Charging Circuit Design

Hello all,

I had some general questions about a project I'm working on for a senior design class.

I found a schematic for a battery charging circuit and modified it somewhat to fit my purposes.



First, I wanted to run this by someone to make sure it makes sense.

In essence, a solar panel will act as an input and be split to 2 LM317s for voltage regulation. On the left, the voltage will power the laser while the panel receives sun, whereas the right regulator will act as a power source to charge a battery.

When the panel is gone, the battery will discharge through the laser to make it operate beyond what the panel alone can do.

First, our design specs call for the battery to charge in about 4 hours while we obtain sun. Is this even possible?

Second, the battery we chose is a rechargeable 9V Duracell. I believe I can modify this by just tweaking the outputs of the regulators if need be.

However, a problem I've faced is that after allowing the battery to discharge (in order to obtain an idea of operational lifetime), it does not charge. I can attach it to my circuit and not read any appreciable change in voltage.

Furthermore, when attaching the battery to a meter, the voltage constantly increases, so I can never be sure what the actual voltage is. I believe the meter is somehow charging the battery, but I'm not sure...

At this point, I'd appreciate any suggestions, ideas, critiques, or anything else you can let me know. I left out a lot of information, so also feel free to ask for anything else you need to know.

Thanks.

That looks wrong to me.

The idea is good but there s a serious flaw in the electrical design, it destroys Q1 and possibly the laser diode if you disconnect the 9V battery or adjust R4 wrongly. Did you intend to connect the collector of Q1 to the ADJ pin rather than the output pin of the regulator?

You don't state what kind of laser diode it is. Some have built-in current regulators some don't. If U1 is used to set the current, consider connecting the cathode of D4 to the cathode of D5 so both sources feed U1. At the moment there is nothing to set the current when running from the 9V battery.

Brian.

The zener diode is 9.1 if its a C its +- 1%, D +-2% else +- 10%, so adding in the Vbe of .8V, give a minimum Q1 switch on of 9.0V. I would look very carefully at the voltage across the zener. I think it would be more cunning to re-wire the circuit so as to sense the current on the INPUT of the regulator using the 50ohms, and use the transistor to pull-up the sense point to shut the regulator down. This way the regulator can deliver max current at its set voltage, and there is no fiddle factor about the volt drop across the 50 ohms.
Frank

Currently, U1 simply serves as a way to provide a constant voltage from our fluctuating input, but using it as a current source is an idea I thought about and dismissed. I will look into it again.

Concerning the laser, the spec sheet I got was very minimal and simply said to operate it between 2.8-4.0 V and 50-65 mA. In actual testing that I've done with it, however, the diode operates around 13-17 mA almost consistently, regardless of how I've used it. I also get a jump from 2.2 V up to about 4.2 V as I scale up the voltage across it, which doesn't allow me in its operating range. It gets brighter with this jump, so I assume it's another portion of the diode becoming fully operational.

Also, Frank, your idea sounds good, but I am not entirely familiar with the process you're suggesting. Could you give me some more information about what that involves?

Thanks again.

The wide voltage range of the laser makes me think it has an in-built regulator. The laser operation only takes place over a very narrow current range so to be specified so wide suggests something inside it giving further regulation. Quite often a laser diode will have a light sensor inside it that stabilizes the light output by controlling the current it draws.

I think Frank is suggesting the same as me, the Zener feedback should connect to the ADJ pin so it can set the output voltage of the regulator, not the output pin where all it does is sink current.

Brian.