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How can I calculate what the current "I" output can i have for determining my Efficiency. I am building a simple linear regulator and is it possible to know during design of the circuit , what the efficiency of the circuit would be?
Should I just assume input current to be the maximum current it can out put, or should I put in a resistor for my calculations and see how much current it will draw, if its the second one, then how do I determine the impedance due to other parts in my circuit?
The efficiency is:
$$\eta = \dfrac{p_{out}}{p_{in}} = \dfrac{V_{out}\cdot I_{out}}{V_{in}\cdot I_{in}}$$
For a linear regulator, assuming the output current is much greater than the regulator circuit's internal currents, we have \$I_{out} \approx I_{in}\$ thus:
$$\eta \approx \dfrac{V_{out}}{V_{in}}$$
The power efficiency of a linear regulator will hardly ever get close to 100%.
\$Power_{out} = Power_{in} - losses\$
The losses come from the volt-drop across the regulator. You can reasonably assume the output current is the same as the input current but the input voltage will be usually 1V to 10V (broad-brush strokes here) greater than the output voltage.
Let's say the circuit was a 5V regulator powered from 10V and delivering 0.5A to the load.
Input power = 10V x 0.5A = 5.0W Output power = 5V x 0.5A = 2.5W Losses are of course 2.5W
Power Efficiency = 50%
i calculate the efficiency assuming that the input and output signals to be current(I) not power.the power efficeincy will be way less than 100 % efficiency because of loss due to circuit components, like you said.
Your circuit efficiency will be the output current over the input current multiplied by 100 to state it in % (percentile form) .how you design the regulator determines the final true efficiency.the input current will be nearly equal to the output current your load consumes. In this case, the efficiency will be around 100 %. In practice, you do not get 100 percent efficiency but you can get close to it. All depends on how you build your system for the desired output.
