But the book is saying about peak to peak value

P = (2*3.41V)^2/(8*8Ω) = 726.75mW

For peak value, we will have

P = (Vpk/√2)*(Vpk/√2)/R = (Vpk²/2)/R = V²/2R = 3.41V^2/(2 *8Ω) = 726.75mW

And for the RMS value

P = (2.41V)^2/8Ω = 726mW

As you can see everything is in order.


Where? in these formulas

How can Pout(peak-to-peak) = Pout(peak) = Pout(rms)?

Is peak not equal to 1/2 of peak-to-peak and rms not equal
to peak times 0.707?

 

How can Pout(peak-to-peak) = Pout(peak) = Pout(rms)? Is peak not equal to 1/2 of peak-to-peak and rms not equal
to peak times 0.707?

I think that you misunderstand the book about the power calculations.
In general, when we are dealing with AC signals we distinguish the power like this:

1)
Instantaneous power p = u*i (power at any instant of time)

2)
Active (real) power P = Vrms*Irms

3)
And reactive, and apparent power. Not important here.

As you can see only Instantaneous power can have a peak value and meaningless RMS value.

So we are only interested in active/real power.

And the book gives you the equation for an active/real power.

P = Vrms*Irms = Vrms^2/R = Vpeak^2/(2R) = Vpeak-to-peak^2/(8R)

Is that clear now?

 

I think that you misunderstand the book about the power calculations.
In general, when we are dealing with AC signals we distinguish the power like this:

1)
Instantaneous power p = u*i (power at any instant of time)

2)
Active (real) power P = Vrms*Irms

3)
And reactive, and apparent power. Not important here.

As you can see only Instantaneous power can have a peak value and meaningless RMS value.

So we are only interested in active/real power.

And the book gives you the equation for an active/real power.

P = Vrms*Irms = Vrms^2/R = Vpeak^2/(2R) = Vpeak-to-peak^2/(8R)

Is that clear now?

Okay, so Pout(active or real) = 726.75 mW.

The 1.45 W calculated by LTSpice is peak which is useless?

 

Are you calculating the power in just the load RL, or the total power including that dissipated in the emitter resistor as well?

 

Are you calculating the power in just the load RL, or the total power including that dissipated in the emitter resistor as well?

Just in RL. I guess if you include RE then the dissipation would be 1.45W. In the
sim it is using Vout(pk) * RL but it comes up with 1.45W.

 

I guess if you include RE then the dissipation would be 1.45W.

Your guess would be wrong. The emitter resistor dissipates nearly 8W average !

 

Okay, so Pout(active or real) = 726.75 mW.

The 1.45 W calculated by LTSpice is peak which is useless?

And if that is the case why the heck would LTSpice do it that way?

 

The 1.45 W calculated by LTSpice is peak which is useless?

This is the plot of a Instantaneous power p = u*i (power at any instant of time)


To plot the instantaneous power dissipation of a component, hold down the Alt key and click on the body of the symbol of the component. The instantaneous power dissipation will be plotted as an expression of voltages and currents. And to obtain the average and RMS value switch to the waveform then move the mouse to the label of the trace, hold down the control key and left mouse click.

 

why the heck would LTSpice do it that way?

Spice calculates instantaneous power and average power. There may be occasions where peak power is important, e.g. in determining if a delicate component of low thermal mass will get fried.

 

This is the plot of a Instantaneous power p = u*i (power at any instant of time)
View attachment 173707

To plot the instantaneous power dissipation of a component, hold down the Alt key and click on the body of the symbol of the component. The instantaneous power dissipation will be plotted as an expression of voltages and currents. And to obtain the average and RMS value switch to the waveform then move the mouse to the label of the trace, hold down the control key and left mouse click.

Thanks!

 

Spice calculates instantaneous power and average power. There may be occasions where peak power is important, e.g. in determining if a delicate component of low thermal mass will get fried.

Thanks!