower cap.

New: http://smd.ru/files/upload/1253/ru/bc847_bc547_ser_6.pdf

for the inductor: I know in the diagram that the symbol is a trim pot but the student version will not allow me to import components.

Thank you for all the help
-Mystery.
 

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C3 and C1 should be the same capacitor :) Oh, I don't know, maybe 1000uF

I'm surprised that you don't get a higher voltage from the generator, but it's not a bad thing.

I do wonder though if you're going to get the power you need. If you put the appropriate load on the generator (say, the matching headlamp), what voltage do you see?
 

Resqueline

Jul 31, 2009
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If the BC547 is ever called upon to do any duty in that circuit it will fail. It's way too puny (0.2W). Use at least a 60V 2A 15W device for fail-safe operation.
A hub generator (like any PM generator) is essentially a constant current device (0.45A), with an off-load voltage proportional to rpm. I doubt your measurements are correct.
Your voltage reading is also in DC. Is that measured after a full-bridge rectifier, & w/ or w/o a capacitor?
If you want to know the full-load current you'll have to short the generator with the ammeter. Remember the short-circuit current is the same as the full-load current.
The capacity of C1+C3 is quite irrelevant to drag. You won't ever notice any difference while biking. I'd go for a big cap.
 

NickS

Apr 6, 2010
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Well if your peak voltage really is as low as you say then the BC547 portion of the circuit is not needed(in my opinion) but if you do leave it in you could protect your puny device with a resistor at the collector, after all you are not looking to hard short the rails. You just want the max voltage to come down into the safe range.

Now As for the switcher I have to disagree on a few points. This is your first switcher thus I already stated that it is unwise to take liberties with the suggested parts yet you are doing it anyway.

(1) Input cap. This is an important part of how a switcher works. If you reduce C1 below the recommended 100uF than you will very likely starve the switcher of the high in-rush current it needs. If its too hard to pedal with that cap(C1) in place then the LM2575HV may not be the right part for you.

(2) I disagree that C1 and C3 should be the same cap. Switcher outputs can be very sensitive to their output capacitance(there is a lot of theory behing choosing the right output cap). It may work with other values just fine, but try it as specified first. Also the input cap needs to be rated to a much higher input voltage than the output. So why use a bulkier more expensive output capacitor than needed.

(3) The inductor is key. I don't know what part you plan on buying but I would very highly recommend you use the one they suggest. If you deviate from that part at least make sure the replacement part has equal or better DC resistance and 10% sat current. Of course the same inductance goes without saying.

(4) I see no need for R1 and R2. The LM2575HV comes in a fixed 5V output so you really don't need to buy the adjustable output and set it to 5V.
 

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NickS

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I attached the wrong pic. Here we go.
 

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If the output voltage of the generator is so low (and the warnings that you can get a shock from it suggest otherwise) then there is little point in using a switchmode regulator, you may as well use a linear regulator (and you'd be wise to consider a low-drop-out version.

However, as Resqueline suggests, the readings do not seem right, and need verification.

I take the point about C1 and C3. It also may be that the part specified for C3 is a low ESR capacitor.

I don't think the size of C1 (within reason) will have any significant effect on the drag produced by the generator.
 

Mysteryname

Sep 6, 2010
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Well if your peak voltage really is as low as you say then the BC547 portion of the circuit is not needed(in my opinion) but if you do leave it in you could protect your puny device with a resistor at the collector.

I'll look for one that can handle the amperage. It just comes down to spending the time to look at data sheets.

NickS said:
Now As for the switcher I have to disagree on a few points. This is your first switcher thus I already stated that it is unwise to take liberties with the suggested parts yet you are doing it anyway.

Don't worry, I'll use the parts that are specified for the inductor and the output/input capacitor, I know the inductor has a massive influence of the output of the circuit.


NickS said:
(2) So why use a bulkier more expensive output capacitor than needed.

The output of the switch-reg will not be changed from the specifications. I was mainly concerned with the input of the current revision of the circuit. I have complete trust that the engineers that designed the circuit have selected the correct components for the regulator.

NickS said:
(3) The inductor is key.

This part will not be changed from the specifications

NickS said:
(4) I see no need for R1 and R2. The LM2575HV comes in a fixed 5V output so you really don't need to buy the adjustable output and set it to 5V.

This will come down to availability. I don't have a extensive shop like digikey in Australia. I'll have to buy parts from multiple vendors. and cost will be a huge factor. (RS components only sold the fixed 5v ones in tubes of 250)

If the BC547 is ever called upon to do any duty in that circuit it will fail. It's way too puny (0.2W). Use at least a 60V 2A 15W device for fail-safe operation.
A hub generator (like any PM generator) is essentially a constant current device (0.45A), with an off-load voltage proportional to rpm. I doubt your measurements are correct.
Your voltage reading is also in DC. Is that measured after a full-bridge rectifier, & w/ or w/o a capacitor?
If you want to know the full-load current you'll have to short the generator with the ammeter. Remember the short-circuit current is the same as the full-load current.
The capacity of C1+C3 is quite irrelevant to drag. You won't ever notice any difference while biking. I'd go for a big cap.

Cheers Res, I'll invest in a higher capacity transistor and a reasonable resistive load.
This was after 3m of low ohm speaker cable. full-bridge rectifier over a 5v LED with 280 ohms in series.
When I get home I'll double check those results. (I'll get a friend to spin the wheel while I'll hold the multimeter to a completely empty circuit. I could be looking at a considerable voltage drop over the speaker cable.

I take the point about C1 and C3. It also may be that the part specified for C3 is a low ESR capacitor.

I don't think the size of C1 (within reason) will have any significant effect on the drag produced by the generator.

So only cost will be considered for the cap then.

-Mystery

EDIT: I retested the circuit. with a friend. It produces a constant current of 559mA at any voltage.
So I'll beef up the Voltage limiter to a 1A model when I find it. Then It should be smooth sailing.
 
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Resqueline

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Ah, yes, that seems more like it. But how high did the voltage go off-load, and with a small load?
Remember, with transistors you have something called "secondary breakdown" and "Safe Operating Area".
That means you can't cut the ratings as close as you'd first believe. There's no reason to not use something like a cheap MJ3055.
And how about a heatsink? If the load goes off-line then 560mA * 30V = 17W...
 
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