Help me build a bicycle phone charger

Hello, I recently built-up a front wheel for my bicycle that has a 6v/

3.5w dynamo. These are designed to power headlamps, but I'd prefer to store the electricity in a battery which could charge my phone. I believe iPhones charge on standard USB.

My first question is about the battery. Of course, any cell or chemistry would do, but I'd like to use NiMH AA cells so that I could easily replace them, and use them in other devices like cameras and radios while I'm bike-camping. From what I've read, charging NiMH isn't easy, so I'd prefer to find something off the shelf to use. Is it possible (or advisable) to use a AA charger as a battery while also charging the cells? I'm quite in the dark on this topic, and would appreciate any guidance I can get.

Next, once I have a battery pack that's being charged by my front wheel, how should I go about charging the iPhone. I will likely use an off-the-shelf iPhone car charger which are often little more than cigarette lighter to USB adapters, but these are made for 12v, so I assume their circuits will be mostly useless to me. What would be involved in creating a USB outlet on my battery pack?

Once those two fundamental questions are answered I'd like to dream a little; It'd be most awesome to have a loud speaker built into the battery pack so that the iPhone could also bump tunes when I'm riding with friends. Assuming I can somehow re-purpose an OTS iPhone dock, I should have access to some audio wires, and run them to a volume dimmer and then to a small speaker. A headlight would be handy too, but I'm not sure that there will be enough juice for all that. Just wanted to throw that out there so y'all will know where I'm going.

And just to put minds at ease, I'll use a good sturdy protector case to house it while it's mounted on the bike. I don't want it to get hurt in an accident, but it'll be handy to have maps available on my handlebars.

This is my first serious electronics project, so I'm all ears.

Reply to
BikePilgrim
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USB gives 5V, so you can probably just plug the iPhone in (with the right cable, of course) to the dynamo itself. The iPhone is almost certainly set up to handle higher voltages during charging. If you are worried, you can take your car to iPhone charger, and see if it is regulated down to 5V. You could also regulate the output of the dynamo down to 5V yourself using a low dropout 5V regulator.

NiMH cells are pretty easy to charge at low current. Just limit the current to 1/10 of the rating of the battery, and you should be fine. They can overheat if they get overcharged, leading to shorter life. However, they won't burst into flames like a LiION pack. However, why not just use the iPhone itself to handle charge management, by plugging the thing directly into your dynamo. That way, you can play it while you ride, and it'll charge when the voltage is high enough.

You'll need a battery pack for this one. You can buy batteries that fit into a bottle holder. Here is one:

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You can build a nice little amplifier for it using an LM384 or something like that. Google for LM384 datasheet, and scroll down. There are schematics for an amp that would work for you. You could probably just tear the OTS iPhone dock apart to get the speakers and plugs.

Reply to
Bob Monsen

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Before you do *anything*, flip the bike over and connect the hub 'dynamo' to an appropriate bulb. Make sure there is *no* way the bike can fall over then spin the wheel fast enough to light the bulb and using a multimeter on DC volts, check the voltage from the 'dynamo' then repeat on AC Volts. I strongly suspect that your 'dynamo' is actually a permanent magnet alternator and outputs AC!

It would also be worth checking the output at the maximum speed you are likely to do with no load on the 'dynamo' (a bike computer to check the speed and a drill with a foam sanding drum with no sandpaper on the outside of it to drive the wheel by pressing it agents the tyre may be some help here). I wouldn't be surprised to see way over 12V on open circuit, downhill conditions.

OTOH maybe Bob is trying to do us all a public service by incinerating your phone on the basis that cyclists operating iPhones while riding although a self correcting problem from a Practical Darwinist perspective, is likely to result in considerable expense to the rest of us in terms of loss of no claims discount and visual clutter consisting of tacky roadside 'shrines'.

--
Ian Malcolm.   London, ENGLAND.  (NEWSGROUP REPLY PREFERRED)
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk
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Reply to
Ian Malcolm

You've blown my cover! :)

Thanks, I was assuming DC. Sorry to the OP.

Regards, Bob Monsen

Reply to
Bob Monsen

If it were me, I'd start up front with efficiency. Forget the I phone car charger since the people that make them may not have efficiency at heart - after all they have an auto to do the peddling and it has a

300+ watt alternator.

Any watts you devote to toys will have to come out of your legs - you produce maybe 50-70 watts with sweat.

I think I'd charge a NiMH battery since they are cheap and easy to deal with and don't explode as easily as lithium.

I'd find the OC voltage of the alternator, and charge a cap to its full voltage then use a high efficiency buck converter to lower the voltage for battery charging and perhaps even run the headlight from the battery pack and put an emergency switch to run the light direct when the battery is discharged.

"Bump tunes when riding with friends?" I wouldn't ride with anyone who takes their music. Weather radio or walkie talkie, maybe, lights for the camp, certainly. You could always get a Winnebago.

Reply to
default

Well, most bicycle lights are powered from AC (and my dynamo measured about 15V with 14 ohms output impedance, but that's variable, the bicycle speed has a major effect). Your iPhone wants 5 Vdc, regulated (i.e. 4.8 to 5.2V allowable range) and with up to 500 mA required.

So, first you have to convert AC (chassis ground) to DC; then you need to use that DC to effectively charge a battery. Then you need to tap regulated power from the battery source (the battery is a very good filter, much safer to get power after the battery than upstream straight from the generator).

I did something similar with a voltage doubler rectifier (capacitor/ two diodes), a lead-acid 6V battery, using the internal impedance of the dynamo as the only current limit. Then a low-dropout regulator set at 5.8V actually ran my headlamp. The linear regulator was a power-wasting component, but it didn't waste much (mainly just kept the AC peaks out of the vulnerable filament).

You'll want to use a switchable 5V regulator of some sort, so it doesn't drain the battery when you aren't docked to the iPod. And you'll want to verify that the batteries you use (NiMH) can reach 5V (it'll probably take five or more cells in series). A standard USB socket (or a pigtail lead with a USB socket) will connect to the standard iPod cables for charging.

Reply to
whit3rd

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Reply to
doudoune78doudou

are you guys saying that it is totally fine to just go ahead and wire up a usb cable to a 6v3w dynamo and use it to recharge an iphone? i have been looking into different projects like this and have not found definitive answer.

thanks in advance

Reply to
mikeinternet

No, don't do that. The consensus is that your dynamo is probably outputting AC, which wouldn't be good for the iPod

Instead, use a 7805 voltage regulator, along with a bridge rectifier and a capacitor, like this: (view with courier font) ____ | | .-----------o-------o------|7805|---o-------- iPod Vcc | | | |____| | | | | | | Shottky - Shottky - | | | ^ ^ |+ | | | _ | === | --- | / \\ | /-\\ | --- o---(~ ~)---o | | | | \\_/ | | 2200uF | | 0.1uF | | | | | Shottky - Shottky - | | | ^ ^ | | | | | | | | '-----------o-------o--------o------o-------- iPod GND (created by AACircuit v1.28.6 beta 04/19/05

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This will limit the input to 5V, and rectify the circuit. The thing in the middle of the bridge is the output from your dynamo.

The diodes are all pointing up (ie, line is at the top). You could use Shottky diodes like 1N5817s, or silicon diodes like 1N4001s. The shottky diodes will waste less power, and let you get more juice out of the thing. You can also get integrated bridge chips, that have 4 terminals, and incorporate the 4 diodes. They are generally marked with two terminals that have a ~ marking, and two that have + and -, respectively. If you get one of these, hook the dynamo to the ~ terminals, and hook the + to the 7805 input, and the - to the GND input.

Sorry about the misinformation earlier; I was thinking DC rather than AC. Please test this circuit with an ohmmeter before you try it on your iPod...

Regards, Bob Monsen

Reply to
Bob Monsen

They will easily deliver 12V RMS at more than about 10kph. The only reason they cannot legally be used to drive a 12V 6W light on a bicycle is that there is a minimum speed that the light has to be up to full brightness. (And I think 6W is too bright for the main light on a bicycle in the UK which is (or used to be) limited to a maximum of 3W to prevent dazzling of other road users)

You probably don't want to use a voltage regulator because you'll be having to output the full 6W or so of power when cycling once the battery is fully charged just to heat up the voltage regulator.

I've considered a bridge charging a cap and then a DC-DC converter in order to get a constant voltage source. If the front light blows it gets very expensive as that often means the end of the front standlight and rear light. Currently I'm using two zeners to clamp the voltage to about 9V (based on 6V RMS having a P-P voltage of about 9V but even with that I've destroyed a front standlight (but the rear light survived) when the front bulb went.

Tim.

Reply to
google

Actually, no, this won't happen, because the regulator will simply cut off the current drawn from the dynamo when the battery is charged (ie, the iPod no longer draws any current.) So, it'll get easier, not harder.

If you do what you suggest here, then the diodes will conduct to ground whenever the voltage gets above 9V. The diodes will be taking a beating. A DC-DC converter makes much more sense here. You could also use the linear regular circuit above with an LM7809 and a couple of 9V DC bulbs.

An LM7805 will accept inputs up to about 30V. If you use a bridge, like the simple circuit above, and the input is 12VAC RMS, then the max voltage it'll see is 12 * sqrt(2) = about 17V, minus the diode drops, which depend on what diodes are used. The linear regulator and capacitor form a buffer which will protect your bulbs from transient spikes in voltage (up to 30V).

Regarding the iPod, after looking at the apple site, they appear to charge the batteries using a fast charge scheme, requiring 1A initially. I'm not sure how this works with USB, which is only supposed to allow 500mA maximum.

So, with a linear regulator, the iPod will charge at 1A at 12V RMS, meaning it'll suck 12W or more out of the OPs legs when charging. Also, it means that the linear regulator will dissipate about 7W, which will get it quite hot unless it is well ventilated (lots of opportunity for that on a bike, I guess!)

On the other hand, a DC DC converter will let him charge his ipod while supplying only about about 6W, i.e., 1/2 the muscle power. Now, since you say that one only gets 12VAC when going over 10kph, it may not be an issue. Not sure.

The main tradeoff is whether the 6W at high speed is worth learning about and debugging a DC-DC converter.

Regards, Bob Monsen

Reply to
Bob Monsen

The diodes shouldn't be conducting anything in normal usage. I've got a 2W4 bulb and a 0W6 rear light being driven from a 500mA constant current source. So I should be getting 6V RMS and I should just be under the (approx) 9V zener clamp. (Actually I'm using a 3W front bulb because the 2W4 bulbs blow much too quickly)

The problem is when the front bulb fails. The hub still want to supply

500mA and so the voltage shoots right up. Unsurprisingly, the bulb tends to go when you're doing 50kph and, on streetlit roads it can be easy to miss the light going anyway. At around 80GBP to replace the front and rear lights I don't really care about hammering the zeners occasionally ;-)

I don't know how high the voltage will go. But driving 2x 6V 3W lights in series is something that many cyclists with these hubs do. So 12V RMS at normal cycling speeds (15-20mph) is easily achievable. Depending on hills you need to consider the open circuit voltage at

30, 40 or even 50mph if you are going to be connecting something like an ipod

No. The hub will only supply 500mA. The voltage may rise to 18 or even

24V RMS open circuit. But if you drive a 12Ohm resistive load then you'll get 6V RMS. (The frequency also changes with speed so inductive loads are harder to analyse)

If you want 6W at 6V then you've got to use a DC-DC converter (or a transformer). You'll draw 12V 500mA from the hub and then down convert to give 6V 1A.

It's not really high speed. It's just that to be road legal the front light has to be fully lit at little more than walking pace. If you're prepared to accept your lights aren't fully lit until you are at a speed where it's possible to balance without weaving all over the road then you can get twice the voltage out of them (but no more current)

Tim.

Reply to
google

The following may be applicable - or I may have missed something in the thread.

Rectify, pre-regulate to ~30V with a zener & hefty pass transistor. Then whatever VR chips to provide the needed outputs. Conceptually:

+---+---+--- ---+---+---[78xx]--- |+ | | \\_/ | | | AC==BRIDGE [C] [R] | [C] | Gnd |- | | | | | | | +----+ | +---[78xx]--- | | | | | | | | | [C] [Zd] | | Gnd | | | | | | Gnd -----+---+---+----+----+---+

You could change it a little to make a shunt pre-regulator if desired.

Feature you can add: You can use the DC to energize a relay when the DC exceeds X volts. With the relay de-energized, the lights can be connected to the dynamo or a battery or disconnected or whatever. When the DC voltage is high enough, the lights can be connected to a VR output.

DC + ---+---[Relay]---+ | | P --- O

Reply to
ehsjr

You probably won't get 30V from the hub if you're trying to draw the full 500mA.

Unfortunately I don't know what you will get and it's hard to measure. Ideally I'd like to strap an oscilloscope onto the handlebars and then ride up and down the road trying different (resistive) loads but I don't have a portable oscilloscope. Another thing I've considered is using two (upside down) bikes, taking the rear tyre off one and the front tyre off the dynamo wheel and then putting a belt between them so that I can try to drive the dynamo at a constant speed using the pedals of the other bike. Maybe a project for a nice summers day.

The power drawn from the hub is 500mA at whatever voltage is necessary to drive 500mA through the load (but with some peak voltage that I've never tried to measure but is definitely in excess of 12V RMS at normal cycling speeds)

That power comes directly from your pedalling and slows you down which means you want it to be as low as possible. (The efficiency might change with the power drawn - again I have no data)

If there is no current drawn at all from the dynamo then there is (effectively) no drag

Some graphs here:

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I'd guess that the "Generator Drag" one is based on a 6V-3W load

There is also some graphs here:

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Here's what I'm looking for:

  1. 100% efficiency :-)
  2. zero weight.
  3. 6V voltage clamped output that will be "on" at lower voltages if that is all that is available at the source (I'd accept it turning on at 6V if the next output (below) is still on at lower speeds)
  4. Second low current (50mA?) 3V output with a largish capacitor so it can continue to provide output for a couple of minutes after stopping. (This isn't critical because there are lights that already provide a standlight facility)
  5. third 6V regulated output that only becomes live once the first output is delivering 6V and even if the third output is short circuited will not drag down the first outputs voltage. (I don't care if a short circuit of output 1 drags the third output down but the system must be able to withstand shorts like this)
  6. The whole thing robust against severe vibration.
  7. Ability to withstand all the outputs going open circuit while the input is being driven by the hub at at least 30mph for an arbitrarily long time. I'd guess, but I don't know, that the hubs voltage is probably clamped at no more than 50V p-p for safety reasons
  8. (not critical) Some sort of LED status output that alerts if any output goes open circuit.

If you can supply that lot then I think the OP's original iPod charger becomes a fairly trivial exercise.

Tim.

Reply to
google

Not really all that hard, I think:

____ | | .------o----o----o------|7805|---o-------- iPod Vcc | | | | |____| | | | | | | | | | | | | | +-+----' z | |+ | | A A 24V A | === | --- IN1 -+ | | | /-\\ | --- IN2 -(-+ LED? | | | | A A | | | 2200uF | | 0.1uF +-+----. | |/ | | | | '--|NPN | | | | |> | | | | | | | | '-----------o----o--------o------o-------- iPod GND

Use a Power NPN Transistor with Heat Sink

(created by AACircuit v1.28.6 beta 04/19/05

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Or, you can use the relay circuit above to cut off the input (not nice at night, I think)

For 100% efficiency, you need the alien technology from Roswell... or those new carbon sheets of thickness 1 molecule...

Instead of the 7805, you could use a DC-DC converter to get slightly better performance. However, you are mostly going to be toodling around with the thing at 9V or less, so this will be fine. In fact, a low dropout regulator might be better.

Regards, Bob Monsen

Reply to
Bob Monsen

If you drive the voltage regulator from a constant current source and draw the full current from the output then what is the input voltage of the regulator? Is it just Vout plus the minimum voltage drop of the regulator or is it higher?

------

500mA>----| 7805 |---+ ------ | | 10R | | ------------+------+

What happens if the output isn't quite drawing the full input current? (replace the 10R with 12R) In this case I'd expect the input voltage to rise as high as the source will go which means we're going to be wasting a lot of our leg power in the regulator.

What about if we're trying to draw more current? (Replace 10R with

8R2)

While an overvolt protection circuit for the lights is nice (and I'm using a couple of zeners to clamp the AC voltage) what I really want is to be able to draw 6V 3W to drive my lights and additionally, when available because I'm going faster, another 6V 3W to do something else with - whether that be drive a second light, charge some batteries etc.

Tim.

Reply to
google

That's good. The circuit above essentially conducts from very low voltage up to about 30 volts DC from the source to the inputs of the 78xx regulators.

The reason to pre-regulate was to ensure that the regulators' input voltage was kept to ~30 volts or less.

Don't want much, do you? Let's see, 100% efficiency,

0 weight ... I give up. Those things are available in the Twlight Zone ... or not. :-)

To play the game, the specs have to be properly stated. The voltages alone don't suffice, we need to know the current that will be drawn from your supplies. You did not say anything about the *second* 6V supply, you only mentioned a first and third 6V supply. And you did not specify the current to be drawn from the two

6V supplies. We also need to know what input power is available, at what speeds.

The best we can do at this point is a pre-regulator feeding a DC-DC converter or linear regulator. Bob gave you a shunt pre-regulator (which would benefit from an added dropping resistor to protect the shunt transistor) which is nice, I gave you a pass pre-regulator, which would also benefit from a dropping resistor (since you insist on 500 ma CC) to protect the diodes and the source. The dropping resistor will waste power, as will the linear that follows. A wide input range DC-DC converter would avoid at least some of that waste.

But playing the game of finding a "wide input range" DC-DC converter without knowing the range is not something many people want to do. You'll just come back with "what about x" .

So I'm gonna stick with my offering of a pre-regulator followed by linears, and whatever warts it carries with it. :-)

Ed

Reply to
ehsjr

OK. But assuming I'm understanding the examples given they're likely to get worse than 40% efficiency for a single 500mA supply.

Maybe there's a typo above that I can't see. But the first supply is to drive the lights (500mA at 6V). The second supply is to drive a standlight (50mA at 3V). And the third supply is any power left.

Maybe I don't understand the circuits well enough (very probably) but the hub will quite happily supply 6V at 500mA or 12V at 500mA. We have

5V at 500mA of useful power. If the hub is supplying 6V then we've got an acceptable 80% efficiency. If the hub is supplying 12V then we've got an unacceptable 40% efficiency. I don't understand why any of the circuits will keep things in that 80% efficiency zone.

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Has some graphs showing output voltage under no load, a 6V 3W bulb and a 6V3W bulb in series with a 6V2W4 bulb.

Just driving my lights, what I need is overvoltage protection. Currently I'm using two 8V2 zeners to clamp the p-p voltage at about

9V. In normal usage they should (approximately) not conduct at all so 100% efficient. If the front bulb goes open circuit then I get a 9V square wave driving the rear light which, so far, has meant that I've not destroyed a rear light in about 2 years after having gone through three in a year. I still have a problem with the front light standlight (which has it's own overvoltage protection but is only designed to be enough for you to stop and change the bulb - I suspect that the front light overvoltage protection is doing all of the work and the front standlight electronics are destroyed before I notice the bulb has gone)

This is the front light:

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Rear light:
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And the weight thing is a bit of a joke - but I've just ordered a new bike and spent an extra 400GBP to have titanium parts that will save me 915g.

Tim.

Reply to
google

Pre-regulator, followed by linear voltage regulator.

Ed

Reply to
ehsjr

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