Designing a simple and low power consuming analog kit.

[SeungminLee]

Hi, there.

I’m developing low-power consumption filter and amplifier devices but have a wondering point.
The system that I’m trying to is consisted of power regulator and Butterworth high pass filter, low pass filter and basic negative amplifiers. But I wander the driving voltage should be. It can be ranged from 1.2 to 4V as choice of battery. Does low battery voltage consume low current that would decrease the power consumption? Or should I have to employ high voltage battery? In this situation, I assume that battery has all same capacity.

Thanks in advance.

 

[tony_lth]

It's bettter to select battery voltage a little more than the Vdd of the amplifier. Since you use LDO, it has dropout voltage, V battery=Vdd amplifier+ V dopout LDO.
If you have more power source, you can use DC/DC converter. Notice that DC/DC converter efficiency is about say 90%.
The system Id should be the sum Idd from every component. So power consumption is least when voltage is as low as possible.

 

[sam185]

I think high voltage & high Ah battery can deliver higher power & low power battery will deliver low power. Hence if your requirement is 1.2 or 4V you can use some 7805 IC with adjt o/p configuration. The battery should be of high power but should be according to device i/p requirement.

 

[SeungminLee]

Thanks, tony_lth. I'll try low voltage battery and check out. thank you again.

---------- Post added at 03:20 ---------- Previous post was at 03:18 ----------

thank you for advice but what I want to know is the choice of battery. designing input voltage does not make a problem.

 

[tony_lth]

Here is a example from this forum, but I forgot from which thread and who is the author, maybe you can search with baterry number etc.
The example said:"
also battery will loose 5 to 10% charge during storage as they grow old.

the power consumption of the circuit is 0.625W+0.5W+1.35W = 2.475W
if it needs to run for 10 hours a day then the rating of the battery is 2.475*10 = 24.75WHr
Divide it by batt voltage of 1.2V you get Capacity in MaHr 20.625Ahr.
Take a 80% safety margin you get 25.78AHr.
so you will need 25.78Ahr/2100mAhr =~ 12Batteries.

one better choice would be to use cell phone batteries, which are avalable in 3.7V and up to 5000mAHr, and are very cheap.

So if you need high power, sam185 is right, when the size of battery is limited. But if you select high power battery, you need many DC/DC step-down converter. The conversion circuit size increase.

Anyway, the baterry voltage is high or low is up to your circuits and the budget of your project and the size limits etc.

Cheers.