[Tutorial] Il metodo del solfeggio impedenziale

Messaggioda elgiovo » 02/06/2011, 22:39

In seguito a questa discussione, mi è venuta l'idea di scrivere e tenere aggiornato un tutorial sul metodo del "sofeggio impedenziale". In particolare, mi piacerebbe mostrare i vantaggi che offre rispetto all'approccio classico basato sui circuiti equivalenti per piccolo segnale. Ci tengo a precisare che non è una mia idea, ma è alla base degli insegnamenti dei corsi di elettronica al Politecnico di Milano, e che si può fare riferimento ai seguenti testi:

- Teoria

- Esercizi
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BJT - Fundamental facts

Messaggioda elgiovo » 02/06/2011, 22:55

I was forgetting one important thing: I'm going to write in English (and all the replies will have to be in this language, or they will be otherwise ignored). This is the second consequence of the aforementioned discussion. :twisted:

Let's start with some important facts on the bipolar transistor, which should already be part of your background if you're studying electronics.

1) The transconductance is given by

\( \displaystyle g_m=\frac{\partial I}{\partial V_{BE}}=\frac{I}{V_{th}} \)

2) The base-emitter-collector currents are related this way:

\( \displaystyle i_e=(\beta+1) i_b \) , \( \displaystyle i_c=\frac{\beta}{\beta+1} i_e=\alpha i_e\simeq i_e \)

3) From the definition of \( \displaystyle g_m \) , in the small-signal regime

\( \displaystyle i_e=g_m v_{be} \)

holds. This very important relation will be used widely, and it states that the availability of current of a transistor is directly proportional to its \( \displaystyle v_{be} \) through its transconductance. It is basically the linearization of the transistor model, which is strongly nonlinear (it follows an exponential law).
Ultima modifica di elgiovo il 02/06/2011, 23:48, modificato 1 volta in totale.
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BJT - Base, Collector and Emitter impedances

Messaggioda elgiovo » 02/06/2011, 23:37

The method we are going to follow to find these impedances is the following:

- apply a voltage or current generator with arbitrary values \( \displaystyle v_s \) or \( \displaystyle i_s \) to the point where we're reading the impedance;
- read the consequent current/voltage, which are the circuit response to the stimulus;
- find the impedance by application of Ohm's law:


\( \displaystyle Z=\frac{v_s}{i_s} \) .

Let's start with the base impedance. We are applying a voltage source:

Immagine

Of course, \( \displaystyle v_b=v_s \) . We need to find the emitter voltage. This is simply done:

\( \displaystyle v_e=R_e g_m(v_s-v_e) \)

where we have used Ohm's law and the fundamental transistor law \( \displaystyle i_e=g_m v_{be} \) . This leads to

\( \displaystyle v_{be}=v_s \frac{\frac{1}{g_m}}{\frac{1}{g_m}+R_e} \) . We are already able to write down

Remarkable fact n. 1

The voltage on the base of a BJT undergoes a partition between the b-e junction and the emitter resistance. Everything acts as if the junction had a resistence \( \displaystyle \frac{1}{g_m} \)

The value of the base impedance is readily found: since

\( \displaystyle i_e=(\beta+1) i_s=g_m v_s \frac{\frac{1}{g_m}}{\frac{1}{g_m}+R_e} \) ,

must be

\( \displaystyle Z_b=\frac{v_s}{i_s}=\frac{(\beta+1)}{g_m}+(\beta+1) R_e \) .

If one adds a series resistance \( \displaystyle R_b \) to the base, the expression becomes

\( \displaystyle Z_b=R_b+\frac{(\beta+1)}{g_m}+(\beta+1) R_e=R_b+r_{\pi}+(\beta+1) R_e \) .

Otherwise stated, the impedance on the emitter of the bjt is multiplied by the current gain \( \displaystyle \beta \) if seen from the base. This is very useful when trying to build an efficient voltage follower, since very high input impedances are needed.
Ultima modifica di elgiovo il 03/06/2011, 09:18, modificato 1 volta in totale.
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Messaggioda elgiovo » 03/06/2011, 00:32

Let's go on with the emitter impedance. This time we apply a current generator \( \displaystyle i_s \) :

Immagine

The base current is \( \displaystyle i_b=\frac{i_s}{\beta+1} \) , the base voltage \( \displaystyle R_b i_b \) , thus

\( \displaystyle i_s=g_m\left(\frac{R_b i_s}{\beta+1}-v_s\right) \)

and

\( \displaystyle Z_e=\frac{1}{g_m}+\frac{R_b}{\beta+1} \)

The effect is then the opposite of the preceding one: looking from the emitter, the base resistance is divided by \( \displaystyle \beta+1 \) .

Last but not least, what is the collector impedance? Without considering the Early effect (which considerably complicates all the equations, but will be considered in a later moment since it is quite relevant), the answer is simple:

\( \displaystyle Z_c=\infty \) .

This is because one cannot drive a transistor from the collector (or from the drain): the current is a function of \( \displaystyle v_{be} \) only, thus any change in collector voltage doesn't produce any change in collector current, which leads to the previous formula.

Believe it or not, these formulas will allow us to find everything we want (voltages and currents) in any circuit made by transistors, so they have to be learnt by heart. Don't worry, this "terrible effort" will be paid off abundantly. :D
We recall them here for convenience:

\( \displaystyle Z_b=\frac{\beta+1}{g_m}+(\beta+1)R_e \) ;

\( \displaystyle Z_e=\frac{1}{g_m}+\frac{R_b}{\beta+1} \) ;

\( \displaystyle Z_c=\infty \) .

For the sake of simplicity, we can approximate \( \displaystyle \beta+1 \) with \( \displaystyle \beta \) quite safely, since \( \displaystyle \beta \) is of the order of \( \displaystyle 10^2 \) :

\( \displaystyle Z_b=\frac{\beta}{g_m}+\beta R_e \) ;

\( \displaystyle Z_e=\frac{1}{g_m}+\frac{R_b}{\beta} \) ;

\( \displaystyle Z_c=\infty \) .
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MOS impedances

Messaggioda elgiovo » 04/06/2011, 16:03

Having analyzed the BJT resistances, the MOS ones are straightforward, since we only need to consider \( \displaystyle \beta=\infty \) in the previous equations. The other fundamental difference is in the value of the transconductance:

\(\displaystyle g_{m,\text{MOS}}=\frac{2I}{V_{ov}}=2\sqrt{k I}=2 k V_{ov} \)

where \( \displaystyle V_{ov}=V_{GS}-V_T \) is the overdrive voltage, \( \displaystyle I \) is the large signal current, \( \displaystyle k=\frac{1}{2}\mu C_{ox}\left(\frac{W}{L}\right) \) is the MOS conductance.

Immagine

The MOS equations are then

\( \displaystyle Z_g=\infty \)

\( \displaystyle Z_d=\infty \)

\( \displaystyle Z_s=\frac{1}{g_m} \)

These are quite simple, aren't they? :D

Also for the MOS, we have a

Remarkable fact n. 1bis

The voltage on the gate of a MOSFET undergoes a partition between \( \displaystyle \frac{1}{g_m} \) and the source resistance, and the corresponding \( \displaystyle v_{gs} \) is the voltage across \( \displaystyle \frac{1}{g_m} \) :

\( \displaystyle v_{gs}=v_g\frac{\frac{1}{g_m}}{\frac{1}{g_m}+R_s} \)

Of course, the fundamental relation of the MOS transistor is now

\( \displaystyle i_d=i_s=g_m v_{gs} \) .
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The CE configuration

Messaggioda elgiovo » 12/06/2011, 11:23

Now that we have learnt the impedances seen from a transistor's terminals, we can analyze the most important circuit configurations. They are quite simple, nonetheless they turn out to be very useful, because of their properties. We start with the common emitter configuration, or as it is usually abbreviated, the CE configuration.

Immagine

Of course, we assume that the circuit is already polarized, and the large signal current flowing in the bipolar transistor is known. Furhtermore, we assume that the polarization grants a voltage of about \( \displaystyle 0.7 \text{V}=V_{\gamma} \) between the base collector and ground, so that the BJT is turned on. After that, we will have to check that it is working in the correct region. In fact, it may be whether in saturation (i.e. both BE and CB junctions turned on) or in active region (i.e. BE junction turned on, CB junction turned off). In analogic applications, we never want the BJT transistor to be saturated, otherwise its parameters are subject to degradation. For example, in saturation it is no more true that \( \displaystyle i_c=\beta i_b \) , in fact the active CB junction injects electrons in the base, and the transistor is forced to satisfy \( \displaystyle i_c < \beta i_b \) .

So, assuming the BJT is in forward active region, we may ask ourselves a few questions:

1) What is the small signal gain of the transistor?
2) What is the input impedance of the circuit?
3) What is the output impedance of the circuit?

Let's answer with order.

1) This is straightforward. Actually, we already analyzed this configuration, studying the BJT impedances. The collector current is, of course,

\( \displaystyle i_c=g_m v_{be}=g_m v_{in} \)

and this is the same current flowing in the load resistor \( \displaystyle R_c \) , so the answer is

\( \displaystyle G_{CE}=\frac{v_{out}}{v_{in}}=-g_m R_c \)

The minus sign is due to the current flowing away from \( \displaystyle R_c \) , which holds for positive \( \displaystyle v_{in} \) . So, CE is an inverting stage.

2) Again, straightforward. What could it be, other than

\( \displaystyle Z_{in}=\frac{\beta}{g_m}=r_{\pi} \) ?

3) This is not totally straightforward, but it is 98% straightforward. Collector impedance is infinite, but it has \( \displaystyle R_c \) in parallel, so...

\( \displaystyle Z_{out}=R_c \) .

That was easy, wasn't it?
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Re: [Tutorial] Il metodo del solfeggio impedenziale

Messaggioda raffamaiden » 22/08/2014, 21:38

Thanks for the tutorial. Images are gone, please substitute it. Please note that you can now use FidoCadJ to insert circuits in your messages. I'd like to mention an article of mine regarding the same subject, applied to transdiode configuration
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