# Question Re LTSpice Simulation

#### elec_eng_55

Joined May 13, 2018
214
Are the ac values in simulations peak values?

Thanks

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#### ericgibbs

Joined Jan 29, 2010
8,887
hi, David,
Note the ' mag' in the listing: ie magnitude , peak values.
You can always double check by running a quick .tran

E

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#### elec_eng_55

Joined May 13, 2018
214
hi, David,
Note the ' mag' in the listing: ie magnitude , peak values.
You can always double check by running a quick .tran

E
Hi Eric:

The sim computes the Pout(pk) = 3.41 V(pk) *424 mA = 1.45 Watts

If you use Pout(pk) = Vout(pk)^2 / 4 * re where re = RE//RL = 4 Ω
Pout(pk) = 3.41 V(pk)^2 / 4 * 4 Ω = 726.376 mW

Pout(ppv) = [Vout(pk) * 2]^2 / 8 * 4 Ω = 1.45 W

That is why I asked this question. I am confused.

David

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#### crutschow

Joined Mar 14, 2008
23,545
What exactly, are you trying to calculate?

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#### Ylli

Joined Nov 13, 2015
799
Looking at the power delivered to RL with the given input:

#### elec_eng_55

Joined May 13, 2018
214
Last edited:

#### ericgibbs

Joined Jan 29, 2010
8,887
hi D,
On your asc sim, make the .tran active and run the sim.

Then place the cursor on the emitter resistor, press down the 'Alt' key on the keyboard, the power plot should appear. [click the left mouse key]

To show the power label, place the cursor on power label on the top of the plot screen and press down the Ctrl key.

E

#### crutschow

Joined Mar 14, 2008
23,545
I am still confused with respect to the formulae.
That makes two of us.

#### elec_eng_55

Joined May 13, 2018
214
hi D,
On your asc sim, make the .tran active and run the sim.

Then place the cursor on the emitter resistor, press down the 'Alt' key on the keyboard, the power plot should appear. [click the left mouse key]

To show the power label, place the cursor on power label on the top of the plot screen and press down the Ctrl key.

E
Thanks Eric.

I am still confused with respect to the formulae. These formulae are taken from
Malvino's text "Electronic Principles" 8th edition. See attached.

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#### eetech00

Joined Jun 8, 2013
1,726

#### Jony130

Joined Feb 17, 2009
5,024
I'm I right to say that you are using AC analysis to determine the peak values?

#### elec_eng_55

Joined May 13, 2018
214
I'm I right to say that you are using AC analysis to determine the peak values?
Yes, Jony they are off the ac analysis of the attached sim.

#### Jony130

Joined Feb 17, 2009
5,024

#### elec_eng_55

Joined May 13, 2018
214
I'm I right to say that you are using AC analysis to determine the peak values?
These formulas should yield the same results as the sim.

Pout(pk) = 3.41 V(pk)^2 / 4 * 4 Ω = 726.376 mW

Pout(ppv) = [Vout(pk) * 2]^2 / 8 * 4 Ω = 1.45 W

The sim says that Pout(pk) = 1.45 W

My question is. Is Pout(pk) = 1.45W or 726.376 mW ?

#### Jony130

Joined Feb 17, 2009
5,024
As I said you shouldn't use the AC analysis resoults in this case. Only transient analysis is valid in thia case.
And for Vin = 5.07Vpeak the output power is given by LTspice is 715.87mW

#### elec_eng_55

Joined May 13, 2018
214
As I said you shouldn't use the AC analysis resoults in this case. Only transient analysis is valid in thia case.
And for Vin = 5.07Vpeak the output power is given by LTspice is 715.87mW
View attachment 173632
The transient graph says 1.45 W and the ac analysis supports that number. The average is 715.87 mW per the graph.

#### Jony130

Joined Feb 17, 2009
5,024
But the average power is the "real power".

#### elec_eng_55

Joined May 13, 2018
214
But the average power is the "real power".
I get that the real power is the "average".

But what good are the formulas in the text if they don't yield the right value?

In some places they use RL in the denominator and in other places they use re (=RE//RL).

RL is the correct value?

#### Jony130

Joined Feb 17, 2009
5,024
But what good are the formulas in the text if they don't yield the right value?
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.

RL is the correct value?
Where?

In some places they use RL in the denominator and in other places they use re (=RE//RL).
Where?