A better circuit for longer battery life?

Discussion in 'General Electronics Chat' started by Fuji, Nov 11, 2014.

  1. Fuji

    Thread Starter Member

    Nov 8, 2014
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    I know everybody who builds a device that is battery charged would always think about making a circuit to preserve battery life. I was just wondering what would be a simple way to save power with 3v coin batteries? I know this also depends on how many components are on the board as well.

    I have 3 3v coin batteries which = to 9v of power. I have my 9v entering 2 capacitors, but its making me think that all the capacitors do is stabilize the voltage instead of preserving battery life. What can I ADD to the line of circuitry before or after these capacitors that can preserve battery life?
     
  2. ISB123

    Well-Known Member

    May 21, 2014
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    Capacitors wont drain your battery.What else do you have in the circuit other than capacitors?
     
  3. Fuji

    Thread Starter Member

    Nov 8, 2014
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    Before the capacitors I have a red LED/resistor to indicate that the power is on. After the capacitors is a resistor and to lower the voltage, then a potentiometer to PIC16F84 and quartz crystal oscillator connected to the PIC. Thats it.
     
  4. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    you add a circuit designed to operate with minimal power usage. One that can power down or enter a standby state when inactive. ect. ect.

    batteries are a defined capacity power source. they have a specific amount of energy. you can only lengthen the life by lowering the amount of power per unit time that is consumed by a circuit.
     
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  5. ISB123

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    You need to get the IC in standby mode or you need pico version of the chip.
     
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  6. MrChips

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    Oct 2, 2009
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    For longest battery life you design your circuit for the lowest average current consumption.

    For example, an LED at the rated brightness might require 20mA which will kill your coin batteries in a short time.
    If you can live with lower LED brightness, high efficiency LEDs will work with less than 1mA.

    Have you ever noticed the LED status indicator on a household smoke detector? The LED flashes very briefly once every minute. That is another way to reduce average current consumption.

    Going even further, ask yourself if you really need an LED as an indicator. Eliminate the LED entirely and save on that portion of the current draw.

    LCD displays consume less that 1μA if you really want a low power display.

    In summary, you can select components and a circuit design that draws micro-amps of power.
     
  7. ErnieM

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    Apr 24, 2011
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    A LED that beiefly blinks on but spends most of it's time off takes far less power then a steady LED. Well worth a PIC pin, plus it not only tells you power is on but your PIC is alive.

    A resistor makes a poor voltage regulator. Why start with 3 cells for 9V when a PIC can directly run off just one cell for 3V?

    The PIC uses power when it runs. The faster it runs the more power. So use the slowest clock you can, or even chankg clocks depending on the task. Sleep mode is the most efficient way to be when you can afford it, using say a timer to wake up, or a change on pin interrupt.
     
  8. MrChips

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    Oct 2, 2009
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    You really ought to tell us what you are building and it helps to provide a circuit schematic.
    At 9V you are wasting a lot of power. Many mcu chips today will run on 3V and consume an average current less than 1μA.
     
  9. wayneh

    Expert

    Sep 9, 2010
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    Using resistors to drop voltage is a wasteful practice, by design. Consider operating at a lower voltage. Use intermittent pulses to save power, or better yet an LCD indicator as mentioned. It's why they use those in solar-powered calculators - they need so little power.

    Since you haven't mentioned that your circuit has any function, you can also save batteries by turning it off.
     
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  10. ian field

    Distinguished Member

    Oct 27, 2012
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    Consider pulsing the LED at spaced intervals to save approaching 20mA

    The simplest circuit I can think of uses a programmable unijunction, small signal thyristor or you can make the equivalent with a pair of transistors.

    A PUT is just a thyristor with the gate at the other end, the basic PUT oscillator can be duplicated by a small TO92 Thyristor such as the 2N5061 etc by turning the basic circuit upside down.

    A complementary pair of transistors can be cross coupled to imitate the 4-layer devices mentioned above.

    In each case a capacitor charges slowly via a resistor until it reaches the 4-layer device breakover voltage, which flashes the LED.
     
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  11. Fuji

    Thread Starter Member

    Nov 8, 2014
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    It is just a little project im doing to output frequencies from a potentiometer and PIC. The LED pulses frequencies. That is it. The reason for a 9v is because I want my circuit to last for a long time, whereas the PIC is around 3v as you said, and an oscillator connected to pin 15 and pin 16 of the PIC (OSC1/CLKIN, OSC2/CLKOUT).

    As for detecting if the batteries are high or low in power, can I just build a circuit where I can just press a momentary tactile button and it will display green or red light to indicate that the power is high or low instead of having the LEDs constantly on? I guess this will have me remove the Red LED/resistor at the beginning of the circuit to indicate that the power is on.
     
    Last edited: Nov 11, 2014
  12. Fuji

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    Nov 8, 2014
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    Thanks for the reply. May I ask however if it is fine to use a resistor or two, OR a voltage regulator to lower the voltage from 9v to 3v for the Microcontroller at least?
     
  13. Fuji

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    Nov 8, 2014
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    Is it possible to use a bi-color LED and pulse it to indicate that the power is high or low? For example, when blue lightens up and pulses, it indicates the battery is high. When the battery is low, the red light pulses. This is with a bi-color LED. Would this work?
     
  14. wayneh

    Expert

    Sep 9, 2010
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    What do you mean by "fine"? A resistor or a linear regulator is cheap and simple but will waste the amount of power. A buck regulator or DC-DC converter can be very efficient.
     
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  15. Fuji

    Thread Starter Member

    Nov 8, 2014
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    Thanks for the reply. That is a good idea.

    As I understood before, a buck regulator is like a switching regulator. It converts power instead of dissipating heat. I'll assume this would be found in a sot23 package for a switching regulator, which I think inside has coil and SMD's.

    Do you think relays may work as well?
     
  16. ISB123

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    They do need external parts coils, caps,etc...
     
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  17. wayneh

    Expert

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    Or you can buy a complete module on e-bay.
     
  18. ErnieM

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    Apr 24, 2011
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    There's no reason why that would not work.
     
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  19. ian field

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    Oct 27, 2012
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    If the circuit can tolerate a bit of ripple on the supply - a hysteretic buck regulator is often a lot simpler to design.
     
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  20. Fuji

    Thread Starter Member

    Nov 8, 2014
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    Well I am currently using MC34063 DC/DC CONTROLLER Boost-Buck Regulator. I see with this regulator that it should be placed after the power button and before the rest of the circuity on the board.

    Is it only after this regulator that resistors would be fine to use???
     
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