# What's the relationship between volts and amperes in batteries?

I recently bought a bag of 100 5mw 4.5V laser diodes (the spec said they have "driver: out"). My idea was to connect two and two of these in series to get something that needs 9V, and then connect about 70 of these diodes in parallel and run it off a 9V battery.

This didn't work as expected.

Just using 6-8 diodes, the light was already getting visibly dimmer. I figured there could be a short somewhere, but it seemed everything was in order. The 9V battery was used, but only for driving a few LEDs for a few hours.

So I decided to try something else. By using a battery pack of 6 1.5V batteries, the laser diodes were shining at full force again. I tried using about 20 of the diodes in the same setup as above, and still everything seemed fine.

Thus it seems I'm stuck with AA batteries for my setup.

Anyway, my questions are:

1. Why is this happening?

2. Are there not enough amperes to drive the lasers?

3. Why is the battery pack able to drive the lasers when the 9V battery isn't?

4. Is there any way I can calculate the approximate number of lasers the batteries can drive?

5. Will an amplifier or extra driver be able to fix this so I can still use the 9V battery?

Bonus question: If the answer to 5 is "yes", can I find such an item on eBay? I just need an example of what to look for since I'm just a hobby electrician.

EDIT:

Since you good folks asked for it, here's all the data I have regarding the laser diodes:

http://www.ebay.com/itm/110887381932

EDIT 2:

A brand new 9V battery was able to drive the diode arrangement after all. There appears to be a small resistor on the diode board, so I took my chances that the diodes would not blow just using them as described above (and I had spent way too much time on this Halloween cannon already to bother soldering 40 or so more things into the setup). Last night I used the cannon for a full entire night, and the lasers were still jolly good by the end of the evening.

Plus the cannon looked damn good in a dark nightclub with some artificial fog :)

• A 9V PP3 is really 6 tiny cells so 6 x AA will deliver a lot more current. To get a good answer on 2-4 you'd need a datasheet or at least a link to the laser and specific batteries used. Answer to number 5 is a DC-DC converter may possibly help a little, depending on the above, but probably not a lot. – PeterJ Oct 23 '13 at 10:24

First thing to say is that whenever using laser diodes (or LEDs for that matter) you need to limit the current into the device. You can't rely on the battery voltage for producing exactly 4.5V x 2 because if it produces 4.6V x 2 then you might find that the current into the laser diodes (2 in series) would go from 50mA to over 100mA or much higher leading to destruction of the laser diode.

Like LEDs, laser diodes have a typical forward volt drop (4.5V in your case) at the rated current (say 50mA) and some may be only 4.4V and if you had a bunch of these they may well have died and you'd be regretting spending the money so use a series resistor or an active current control mechanism.

Without seeing the data sheet I can imagine that full 5mW laser output is achieved with round about 50mA of current and that your battery arrangement has just enough internal series resistance to limit the current to safe levels.

If a pair of lasers in series took 50mA then 4 pairs would be taking 200mA and already a PP3 battery would start to reduce its output voltage and dim the lasers. Look at the discharge curve for an energizer 522 9V battery: -

With 200mA load after 2 hours the voltage will be at 4.8V. You could estimate that after 40 minutes it would be 7.6V and well below the 2 x 4.5V needed for the laser diode. You could estimate that each minute that passes the battery terminal voltage drops 35mV and after 2 hours this drop is 4.2V hence only 4.8V left at the terminals!!

In fact, if the laser diodes were exactly 4.5V then after a few minutes there would be not enough voltage to activate them. And this was a battery that had been used to drive a few LEDs for a few hours!

The battery pack made from several 1.5V batteries can generate much more current for a longer time period. All the same, if I was using a battery pack I'd make it 6V and have individual current limit resistors for lasers i.e. don't wire in series.

To get more out of the 9V battery (and you won't get a whole lot more) you could drive each one individually from the 9V with a current limit resistor. You need to find out what the forward current of the laser is. I can guess at 50mA and this would mean a dropper resistor would be: -

$\dfrac{9V - 4.5V}{50mA} = 90 ohms$

• As you said hard to know without seeing a datasheet, but I've seen a lot of things sold on e-bay etc as 3V or 4.5V laser diodes and they also include a current limit resistor so it's not really the forward voltage drop. Connecting two in series probably isn't helping in that regard either if that's the case. – PeterJ Oct 23 '13 at 10:48
• @PeterJ I am just putting up an answer to help the OP understand what he has to do. Without data sheets on the laser diode (and I've never seen anything of any substance on ebay items sold for that matter) it's just guess work. – Andy aka Oct 23 '13 at 10:56
• Not arguing with your answer in any way - was just commenting a lot of things sold on e-bay and similar discount stores described as laser diodes aren't really what we'd consider diodes, more laser diode assemblies. – PeterJ Oct 23 '13 at 11:03
• Thanks for an elaborate answer. I've updated the question with all the information I have regarding the diodes. Hope this coincides with your interpretation of the problem. – Pedery Oct 24 '13 at 8:01
• @Pedery I checked the link and searched on the manufacturer's website but they had no more information other than it can be powered from 3 series AA or AAA batteries. This probably reinforces the general feeling that running from a PP3 style battery is going to drain it quickly (as you have seen). Neither could I find any mention of how much current the device took! – Andy aka Oct 24 '13 at 8:13

A battery is a collection of cells; the voltage of each cell is determined my its chemistry, plus a small factor dependant on how fully charged it is.

Current delivery is limited by the internal resistance of the battery. This depends mostly on the surface area of the cell electrodes, which is why larger battery packs are inevitably needed for high current applications.

You can measure the internal resistance of a battery by looking at how the voltage drops for an increased current draw then applying Ohm's law. That would enable you to answer (4).

70 diodes drawing 50ma each would be 3.5 amps, which is quite a lot. For that level of current I would look at R/C battery packs; "3S" ones have a nominal voltage of 11.1V. This is slightly above the rated voltage of the LEDs, so add a 47 or 36 ohm 1/4 watt resistor in series with each pair of laser diodes.

• Thanks for an elaborate answer. I've updated the question with all the information I have regarding the diodes. Hope this coincides with your interpretation of the problem. – Pedery Oct 24 '13 at 8:02

An ideal battery would have 9V output, and would keep this 9V output with 1, 10, 100 lamps (or laser diodes) for 1-2-10-1000 hours and forever.

Unfortunately, real world batteries are different.

• each battery comes with some series resistance - so as you add more and more lamps (or lasers) more and more power is wasted on this series resistance and your lamps will glow dimmed

• each battery works on a chemical reaction, and because there is no endless source of "fuel" in the battery, the battery will eventually drain out. Depending on the load, this drain-out time can be acceptable (for example, a garage door opener can run on a battery for 2 years), but sometimes unacceptable quick.

It is very hard to find exact numbers about consumer grade batteries. The rule of thumb is that an 1.5V AA cell is 1500mAH. This very roughly means that it can run a load consuming 1500mA for one hour (this is theory), or can run a 150mA load for 10 hours, or can run a 15mA load for 100 hours.

Consumer batteries are designed and optimized for specific loads. For example the 9V battery is optimized for low power consumtion things such as transistor radios, and works best if the load is 15mA or less. If you load it more, the relation above won't hold. If you want to get out 1000mA current, it won't happen for too long.

When picking batteries, look for the typical drain - your circuit shall draw not more than this if you want to stay in the safe area. Then depending on your target run time, pick a mAh which matches. For example, a 9V battery can serve up to 15mA normally, and it is 500mAh, so you can get a 33 hour run time.

Note that better chemistries will give much better result. Alkaline, Lithium, etc. makes a difference here, but in consumer grade batteries it is very hard to get real numbers behind the marketing bullshit. Therefore it is common to use industrial batteries with Lithium Ion or Lithium Polymer chemistry, which give you a lightyear better results, and come with straight and detailed documentation about the real numbers.

• A bit of necromancy, but thanks for the elaborate answer :) – Pedery Mar 10 '16 at 17:54
• OMG :) True, I didn't check the timestamp. StackExchange just shown me your question, probably because it has no accepted answer. Anyhow, it was a nice experience to practice my non-native English :) – Gee Bee Mar 11 '16 at 0:31
• Really? Your English is excellent! As you see above, I did manage to string the project together, and it works like a charm. Of course I cannot power 72 lasers for a long time on a single 9V battery, but it's used as a gun and works nicely for an entire night. Plus it kicks ass on Halloween ;) – Pedery Mar 11 '16 at 13:37
• Haha, thanks! Just a moment ago I had to look up a word :) Learning English is a full life-time challenge... Nice lasergun project, I loved the picture too! Keep up the good work! I hope next time I can really give a helpful answer. :) – Gee Bee Mar 11 '16 at 17:27