I've never designed a battery pack for a device before so please be patient with me if this sounds trivial. The task is to build a battery pack for a device that runs off 12V, draws 1-10 mA (which I measured using a .1-ohm sense resistor while it was operating), and must run for 4~5 hours.
The device is a radio transmitter. I don't have its internal schematics but know that it runs off of 12V fed through a DB25 RS232 connector.
I read this article (1.5V vs 9V) which concluded that 1.5V batteries have higher energy density than 9V batteries. This article (the capacity of 1.5V batteries) suggests that capacity vary with current draw.
With those two articles in mind, I thought of two solutions:
Option 1 is to get 12V from eight 1.5V batteries. Hooked up in series.
Option 2 I had in mind was to go with (2) 9V batteries (in series) with a voltage divider to get the 12V. The additional resistor means wasted energy but I don't care about efficiency -- the device only needs to run for a demo for a couple of hours.
I don't think I have to take load-distribution into account (because this is a quick one-off project and not part of a device to be mass-produced). Are there other things to consider?
The voltage dropped across your resistor depends on the current drawn. For an active circuit the current is not constant. You can do this with a single component like a LED or a lamp but not for an active circuit.
If it is just for demo, there are small 12V batteries like the 23A.
You can buy battery holders for AA and AAA batteries in almost any configuration you can imagine. 8 in a 2x4 brick or long-and-skinny or anything.
Many battery holders even include a switch.
Since you only need a small capacity, you could use a small number of batteries with a boost converter to reach the required voltage. Since these converters are relatively new, your highschool physics textbooks won't tell you about them.
MorganS:
You cannot use a voltage divider for power.
I see why 'active device' (a device which draws variable current) is an important qualifier now. I watched this video and realized that the voltage divider only gives you the right voltage if the current is constant. If you change the current flowing through it (for instance, by loading the voltage divider circuit with another source of impedance (like my device) at that divider junction), the voltage-divider output voltage will change.
Books that teach you about the voltage divider really need to include this under 'trap for young players'.
MorganS:
Since you only need a small capacity, you could use a small number of batteries with a boost converter to reach the required voltage. Since these converters are relatively new, your highschool physics textbooks won't tell you about them.
MorganS:
You can buy battery holders for AA and AAA batteries in almost any configuration you can imagine. 8 in a 2x4 brick or long-and-skinny or anything.
Many battery holders even include a switch.
In my naivete, I thought a voltage divider would do the trick but I definitely overlooked an important fact about loading a voltage divider
Circuit theory tells me that 8 batteries connected in series is effectively 1 voltage source in series with 1 resistor (whose resistance value is the sum of the internal resistances of each battery). The voltage at the device is 12V - (Voltage drop due to internal resistances).
Does this mean I can just use a battery holder without a boost converter?
The only drawback I see is that if one battery drops below nominal voltage, I won't get 12V anymore.
And yes, the voltage will slowly drop over time from 12V as the batteries convert their chemical energy into electrical energy. But because they are in series, they all will slowly drop.
It’s unlikely your circuit will studently stop working when the voltage drops below 12V. But at some point, it will fail, just like any other battery powered thing, even Voyager 1 which has plutonium battery.
Klaus_K:
Yes, you can just use 8 batteries in series.
And yes, the voltage will slowly drop over time from 12V as the batteries convert their chemical energy into electrical energy. But because they are in series, they all will slowly drop.
It’s unlikely your circuit will studently stop working when the voltage drops below 12V. But at some point, it will fail, just like any other battery powered thing, even Voyager 1 which has plutonium battery.
Thanks for blessing my logic, Klaus XD!
By the way, does the subject of designing battery circuits to provide power for devices fall under 'power supply design'? My ego is paranoid about the things I don't yet know.
Power supply design and battery circuit share common features. But then they have a lot of separte issues. Especially batteries have this added component of chemical science.
