# Question for a formula related to electrical power?

#### samy555

Joined May 24, 2010
116
Hi
http://www.electronics-tutorials.com/amplifiers/negative-feedback.htm
I found this formula for calculating the output power of RF amplifier:

2 * Po = [Vcc - Ve][SUP]2[/SUP] / R

Also found the same formula in the book: ‎ARRL - Experimental Methods in RF Design, where:
Po: the output power
VCC: the battery voltage
Ve: the emitter voltage
I think that number 2 came from a conversion of the Vp to rms

My question is: Why [Vcc - Ve][SUP]2[/SUP] and not simply Vc[SUP]2[/SUP] ; the ac collector voltage??
Thank you

#### GopherT

Joined Nov 23, 2012
8,012
Hi
http://www.electronics-tutorials.com/amplifiers/negative-feedback.htm
I found this formula for calculating the output power of RF amplifier:

2 * Po = [Vcc - Ve][SUP]2[/SUP] / R

Also found the same formula in the book: ‎ARRL - Experimental Methods in RF Design, where:
Po: the output power
VCC: the battery voltage
Ve: the emitter voltage
I think that number 2 came from a conversion of the Vp to rms

My question is: Why [Vcc - Ve][SUP]2[/SUP] and not simply Vc[SUP]2[/SUP] ; the ac collector voltage??
Thank you

Some amplifier power supplies have, positive and negative rails with a common (0V) at near the midpoint. Since mid-point is not exact, and other reasons, they just say you should use the total voltage swing positive (max) to negative (min).

Your illustration is hard to read because SUP doesn't work on this forum.

#### studiot

Joined Nov 9, 2007
4,998
I don't think you have this formula quite correct.

The derivation is as follows:

The maximum sine wave that can be fitted between two rails is a rail to rail sine wave.
The rails may be ± about earth or they may be + and 0.
I will take the value to be + and 0 for simplicity. (the zero is usually Ve and the plus is usually almost Vcc)

So let the rail to rail voltage be V

Then this the the peak to peak voltage

So the peak is half this ie V/2

So the RMS value is V/(2√2)

If we apply this to a load of R ohms then the RMS current is V/(2√2R)

So the RMS power = Irms x Vrms = V/(2√2R) * V/(2√2)

$${P_{rms}} = \frac{{V^2}}{{8R}}$$

This is for a single ended output, often called a half bridge.

A full bridge circuit, often called an H bridge, produces four times this.

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#### samy555

Joined May 24, 2010
116
Some amplifier power supplies have, positive and negative rails with a common (0V) at near the midpoint. Since mid-point is not exact, and other reasons, they just say you should use the total voltage swing positive (max) to negative (min).

Your illustration is hard to read because SUP doesn't work on this forum.
thank you GopherT

Why is the biggest Up swing of the signal is to Ve??

How this is done, I hope that you explain to me that, if possible

I don't think you have this formula quite correct.

The derivation is as follows:

The maximum sine wave that can be fitted between two rails is a rail to rail sine wave.
The rails may be ± about earth or they may be + and 0.
I will take the value to be + and 0 for simplicity. (the zero is usually Ve and the plus is usually almost Vcc)

So let the rail to rail voltage be V

Then this the the peak to peak voltage

So the peak is half this ie V/2

So the RMS value is V/(2√2)

If we apply this to a load of R ohms then the RMS current is V/(2√2R)

So the RMS power = Irms x Vrms = V/(2√2R) * V/(2√2)

$${P_{rms}} = \frac{{V^2}}{{8R}}$$

This is for a single ended output, often called a half bridge.

A full bridge circuit, often called an H bridge, produces four times this.
thank you very much studiot
I need to know about the maximum swing

#### t_n_k

Joined Mar 6, 2009
5,455
At some instant the capacitor might be charged to (Vcc-Ve) with the transistor saturated. At resonance the capacitor voltage can fully reverse polarity at some later instant - say as the transistor goes into cut-off. So the capacitor voltage will have swung nominally between ±(Vcc-Ve) over a full resonant sinusoidal cycle. If the load R sits across this voltage swing then the average power would be

$$P_{av}=\frac{{\(\frac{(Vcc - Ve)}{\sqrt{2}}$$}^2}{R}=\frac{{(Vcc - Ve)}^2}{2R}\)

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#### t_n_k

Joined Mar 6, 2009
5,455
Hi samy555,

With respect to your posted schematic one needs to first keep in mind that the load power is also determined by the input signal drive.

The value given by the formula assumes that the input drive is sufficient to ensure the load voltage swing is the maximum achievable for the given circuit conditions.

As shown, the amplifier is probably intended to operate in Class A mode. However, increasing the input signal drive level will ultimately force the operation into a less linear mode which may be difficult to clearly define without actually resorting to building the circuit. In fact, it may require some over-driving of the amplifier into partial non-linear operation which ensures one maximizes the load voltage swing.

In your schematic Ve will have a mean value - presumably that value which one would plug into the formula. But since the mean emitter voltage also has a superimposed AC component (approximately equal to the AC input signal voltage) the exact value of Ve at the corresponding load voltage minima won't be Ve_mean. The difference won't be that significant but observable.

In any event the formula will give a good starting estimate for the achievable maximum load power - provided the circuit is well behaved in other respects.

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