# Common Emiter-Emitter Follower to speaker

Discussion in 'Homework Help' started by Joanna91, Jun 18, 2012.

1. ### Joanna91 Thread Starter New Member

Jun 18, 2012
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0
Hello =) I am new to Electronics and i'm struggling with this exercise, I hope somebody can help~

http://tinypic.com/r/2j5c1l4/6

It reads: the Common Emitter has Voltage gain 20 without external load. a)Calculate the voltage gain when the common emitter leads to a 8Ω resistance speaker (left circuit)
b)Calculate the voltage gain with a common collector betwen the common emmiter and a speaker (right circuit)
Va=inf, b=100, I1=5mA

I tried the first one, and I came up with a 17,7 Voltage gain. If the CE has a Voltage Gain of 20=> uo/ui=20 <=> (-gm*ub*Rc)/(ib*uπ)=20 (1)

We want to calculate A'=uo/ui=(-gm*ub*(Rc//8))/(ib*uπ) (2)

(1)/(2) A'=17,7. I am not sure if it's correct so I would love some advice.

I'm stumped about the second part, mostly because I'm not sure what to do with the current source.

Last edited: Jun 18, 2012
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,174
1,188
If the voltage gain Av is equal 20 for no load resistor.
And this means that re is equal to

Av = Rc/re

re = Rc/Av = 1K/20 = 50Ω

re = 1/gm

And now if we connect the load resistor RL = 8Ω the voltage gain will drop to:

Av = (Rc||RL)/re ≈ RL/re ≈ 0.16V/V

But if we use the emitter follower as a buffer the voltage gain will drop to

Av = (Rc||Rin)/re

where Rin ≈ (β+1)*(Rsp + re2) = 101*(8Ω + 5.2Ω)≈ 1.34kΩ

Av = Av = (Rc||Rin)/re = (1KΩ||1.34KΩ)/50Ω ≈ 572.6Ω/50Ω = 11.4V/V

Last edited: Jun 18, 2012
3. ### Joanna91 Thread Starter New Member

Jun 18, 2012
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0
Thank you very much for your answer! If you would be as kind to answer some questions, since I'm a huge noob xD How does the Av = Rc/re equation come up?

Thanks again!

Last edited: Jun 18, 2012
4. ### WBahn Moderator

Mar 31, 2012
20,226
5,750

Have you been through small signal models for transistors, both where they come from and what they mean?

That first circuit is a really bad design because it isn't biased. It will be highly nonlinear, not responding at all (or only in a tiny miniscule way) for negative portions of the input signal and very nonlinearly for the positive portions and becoming a short-lived space heater for any signal that goes higher than about 0.75V.

An emitter resistance, re, implies a bias current in the emitter of about 0.5mA, but there is nothing in the circuit to establish that.

Now, lot's of authors are, unfortunately, really, really sloppy and do not make a clear distinction between the DC bias circuit and the small-signal model and so draw circuits that are hodge-podge mixtures of the two. That may well be what is happening here.

Last edited: Jun 18, 2012
5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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As you shoudl know the voltage gain of a Common Emitter we can find by using a small signal analysis. If you do this kind of analysis you will end up with this equation for the voltage gain.

Av = gm * Rc = Ic/Vt * Rc ≈ Ic/26mV * Rc ≈ 40*Ic*Rc

But sometimes is it easier to look at this equation from a different angle.
If we replace gm by using a re - transistor's internal emitter resistance, approximately equal to re = 1/gm = 26mV/Ic .
We now can change our voltage gain equation to this more elegant form .

Av = gm * Rc = 1/gm * Rc = Rc/re

As for re2,

re2 - common collector transistor internal emitter resistance.

re2 = 26mV/Ic = 26mV/5mA = 5.2Ω

6. ### Joanna91 Thread Starter New Member

Jun 18, 2012
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0
I have the basic knowledge of bjt transistors, and I did the small signal analysis using the π-hybrid model, that doesn't have an re resistor so I was a bit confused.

Thank you for shedding some more light on the subject!

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But you have $r_{\pi}$ resistor instead ??

$r_{\pi}$ - is a small signal resistance seen from base terminal looking into the BJT with emitter shorted to ground.

$r_{\pi} =( \beta + 1)*re$

$re$ - is a small signal resistance looking into the emitter terminal with base shorted to ground.

$re = \frac{Vt}{Ie}\approx\frac{26mV}{Ic} = \frac{1}{gm}$

8. ### WBahn Moderator

Mar 31, 2012
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5,750
When you went through the small-signal analysis, was it made clear that all that was being done was that the principle of superposition was being applied such that all of the supplies (power supplies as well as signal supplies) were being divided into two groups. The first group contains all the static (i.e., DC) supples and the second group contains all the time-varying supplies (normally just the input signal of interest).

By convention, we use lower case variable with uppercase subscripts to denote the total solution, all uppercase to denote the quiscient (a.k.a. DC, bias, operating, large-signal) solution and all lowercase to denote the small signal response. Hence:

$
i_E = I_E \ + \ i_e
\
v_{BE} = V_{BE} \ + \ v_{be}
$

The circuit is then analyzed with the DC supplies turned on and the others turned off and the "operating point" of the transistors is found. Before we can do the second analysis, with the DC supplies turned off and the signal supplies turned on, we have to deal with the fact that transistors are highly nonlinear devices and that superposition is invalid for nonlinear circuits. So, instead, we make the assumption that the applied signal is sufficiently small such that the circuit can be reasonably modelled as a linear circuit about its DC operating point. This then leads to creating a model of a transistor that reflects the differential behavior of the transistor -- i.e., how much the collector voltage changes in response to a change in base emitter voltage, and so on. There are two equally valid models that you can come up with -- the T model and the hybrid-pi model. Some analyses are easier with one than the other and you get to pick and choose which you use and you can change between the two in midstream if you want.

In the T-model, there is a emitter resistor, re, between the base and the emitter; there is also a current source from the collector to the junction of the base and the emitter resistor. A resistor relates the voltage across it to the current through it. So this resistor relates the base-emitter voltage to the emitter current (it does NOT relate it to the base current because the base current is not the current through the emitter resistor due to the node connecting it to the current source as well). Since this a differential resistance, it is actually relating the change in one to the change in the other. So:

$
r_e \ = \ \frac{dv_{BE}}{di_E} = \frac{v_{be}}{i_e}
$

You then use the constitutive relation for a BJT transistor, namely

$
i_C \ = \ I_S \left( e^{\frac{v_{BE}}{V_t}} \ - \ 1 \right)
$

In the active region, the "-1" is insignificant, so you have:

$
i_C \ = \ I_S \ e^{\frac{v_{BE}}{V_t}}
$

You also have

$
i_E \ = \ \alpha i_C
$

so

$
i_E = \alpha I_S \ e^{\frac{v_{BE}}{V_t}}
$

Taking the derivative with respect to v_BE, we have:

$
\frac{di_E}{dv_{BE}} \ = \ \alpha I_S \ e^{\frac{v_{BE}}{V_t}} \left( \frac{1}{V_t} \right)
\
\frac{di_E}{dv_{BE}} \ = i_E \left( \frac{1}{V_t} \right)
$

Since we are assuming that the change in the emitter current is relatively small, we can estimate it as simply being the same current, I_E, as in the DC solution. This gives us:

$
\frac{di_E}{dv_{BE}} \ = \ \frac{i_e}{v_{be}} \ = \ \frac{I_E}{V_t}
\
r_e \ = \ \frac{V_t}{I_E}
$

Do yourself a really big favor and don't attempt to memorize these models and how the parameters relate to the quiscient (i.e., DC) solution. Instead, take some time to comprehend them well enough so that you can draw the two models (T and hybrid-pi) and, from there, derive the relationships as I have done for r_e directly from the constituitive relation. Once you can do it for a BJT, then you should be able to do it for a FET without looking at anything (other than perhaps looking up the constituitive relation). As you use the models to solve problems, the relationships, like Ohm's Law, will become second nature. But if you just memorize a bunch of formulas, you will likely never gain the necessary comprehension to do anything more than be a circuit monkey (someone that blindly throws equations at problems hoping that they happen to grab the right one since they don't really know when those equations do and don't apply).

9. ### Audioguru AAC Fanatic!

Dec 20, 2007
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911
Nobody makes an amplifier like that so why are you talking about those horrible circuits??

10. ### Joanna91 Thread Starter New Member

Jun 18, 2012
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0
Because I thought this was the homework section, and this is my homework.

Thank you Jony130 and WBahn! I will keep studying to gain a better understanding of circuits!

11. ### Audioguru AAC Fanatic!

Dec 20, 2007
9,454
911
Didn't you study about the classes of amplifier outputs?
A class-A amplifier like you show is a HEATER that wastes a lot of current and makes heat all the time even when the amplifier is not playing.
A class-B amplifier has no current when it is not playing but has a lot of horrible crossover distortion.
A class-AB amplifier is used by 90% of audio amplifiers and has a very low current when not playing and has very low distortion.
A class-C amplifier is used for some radio circuits that are tuned and not for an audio amplifier.
A class-D amplifier has the output transistors switching completely on and off (at a high frequency) so they remain cool. Pulse-width-modulation modulates the width of the pulses which are filtered into audio. Many new audio amplifiers are class-D today.

12. ### WBahn Moderator

Mar 31, 2012
20,226
5,750
That's irrelevant here. The point isn't whether it is a good application of this circuit or whether or not she knows whether it is a good application of this circuit. The point is that, good or not, it IS the homework that she has been assigned to do.

I don't know where she is in her curriculum or how her curriculum is structured, but it is very likely that they are just starting to get into amplifier circuits at all and you have to start somewhere.