Dynamo Charger Kit

scraynes

Über Member
Location
Jersey
Hello,

I'm wanting to produce a box that will take the output from my dynamo hub and use it to charge 2 AA batteries or something via USB.

I understand that the hub output is about 6v AC and I will need 5V DC for charging purposes.

I've found this little kit at maplin that looks like it might do the trick.

Just need to add an AA case and a USB socket, put them both in pararel from the output and only use it to charge via the USB or the AA case.

I know that the circuit will need to be set to provide the 5v output which is fine, I'm just wondering if the circuit will provide 5v output from a 6v input. Is the split between the two volatges large enough?

I've found another circuit that uses a difference regulator, which will work, but I doesn't come as a kit- and I prefer that idea.

Anyone out there got any views?


Thanks,

Spencer
 

WJHall

Über Member
The spec sheet for the LM317 (1) referred to suggests a drop out voltage of about 2 V, which implies that you could have difficulties. Also the 6 V charger example in the spec sheet shows an input voltage of 9 V. The resistor alleged to set the current in the spec sheet example seems to be absent from the circuit diagram for the K1823.

The other thing that you need to think about is that the dynamo output voltage will only be about 6 V if it has regulator diodes, or is working into a 12 Ohm load. An open circuit dynamo can give about 50 V rms if you go fast enough. The voltage across a load is determined by the division of this emf between the internal resistance and inductance of the dynamo and the external load. Because the impedance of the inductance rises with frequency there is a natural regulating effect, which keeps the load voltage at about 6 V for a 12 Ohm load.

For regulated dynamos, some manufacturers seems to prefer you not to run them open circuit, presumably because all the current generated by the higher voltage is being dissipated in the regulator diodes.

To estimate what the input voltage to the K1823 might be, you need to find out its input impedance. In the charger example shown in the spec sheet, it could be delivering 1 A at 6 V so would be drawing about 0.7 A at 9 V, which happens to be about 13 Ohm, which implies that a dynamo would actually produce about 6 V, so the necessary 8 or 9 V would not be available ... I think.


(1) www.national.com/ds/LM/LM117.pdf
 
OP
scraynes

scraynes

Über Member
Location
Jersey
Thanks for the quick reply.

I was thinking that the drop would be too small to provide the required 5V.

I think I'll just use the following circuit and make it up on stripeboard. Not as neat and tidy, but at least it should work.

I'm on the mainland this weekend - so will get the parts then!!
 

WJHall

Über Member
On the other hand, if you do not require 1 A out, which you presumably do not, then the input impedance ought to rise, corresponding to drawing less current, giving you a higher input voltage... perhaps.

But I still suspect that a device intended for use fed from a mains transformer may not suitable.

Interested to hear how you get on.
 

WJHall

Über Member
It is an interesting question whether reducing the current drawn will raise the input voltage enough to get over the drop out voltage and the bridge rectifier diode problems. There is a published circuit for charging a battery (2) and working the lights but this uses a doubling rectifier. However he is taking the full generator load, so that might not be necessary at a lower load, if you could cut the current drawn to 0.25 A, in principle the input impedance might rise so that the generator feeds 0.25 A into 12 V.

The use of the doubling rectifier also reminds us about the generator earthing issue, the K1823 has bridge rectifier input so may need both input sides insulated from earth.

(2) http://www.gobike.org/feat_lights.php
 

WJHall

Über Member
Acutally, since your load is not earthed the earthing question may not be relevant.

Another way of looking at it may be that if you put the batteries in parallel then you may only need about 1.5 V to charge them, which might just get you within the voltage drop specification, bearing in mind the drop over the bridge rectifier.

You are probably still better adding the current control components as in the spec sheet, which may improve the charging characteristics, although these will not be optimum so your battery life may be affected.
 
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