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Metaphysical question

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Skyknight

New Member
Please, everyone, give your opinion :D

Which batt will last longer? :?:

1) The green one
2) The blue one
3) Both the same

Waiting for your answers... :mrgreen:
 

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bogdanfirst

New Member
wel, it depends on the capacity of it. so considering that the current is about the same, the battery with a larger capacity will last longer. if you use a normal 9V battery, wich has a capacity of about 100-200mAh, it wil last less then 5V from some rechargable battery. this is a rough theory. but it is more complicated in real then you think, and i might be wrong.
the regulator will maintain 5V as long as it has more then 8V as input. fo while the battery has a voltage >=8V, the current will be constant in the circuit. as for the 5V its voltage will drop from 5->4.9->4.8......and so on and the current will decrease. so it is not that simple to tell. you need more data do determine wich one will last more. you need the voltags(time) graphical curve of the battery and some integrations. also, it depends when you consider the battery discharged completely, wne it has 0V? and another thing, the 7805 is not a passive element so you need some graphics too. because the voltage will remain constant as long as you have a minimum drop of 3V. the probelm could me more simpler if you have a switching power suply. with a constant efficency.
 

Skyknight

New Member
A 7805 is a voltage regulator. It has an input and an output (and a ground pin too) if the input is higher than the output (better if, at least, 1.4V higher) the output locates exactly on the last ic number (5V DC in this case because the ic is called 7805; a 7812 would give 12V DC). One good point is that it clears any ripple there could be before this device.
 

Gene

New Member
Good question. Remember that a correct answer requires the reader to not only know the current being drawn by your regulator and load BUT the current capacity of the sources.
 

Sebi

Active Member
Definitely the green,because the 7805 suck also a (small 2...3mA) current.
The example better with 7806, because 5V battery not existiert....
 

bogdanfirst

New Member
yes, the greem one has a better chance to last longer. but i dont think it is due to the 2-3mA that goes trough the ground reference pin. if i could get a 5V battery i will try to make this experiment.......but where do i get a 5V batt?
 

Skyknight

New Member
Ok, lads! You're right! :!: :D
There's no 5V avaliable in the market. I've only put that voltage because I use the 7805 device very often and I would like to know. Really, my interest is comparing a real 9V batt against a 4.5V (quite big) or 3*1.5V batts (better the smallest as possible). I think almost any circuit that works with 5V will work with only 4.5V too. So, the question is still alive:

Which "system" will last longer? Thank you all for your interest :mrgreen:
 

Scubasteve

New Member
I don't know what to tell you and I don't know the mystery in solving this. There are advantages and disadvantages to both. The 7805 regulates voltage a bit better then the straight 5V source, but batteries are pretty much regulated anyways.

The 7805 obviously wastes power when it is dropping the 9V down to 5V. Also, as stated, the current through the reference pin is another factor. These two things are obvious and they will cause the straight 5V (or 4.5V) source to be more efficient. If you are powering a large load, then the 7805 will not be able to support this because of its limited current capabilities.

Steve
 

bogdanfirst

New Member
the way i see it the 5V shoud work with 4.5v but not with less, so when the battery is discharged just a little bit the circuit will not work. but as energy, the 4.5 battery has a lot more then the 9V battery. the 9V batteries are low capacity, if you make 9V out of 6 1.5V batteries, you will probably have 5 times, more or less, times more energy. so the 5V(virtual) battery will last longer, but it depends on the capacity.
 

Gene

New Member
The blue circuit draws more current from the power supply because it has an extra component. The regulator has a energy loss that goes into heat energy. Since we don't know what the load is, we can only assume it is in the neighborhood of a couple of watts. But, in any event, both the load and the regulator use current.

The question was which battery will last longer so it is possible to gain life with a BIG battery - even though our loss from the bule circuit is greater. In fact the opposite is true here. The standard 9 volt battery used in the blue circuit gives us 625 mAh to draw from (Everyready Energizer #522). Using three 1.5 AA cells gives us 2850 mAh (Everready Energizer #E91).

So, with a lower capacity battery and more components drawing current, the blue circuit will fail first.

However: I still stand on my original disclaimer that there is not enough information provided to make a firm statement without assumptions.
 

Skyknight

New Member
I didn't know anything about the capacity of a battery, but you're telling me lots of interesting things. :)

The load will be all the components of the clap switch system (see the related topic: "Clap switch headache" where you'll find the FINAL circuit, really really incredible! :!: ) and some add-ons like a blue SMD LED (because, all the clap switch system will be packaged into a transparent polymere box, to give some ambience). 8)

:idea: :D
 

Gene

New Member
To calculate battery life, the Everready Energizer site has the mAh numbers you will need.
 
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