Emitter Follower Help: Understand Physics & Art of Electronics

[sophiecentaur]

Gv = RL/(RL + re) is almost unity, which is what you'd expect. This, of course, is Voltage Gain; (the Power gain is much greater - relating to the current gain). I think the value of Vi, used for that formula just relates to the Voltage on the base. If you include a finite series resistance between input volts and the base then this will affect the gain (potential divider effect) Perhaps this is where your problem lies - but you are right to consider the source resistance for many practical applications. Let's face it, you use a buffer when the source impedance is high, dontcha?

 

[NascentOxygen]

Hello again. I'm not sure If I should post my Emitter Follower question here or make a new thread. I will try posting it here first. So in "The Art of Electronics" it says the gain for an emitter follower is Gv = RL/(RL + re), where RL is the resistance of the load and re is the impedance looking back into the emitter.

That's what theory predicts, from the small signal equivalent circuit.

Isn't the gain Vout/Vin where Vout is the voltage across the Load and Vin is the voltage applied at the base?

That's the universal definition of voltage gain, sure. That's how you measure it, but gives no inkling of what theoretically it should be.

My approach was (which is obviously wrong) Vo/Vi = [Rs/(hfe + 1) + re] / [RL(hfe + 1) + re]
Where Rs is the Impedance of the source.

Rs can't enter into the equation for gain, if Vi is the voltage at the base. Whatever voltage is dropped across Rs is immaterial. The loss due to source resistance is excluded by the decision to measure Vi at the base, and not as Vs, for the gain calculation. (So set Rs=0 here.)

[PLAIN]https://www.physicsforums.com/images/icons/icon2.gif Vi and Vo are the small signal voltages—the alternating component sitting atop the DC bias level.

 

[perplexabot]

Thanks for the replies. Such great help.

Gv = RL/(RL + re) is almost unity, which is what you'd expect. This, of course, is Voltage Gain; (the Power gain is much greater - relating to the current gain). I think the value of Vi, used for that formula just relates to the Voltage on the base. If you include a finite series resistance between input volts and the base then this will affect the gain (potential divider effect) Perhaps this is where your problem lies - but you are right to consider the source resistance for many practical applications. Let's face it, you use a buffer when the source impedance is high, dontcha?

Your right. I think I didn't take into consideration the voltage divider effect.

That's what theory predicts, from the small signal equivalent circuit.

That's the universal definition of voltage gain, sure. That's how you measure it, but gives no inkling of what theoretically it should be.

Rs can't enter into the equation for gain, if Vi is the voltage at the base. Whatever voltage is dropped across Rs is immaterial. The loss due to source resistance is excluded by the decision to measure Vi at the base, and not as Vs, for the gain calculation.

[PLAIN]https://www.physicsforums.com/images/icons/icon2.gif Vi and Vo are the small signal voltages—the alternating component sitting atop the DC bias level.

Ok, now I know why Rs can't enter the equation, because the voltage at the base is the voltage that is left after voltage division with Rs.

 

[Jony130]

The AC voltage gain of a Emitter follower we can find using small-signal analysis.
The small-signal T model of emitter follower look like this:

And I hope that you see that Vout = Ie*Re and Vin = Ie *(re + Re)

So the voltage gain

Vout/Vin = Ie*Re/ (Ie*(re + Re)) = Re/(re +Re) (voltage divider equation).

And when we connect the load resistor parallel to Re the gain will change to

Vout/Vin = Re||RL /(re + Re||RL) and if RL<<Re we have RL/(re+RL)

 

[perplexabot]

The AC voltage gain of a Emitter follower we can find using small-signal analysis.
The small-signal T model of emitter follower look like this:

And I hope that you see that Vout = Ie*Re and Vin = Ie *(re + Re)

So the voltage gain

Vout/Vin = Ie*Re/ (Ie*(re + Re)) = Re/(re +Re) (voltage divider equation).

And when we connect the load resistor parallel to Re the gain will change to

Vout/Vin = Re||RL /(re + Re||RL) and if RL<<Re we have RL/(re+RL)

Hmmm. Interesting model. I have not seen it before. Very easy to understand. Thanks for the full analysis.

 

[Jony130]

If you want included Rs resistor we modify the diagram

And we can find overall voltage gain of this circuit

Av = Vb/Vin * Vout/Vb = Rin/(Rs + Rin) * Re/(re+Re)

where Rin is a Emitter follower input resistance.

Rin = (β+1)*(re + Re)

 

[berkeman]

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