My simple circuit does not seem to work

Hi,

I am working on a project to charge my cell phone while I am riding my bike. I have a small bike dynamo that outputs 15 vac at about 6w (obviously very dependent on speed). This is fed into a full bridge rectifier (4-diodes), then through a shunt capacitor to smooth the voltage. this is then fed through a voltage regulator (LM 309). This generates 5vdc which is fed into the power pins of a usb port (the data pins are shorted together to comply with usb charging standards). Unfortunately when I plug my phone in it reads that it is charging, but does not actually charge.

I feel that this is because the output current is too low. I get a reading of about 350 mA and I think 500 is required to charge. I was thinking about putting a 2:1 transformer before the rectifier to boost the current and drop the voltage, and not burn off so much energy in the rectifier. I thought this might work:
**broken link removed**, but I am not sure I am even on the right track.

If someone could give me their opinion that would be great.

i know that I probably should be going with something more efficient, but for a switched rectifier you need to design the controller, and I was hoping to just keep it simple.

Hi,

the transformer might help if the current is not enough, because twice the power is burned with the LM309(15V to 5V). Have you measured the actual current when the mobile phone is charged on a normal USB port? Maybe the mobile phone instruction manual give you also the current need. It could be better to checkout the power generator with a load resistor(10 Ohm) first to see if volage and current is o.k., before it is connected to the mobile phone.

Sounds like a good idea to me. The lowest voltage characterized on the datasheet for the LM309 is 7.4 volts input.

John

Transformers work in a given frequency range. The bike generator is an alternator: the frequency should change with speed becoming low for low speed.
If the reactive part of the impedence of the primary is too low the whole voltage is applied to the winding resistance that could be damaged. Also the generator is driven to work close to a short circuit.
It is true that for low speed you have both low frequency and low voltage, but you should be careful not to damage the dynamo bike or the transformer.

These are General Purpose Pulse Transformers
not suitable for near sinusoidal voltage about 20...200 Hz
from bicycle-generator .

First thing to test is to try bigger capacitance before
regulator IC.

Second test, find out the real output power with
different speeds .
Try to find out , what is the load resistance ,
when output current is 0.5A and what is the voltage
when you get 0.5A out from dynamo.

Now you can see it is possible to take out
0.5A and needed voltage = about
5V+ 2x0.8V+2V = 9Vmin.
(regulator output + 2x diode volt. drop + regulator volt. drop )

I should have posted some more details in the original post I'm sorry. I have a large capacitor on the output (~5mF) I knew that usb voltages had to be very steady. Second the 350 I mentioned in the post came from a multimeter first, and then was verified by an app that I downloaded which showed charging voltage. The frequency response was really my point of concern. I really had no idea of the frequency put out because i had no idea of the number of poles in the dynamo. Does anybody have other sugestions as to other things I could do to boost voltage besides a transformer? At frequencies that low in this power range it is going to be hard to find a transformer that will work...I think.

How about using a switching regulator for improved efficiency from the rectified AC? There are numerous ones on the market. Here is one example I had in my files: https://www.onsemi.com/pub_link/Collateral/MC34167-D.PDF

John

That is a wonderful idea. Sometimes when I attack a project I have a tendency to try to reinvent the wheel instead of just using readily available rebuilt components. thanks john.

You could charge a big cap (1mF / 25V) with the rectified output of your dinamo. Then you could use a switching regulator that has a voltage sense input, so when the voltage is over, by example, 20V, the switching regulator starts to work and gives 5V output at full rated current until the voltage of the capacitor gets discharged. This way you can improve a lot the efficiency of your system. Take a look at the Analog Devices website, also at Linear, they have lots of this kind of ICs.