Power Supply and battery life

[rbrainard]

Ok, to start I would like to mention that I am VERY new to building circuits. Now that that is out of the way, I have some questions for anyone interested.

I am planning on building the step counter project http://www.electronics-lab.com/projects/misc/007/

My first question is how can you estimate how long the battery for this project will last? Is there a formula or is my best bet to turn it on and see?

My second question has a little less to do directly with the project. I have noticed a lot of the ICs that everyone used specify using 5/10/15V. Now, suppose I want to make a project operate on battery power. How does one find a battery that is compact and can supply 5V? I want my projects to be small but I can't think of any battery combinations that add up to 5V. Is there a small power inverter I can create?

Any help would be greatly appreciated. Thanks in advance.
Ryan

 

[audioguru2]

Hi Ryan,
Welcome to our forum.

To calculate battery life you need to know how much current that a circuit is drawing. That counter project uses ordinary Cmos ICs that draw nearly zero current. But its LED display draws a huge current. You must figure how long the display will be lighted.
A good site for battery info is www.energizer.com and click on Technical Info at the top right. You will see on the datasheets how a battery cell's voltage drops with use and how a very expensive AA Lithium cell works much better than a cheap AA Alkaline cell.

Ordinary Cmos 4xxx ICs are spec'd with operating voltages of 5V, 10V and 15V but are guaranteed to work from 3.0V to 18V. You certainly don't need 5.0V for them like old 74xx TTL ICs needed.
High-speed Cmos 74HCxx ICs are guaranteed to work with a supply voltage from 2.0V to 7V. I have them in some of my projects and they work fine with a supply voltage less than 1.24V.

That counter project uses ordinary Cmos ICs that are not guaranteed to work if the supply voltage drops below 3.0V and its display will be very dim. We have another project that uses the same 4093 IC where its author admits that many of them don't work properly with 3.0V. I think it is the same author. This counter project's recommended two AA cells will drop to 2.0V over their life. I recommend using three AA cells with it so that the voltage starts a little above 4.5V then drops to 3.0V when they will need replacement.

 

[rbrainard]

First off, thank you for the quick reply. You have been most helpful.

Secondly, it seems like the HC74xx series would be a better choice. Is there any counters or decoders in that family? My idea is to use a Lith-Ion coin battery (3v 1000mAh). So, it would seem more stable if I were able to use a chip that could handle the lower voltage? Any ideas on alternative chips?

Thanks again. I really appreciate your help. This is a field that seems to have high barriers of entry sometimes.

Ryan

 

[audioguru2]

Hi Ryan,
Nobody seems to make a 74HC4026 which is a low-voltage version of the IC in our counter project. Since you are a Newbee, why not make our project as it is. If it needs more voltage than your coin cell, you could add an 8-pin and 2 capacitor ICL7660 or MAX1044 voltage doubler IC, or just a 2nd coin cell.

 

[rbrainard]

Perhaps you are right. I should just try the project and make sure I learn the process before I jump in head first. Again, thanks for your help.

Ryan

 

[MP1]

Ryan,
The display is turned on by switch P2 so that the current consuming display is not always on. The circuit uses very little current. I think you will have the hardest time finding the mercury switch since most companies have quit making them.
You can directly contact the author of the project on his website at www.redcircuits.com
Perhaps he has already performed power consumption testing on this project.

MP

 

[rbrainard]

Thanks for the ideas MP. I thought about the mercury switch idea and have been leaning towards using some type of Hall effect switch. I think that would fit my application a bit better anyhow. I will also contact him. Who better to ask right? Thanks again. See you in the forums.

Ryan