Actually i have a source volt of 7.4v 2200Ma Battery... and attached lm7805, 5v voltage regulator and connected my MCR and lcd to the output of lm7805.
Finally, a tidbit of information critical to the overall system design.
Actually, this has been a major issue I've had with this thread since it's inception, not enough design and application specifics, which has also been voiced by FvM and other participates of this thread.
You've went to a lot of effort to save at most 4mA, in reality probably closer to 2mA, by powering down the LCD, however up until now, have made no mention of actual power source and regulator specifications, other than briefly mentioning it was a battery of some sorts, not to mention what steps have been taken to conserve power with the microcontroller. Both of which could likely far exceed the savings of 2mA to 4mA by powering down the LCD.
When attempt to conserve power, linear voltage regulators and batteries, are typically not a good mix, particularly when the linear voltage regulator has a relatively high Dropout Voltage, like the 78xx series.
Where Dropout Voltage is defined as, the minimum voltage drop from input to output of the voltage regulator required to maintain the specified level of voltage regulation. Dropout Voltage becomes critical when calculating power efficiency as shown below.
Linear voltage regulators are overall inherently inefficient, in comparison to switch mode techniques, and this inefficiency is largely due to the required Dropout Voltage.
Efficiency, of any power regulator, is defined as:
Efficiency = P
o/P
i × 100% = (V
o × l
o)/(V
i × I
i) × 100%,
where in the case of a linear voltage regulator, I
i~I
o, which simplifies the equation to:
Efficiency = V
o/V
i × 100%, where V
i - V
o >= Dropout Voltage for specified regulation to be maintained
The Power Loss is approximately:
Power Loss ~ (V
i - V
o) x I
i or a minimum of (Dropout Voltage) x I
i during normal operating specifications.
For example, with the battery fully charged and with the 7805 drawing/supplying 100mA:
Power Loss ~ (7.4V - 5.0V) x 100mA = 240mW, primarily dissipated as heat
In comparison, the power consumption of the LCD:
Maximum Power Loss ~ 5.0V x 4mA = 20mW, in reality more in the range of 5mW to 10mW is typical.
Consider effort underwent to save, at most, 20mW, in comparison to the 240mW expended as heat, do you get the picture?
Also note, with a nominal Dropout Voltage of the 7805 at 2V, once the battery discharges to a level of below 7.0V, the 7805 can no longer maintain its specified level of voltage regulation.
At the very least, a linear voltage regulator with a Low Dropout Voltage (LDO) should be substitued for the 7805 and if greater efficiency is sought a switch mode regulator, like a buck topology, should be considered. Switch mode voltage regulators, unlike their linear counterparts can reach efficiency levels in the 90% range.
As I indicated, you still have not mentioned what steps/techniques you taken to conserve power at the microcontroller, which also would also likely exceed the savings of 2mA/4mA by powering down the LCD. Typically, putting the microcontroller to sleep between tasks, can achieve a surprising level of power conservation.
yes you are right.... but as i told i am new with the MOSFET part.... i was not aware of it of how to use and it functionality.... and also gone througg the datasheet.... but did not understood anything.... so in that case i just made it a try... and all it worked fine....
But will it create any problem or will work fine....????
The answer to your question depends largely on whether or not this is a hobby project, i.e., educational experience, or is the ultimate goal to produce and market your design as a commercial product. If the latter, then by all means use what is available to you, however if eventually the goal is to produce and market your design as commercial product, certainly choosing a P-channel MOSFET more suited to the task would be prudent and far more economical.
I would suggest, now that you've had some experience working with these devices, to spend two or three dollars and purchase a nice assortment of BJTs and MOSFETs of the smaller variety, they will come in handy and the small devices are a fraction of the cost of a single IRF9540. An assortment of BJTs and MOSFETs should be considered a standard staple, just as an assortment of resistors, capacitors and crystals.
I would urge you, as FvM and others have previously, to post a fully and detailed description of your design's purpose, required tasks, along with a schematic or at least a block diagram, rather than continuing to occasionally dribble out a tidbit or two. Doing so will not only aid those attempting to render advice, but would also help organize your thoughts and design.
BigDog