efficiencyvoltage-regulatorzener
How to calculate the power efficiency of zener diode regulator?
Is it correct to calculate it as the ratio of power delivered to load to the total power drawn from input source?
If I am correct, would the efficiency of circuit shown in the schematic below is 0.5% ?
EDIT
The maximum current zener can handle in this case is around 240mA.
Based on this,
$$ \frac{Vin – Vz}{Imax} = \frac{12-3.9}{240} = 32 Ohms $$
Am I correct?
Is efficiency Vout/Vin or Pout/Pin ?
I believe I may need a buck-boost however, because the components will need at least 1.8 V and 3.3 V, and my 3.6 V battery could drain below that
First hit on Linear Technology might be useful: -
Try going to LT to get an idea and use this parameter search engine.
for the 1.8 V rail it's 1.154 mA
Use a linear regulator - it's just not worth the effort I believe to use a switcher from 3V3 to 1V8. However, if current was over 50mA I might start considering a switcher.
Generally, you can assume a buck regulator will give you about 90% power efficiency near full load and a buck-boost will be about 85%. Ball-park figures.
Why don't you use a buck-boost regulator that will produce 5V from an input of below 5V to above 12 volts?
Here's one from Linear technology: -
Best Answer
I gather you're not impressed by 16mW to the load while taking 3 watts from the battery? Yes, efficiency is Pout/Pin.
Shunt regulators are not used for efficiency, they are used for their simplicity.
Most people wouldn't use such a low value resistor as R1 for that load. How did you choose the value? Think about what the constraints are on the value of R1.
Use a three terminal series pass regulator if you want to waste less power. With a 4v load, and a 12v source, an ideal 3 terminal regulator would not exceed 33% efficiency.
To approach 100%, you would need a switching regulator.