Calculating average power

Example

Let’s consider the following circuit. Determine the average power appearing in the resistor R, sensed by the voltage labeled Out, as a function of time.

Solution

The instantaneous power in the resistor is:

The average power,

And,

We will compute and draw the power function with the Interpreter. The u(t) function in the integral (the voltage across the resistor) will be picked up from a previous TINA result produced by transient analysis if we reference it correctly, as TR_result1.

The solution as found via the Interpreter,

Function P(x);

begin

P := Integ( Sqr( TR_result1( tau ) ), 0, x, tau );

end; {P}

Function Power(x);

begin

Power := P(x)/(R*x);

end;

Draw( Power(t), PowerRes )

Verification

For verification we will use the fact that in this case the instantaneous voltage across the resistor settles down to an average amount with a ripple component, and the instantaneous and average values come out very close.

Using the TINA transient analysis output, let’s check the result computed by the Interpreter.

In the TINA transient analysis time diagram, we see that after awhile, the voltage across the resistor doesn’t change much. This will let us use an average value which we will read off the curve in the Diagram Editor. Press the ‘Cursor a’ button, and the cursor will change into a +. Click this new cursor on the output curve, position it to around 80m, and read the voltage:

The average power at this time is:

Using the Interpreter we see that

Power(80m)=[72.7884]; that is, 72.79mW

We find just a small difference of 2.74mW.