Question Re LTSpice Simulation

Thread Starter

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
 
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
 

Thread Starter

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.
 

Attachments

Last edited:

eetech00

Joined Jun 8, 2013
1,726

Thread Starter

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
12.png
 

Thread Starter

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?
 
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