i was wondering what happen to the rest battery voltage while using regulator
does that will make the discharge faster ?

 

Assuming that you are referring to a linear voltage regular, there is no change in the discharge time of the battery.
If the load needs 1A, the battery still has to deliver 1A at the battery voltage. Any excess energy that is not used by the load has to be given up as heat in the linear regulator.

 

i was wondering what happen to the rest battery voltage while using regulator
does that will make the discharge faster ?

Hello,

Do you actually mean the rest voltage or the "rest of" the voltage.

If you mean the rest voltage, then it drops because the regulator always draws some current. So the self discharge would appear to increase which would discharge the battery faster than without a regulator. Sometimes this effect can be minimized by using a regulator appropriate for the application.

 

so there a power loss .
if i have 12v battery and regulator 5v if i have load 5v @1A the run time is the same if i have 12v load @1A ? is that right ?

 

is that right ?

Essentially. Voltage regulators will have a small current that isn't delivered to the load.

 

if i have in the output 4w load how much the power consumption from the battery ?

 

if i have in the output 4w load how much the power consumption from the battery ?

About 9.6W; the difference is dissipated in the voltage regulator.

 

About 9.6W; the difference is dissipated in the voltage regulator.

how did you calculate it ?

 

how did you calculate it ?

Current @5V = 4W / 5V = 0.8 A

Wattage @12V = 0.8A x 12V = 9.6W

 

how did you calculate it ?

4W from a 5V supply = 0.8A. 12V*0.8A = 9.6W

 

For better efficiency you can use a switching regulator which are generally 85-90% efficient.

 

if i use diode zener regulator its not efficient as linear regulator ?

 

if i use diode zener regulator its not efficient as linear regulator ?

If you want efficiency, use a switching regulator.

 

i want to know the efficient of the diode zener how i calculate it ?

 

i want to know the efficient of the diode zener how i calculate it ?

Post a schematic so we can see how the zener diode is being used.

 

if in the output there load 5v @2A
VZ=5v
input 12v
how to calculate the efficient

 

No, it doesn't change if you use a zener diode. The excess power must be given up as heat in the zener diode and in the series resistor.
The battery still has to deliver the full current at 12V.

 

how to calculate the efficient

The rule of thumb is that the current in the zener should be 10 times the load current. With that information, can you calculate efficiency?

We can't because you don't provide enough information; unless you want it expressed algebraically.

 

No, it doesn't change if you use a zener diode. The excess power must be given up as heat in the zener diode and in the series resistor.
The battery still has to deliver the full current at 12V.

i cant know the efficiency ?

 

i cant know the efficiency ?

Of course you can.
It's just the output load power divided by the battery power.