as a part of my project , i have to develop +5v from an 9v carbon-zinc battery with 7805 regulator.it is given to microcontroller and some other circuitries .it consumes nearly 40ma. but,the problem is that " battery voltage drop fastly to less than 6v for continuous usage,nearly 20 minutes".then 7805 wont work and entire system will shut down. after,removing the battery,battery voltage increases from 6v to 8.5v and again performs well for next 20 minutes. i would like to use at least 1 hour continuously,without effecting performance.can you suggest a proper solution?
Electronic – 9v battery voltage drop and regain its voltage
batteries
Related Solutions
For simplicity you could power the Arduino from a 5V low-drop-out regulator. You'd bypass the normal regulator (it isn't low drop-out) and get more usage from the 7V battery. I don't know what current is required by the uno so I can't definitely recommend something but maybe a 1A LDO like the L4941 would suit. It drops 450 mV at 1A and this means your battery can droop down to 5.45V before things go out of spec. You might need a heatsink - it all depends on how much the uno takes.
Or you could go the whole hog and have a buck-boost regulator like this one: -
This takes any voltage from 2.7V to 15V and gives you 5V and up to 2.5A. Or maybe just a buck regulator would suit. This one will work down to an input voltage of 5.5V and still give 5V out at 2.5A:-
Specifications here: http://www.uavfactory.com/product/69
Energy = power x time (Wh)
Duration = Energy / power hours or seconds
The battery is said to have 640 Whr capacity
(actually this is the discharge rate)
No. Capacity is capacity and is correctly stated in Wh = Watt-hour
Capacity = Energy available = Power x time = (here) Watts x hours.
the load being a 2700 W motor.
The missing link is that they are claiming a MAXIMUM (implied) motor power of 2700 Watts.
Your overall problem is not appreciating the various units used and how they relate.
Power = instantaneous rate of doing work. Unit = Watts.
Energy = Sum of Power x time products. Unit = Watt.seconds = Joule, or watt hours.
A motor that runs at a POWER level of 100 Watts for one hour (3600 seconds)
uses 3600s x 100W
= 360,000 Watt seconds of energy
= 100 Watt hours.
So if
the available ENERGY is 640 Wh
and the duration = 110 minutes
= 110/60 ~= 1.83... hours
then IF the figures are correct at the power level involved then
MEAN power = Energy/time_operating
= 640 Wh / 1.83h ~= 350 Watts.
The actual Watts will vary across the operating cycle and
the battery capacity will vary with discharge rate
and load is affected by windspeed and air density and flight profile and ...
and some of these vary with time of day and location and ...
so such figures are an approximation.
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Best Answer
Most batteries under load will drop their voltage and the voltage will then go back up when the load is removed. The voltage will drop especially quickly if the load is high or the battery is close to being dead.
So you likely have one or both of the following:
1) Your load is way higher than 42 mA
2) You are using a battery that is already exhausted or perhaps you are using a carbon zinc battery or "heavy duty" battery? They only have a capacity of about 50 mAh! Even Duracell 9 volt only have approx 300 mAh, and some energizer are closer to 450 mAh.
As mentioned in another answer, using a battery holder with 6 AA batteries with last a lot longer as their capacity would be in the 2000+ mAh range.