Trouble understanding output impedance of common collector

Question

I'm self-studying electronics. I'm currently following "The Art of Electronics" (p. 84, 2.2.5 Emitter follower biasing), helping myself with additional videos and tutorials as I go.

I'm stuck trying to wrap my head around the output impedance of the following circuit:

The input impedance looks really simple: It's just the resistance that the AC signal feels when getting into the circuit. I just have to follow the orange path in the following image to understand that it's going to be a parallel circuit made of \$R1\$, \$R2\$ and \$\beta\cdot Re\$ (because that's the impedance of a common emitter). I have come across many sources that agree with this result.

Continuing with that idea, I would imagine that the output impedance is the resistance between the output and \$Vcc\$, because that's what limits the current that can be feed into the output. I have drawn this path in blue.

However, I see that in my book, it is stated that the output impedance would be \$Z_{out}=R_e||(\frac{Z_{in}||R_1||R_2)}{\beta})\$. I have not found sources on the internet that could clarify exactly how or why this result shows up. I have had trouble even finding articles about calculating $Z_{out}$ in this kind of circuit.

The question: What is the intuition behind this result and how can I get there on my own?

\$Z_{out}=R_e||(\frac{Z_{in}||R_1||R_2)}{\beta})\$

EDIT: fixed equation transcription error as noted in comments.

Answer 1

KCL Analysis, sans Intuition

Let's start by ignoring intuition for a moment and just working through the problem. To begin, the schematic:

^{simulate this circuit – Schematic created using CircuitLab}

(For those interested, I've provided the OP's fuller context at the end, below.)

Normally, for small signal purposes, you might also insert \$r_e\$ in the above circuit right at the tip of the emitter of \$Q_1\$. But the textbook is ignoring its value until section 2.3 and considering it absent for now.

You know that \$V_\text{B}-V_\text{E}=V_\text{BE}\$ and that for small signal purposes, absent \$r_e\$, this is a fixed voltage difference. This allows us to substitute one for the other. Also, note that \$I_\text{E}=\frac{V_\text{E}}{R_\text{E}}\$.

Assuming \$I_x\$ is an arbitrary current sink load that will be either \$0\:\text{A}\$ (no load) or \$1\:\text{A}\$ (loaded), then the KCL is:

$$\begin{align*} \frac{V_\text{E}+V_\text{BE}}{R_1}+\frac{V_\text{E}+V_\text{BE}}{R_2}+\frac{V_\text{E}+V_\text{BE}}{Z_\text{IN}}+\frac{\frac{V_\text{E}}{R_\text{E}}+I_x}{\beta+1}&=\frac{V_\text{CC}}{R_1}+\frac{0\:\text{V}}{R_2}+\frac{V_\text{IN}}{Z_\text{IN}}+\frac{\frac{0\:\text{V}}{R_\text{E}}}{\beta+1} \end{align*}$$

(In the above, I've placed the outflowing currents on the left side and the inflowing currents on the right side. Although I've written about this novel approach to KCL many times, a more recent example is shown here.)

So the above equation says

"The current flowing out of the base node through \$R_1\$, plus the current flowing out of the base node through \$R_2\$, plus the current flowing out of the base node through \$Z_\text{IN}\$, plus the current flowing out of the emitter node through \$R_\text{E}\$ as seen by the base node is equal to the current flowing into the base node from \$V_\text{CC}\$ through \$R_1\$, plus the current flowing into the base node from ground through \$R_2\$, plus the current flowing into the base node from \$V_\text{IN}\$ through \$Z_\text{IN}\$, plus the current flowing into the emitter node from ground through \$R_\text{E}\$ as seen by the base node."

If you solve the above for \$V_{\text{E}\left(I_x\right)}\$, then you can work out: \$Z_\text{OUT}=\frac{\Delta \,V_\text{E}}{\Delta\,I_\text{E}}=\frac{V_{\text{E}\left(I_x=0\right)}-V_{\text{E}\left(I_x=1\right)}}{1\:\text{A}-0\:\text{A}}\$:

$$Z_\text{OUT}= \frac{\frac1{\beta+1}\left(\beta+1\right)R_\text{E}\,R_1\, R_2\,Z_\text{IN}}{\left(\beta+1\right)R_\text{E}\,R_1\, R_2+\left(\beta+1\right)R_\text{E}\,R_1\,Z_\text{IN}+\left(\beta+1\right)R_\text{E}\, R_2\,Z_\text{IN}+R_1\, R_2\,Z_\text{IN}}$$

That's exactly the same result you'd get if you took \$\frac1{\beta+1}\left[R_1\mid\mid R_2\mid\mid Z_\text{IN}\mid\mid \left(\beta+1\right)R_\text{E} \right]\$ or, multiplying \$\frac1{\beta+1}\$ through:

$$Z_\text{OUT}=\left[\left(\frac{R_1\mid\mid R_2\mid\mid Z_\text{IN}}{\beta+1}\right)\mid\mid R_\text{E} \right]$$

The only difference here from the textbook being that the authors chose to use \$\beta\$ as an approximation for \$\beta+1\$.

Intuition

Look back at the original schematic. There, you can readily see that \$R_1\$, \$R_2\$, and \$Z_\text{IN}\$ are all tied from a voltage source (assumed ideal) to a shared node at the BJT base. From the point of view of the base, looking at those three impedances from an AC standpoint, they are all in-effect "grounded" and therefore "in parallel" with each other.

Now, since tiny current variations at the base imply much larger current variations at the emitter, the parallel resistance seen at the base will look \$\beta+1\$ times smaller at the emitter. This is then taken in parallel to \$R_\text{E}\$.

So that's where an intuitive view comes from.

The Art of Electronics, 3rd edition, Page 84

The worked problem you cite has \$V_\text{CC}=+15\:\text{V}\$, \$R_1=130\:\text{k}\Omega\$, \$R_2=150\:\text{k}\Omega\$, \$Z_\text{IN}=10\:\text{k}\Omega\$, \$R_\text{E}=7.5\:\text{k}\Omega\$ and \$\beta=100\$. With those values, you should find that \$Z_\text{OUT}\approx 85.59\:\Omega\$ and \$A_v\approx 0.86446\$. The book writes that \$Z_\text{OUT}\approx 87\:\Omega\$, which is quite close enough.

As the book also points out, since the design is for \$I_\text{E}\approx 1\:\text{mA}\$, then the dynamic Ebers-Moll AC impedance value they will discuss later (\$r_e\$) will be about \$26\:\Omega\$. (They say \$r_e\approx 25\:\Omega\$.) This is added, in series and will increase \$Z_\text{OUT}\$ to \$Z_\text{OUT}\approx 112\:\Omega\$. (The book writes it as \$110\:\Omega\$ using their slightly smaller value.)

Getting overly precise is pointless, so the textbook is handling this just as you should: showing at most two digits of precision.

Fuller Context taken from The Art of Electronics, 3rd edition:

The OP failed to provide the worked example that was at question, I think:

They are relying upon a simpler BJT model that does NOT yet include \$g_m\$ and upon earlier discussions about \$Z_\text{IN}\$ and \$Z_\text{OUT}\$ that also help frame the above discussion in the textbook.

Answer 2

(Supplement (another simple approach) at the end)

There is no magic "intuition" behind this problem. Perhaps it is easier for you to apply another view for solving the problem?

At first: There is a severe error in your formula (and in the quoted text): The value (R1||R2)/beta has to be added to the rest (not considered in parallel). Otherwise, the input resistance would be zero in case of a coupling capacitor at the base.

Correction of the wording: The "error" is that the quoted text completely forgets (neglects) the input resistance 1/gm of the BJT alone (at the emitter node). This has led to a kind of misunderstanding on my side because I did not see any addition of two parts (1/gm + ......).

My calculation: Now - you need the input resistance at the emitter node - hence, you can try to find the input resistance for the common base configuration. This will give you the correct answer because - also in your case (common collector) - the coupling capacitor (3µF) will cancel the influence of R1||R2.

So - what do you expect looking into the emitter node when a certain small-signal test voltage v_in=v_e is applied? What will be the corresponding current ? It will be the well-known emitter current i_e. As a first step let us neglect the external resistor RE - at the end, it will be considered in parallel.

Using the transconductance gm=i_e/v_be with v_be=v_b - v_e=-v_e (base grounded) we can solve for the emitter current i_e=gm * (-v_e) and arrive at the input resistance at the emitter node:

r_in=v_e/-i_e=1/gm

Comment 1: Note that we write (-i_e) because in our case the current i_e is going into the emitter node.

Comment 2: As you can see, in your task description the quantity "Zin" is identical to the inverse transconductance gm.

Comment 3: When the base node is NOT grounded (no coupling capacitor) the resistors R1||R2 have to be considered (added). The corresponding value of the parallel combination will increase the input resistance - because this resistance provides signal feedback. However, it is only the small base current i_b which causes a feedback voltage v_e at the base. Therefore, this resistance (R1||R2) enters the expression for r_in reduced by the factor 1/(beta+1) because i_b=i_e/(beta+1).

Final result (no coupling capacitor): r-in=(1/gm) + (R1||R2)/(beta+1)

---

EDIT/Supplement

In the following, you will find another - very simple, intuitive and system-oriented - approach for finding the output resistance re for a common-collector stage.

For calculating the output resistance re, we connect a test voltage ve at the emitter node. The next step is to use only the basic formula ie=gm*vbe and to represent this relation as a small-signal block diagram using vbe=-veb=ve-vb:

^{simulate this circuit – Schematic created using CircuitLab}

(Note: I have used the commonly agreed sign convention for the currents: ib into the base and ie out of the emitter).

Case 1: When the base is grounded (example: source resistor Rout=0), we have no feedback loop (vb=0) and we find the ratio (as expected) re=Ve/(-ie)=1/gm .

Case 2: For a finite value of Rth the feedback loop ist closed (vb finite). From system theory we know that the input resistance is increased due to feedback by a factor (1-loop gain). From the diagram, we immediately can derive the loop gain expression:

Loop gain= - Rth[gm/(1+beta)]

Therefore: re=Ve/(-ie)=(1/gm)[1+Rth*gm/(1+beta)]=(1/gm)+Rth/(1+beta).

Of course - as the last step, the ohmic emitter resstor RE is to be considered in parallel to to re.

Answer 3

Following with interest the heated discussion among my esteemed colleagues, I am once again convinced of how such a brilliant idea can be lost among the many considerations of accurate quantification.

Before you can find the intuition behind the formula, you have to find the intuition behind the circuit solution... and only then continue with the formula... Let's try to do it.

Such ingeniously simple circuit solutions from the 20th century should be explained by even simpler solutions. So let's first clear the circuit of "redundant" (at this initial stage of intuitive understanding) elements - Vin, Zin, C1, C2 and RL. Thus the voltage divider R1-R2 is a source of DC input voltage and the resistor Re plays the role of a load. In other words, this is an emitting follower driven by a constant voltage.

Now imagine that the base-emitter junction of the transistor is a sensitive voltage input (like a galvanometer) that controls the "resistance" Rce of its collector-emitter section (like a "rheostat*). What famous electrical circuit does this look like to you?

Of course, this is the famous Wheatstone bridge from the 19th century... and in particular, a balanced bridge. Its idea is extremely simple. It consists of two voltage dividers: the one of them (R1-R2 on the left) is fixed and it produces Vin (Vb); the other (Rce-Re on the right) is variable and it produces Vout (Ve). The transistor input is connected vetween their outputs like a bridge; hence the name of this topology. Notice something very important here - the resistors on the left divider have a much higher resistance than the resistors on the right divider.

The operation of this bridge is extremely simple and well known. The transistor senses the bridge imbalance through its input (base-emitter junction) and regulates its output "resistance" so as to zero the difference between the two voltages. As a result, the (output) voltage of the right divider follows the (input) voltage of the left divider.

The two voltages are (almost) equal but the currents are very different. So the divider's output (Thevenin) resistances are different... and we use the lower of them to drive the external load. This is the ingenious idea of ​​this famous circuit solution - a low resistance divider copies the output voltage of a high resistance divider.

For example, if R1 = R2 = 100 k and Re = 1 k, then the transistor will initially adjust its collector-emitter "resistance" Rce = 1 k... and the right divider output resistance will be only 0.5 k (versus 50 k of the left divider). Then, if some (input or output) quantity varies, the transistor will vary its Rce so that to keep relatively constant emitter (output) voltage; Re stays constant.

Thus the extremily low output resistance (regarding the signal changes) is due to the extremely low dynamic Rce. Indeed, it sounds strange since all we know that the dynamic output resistance of the transistor is very high... but here it is modified (decreased) by the voltage-type negative feedback.

Looking from the side of the external load, we see two cascaded voltage dividers in parallel... and the one with low resistance dominates. Actually, all their resistances are in parallel as the formula says. Note that the low Rce is represented by the term of (beta + 1) in the denominator.

---

I assume that you will not appreciate my story but will include it among the many other explanations on the web. But let me still give some clarification.

I first came across this circuit in the late 60's when, in technical school, they "explained" it to me with complex formulas... but I needed such an explanation. Later, at university, they explained it to me with even more complex formulas... and I was still looking for such an explanation.

Even later, as a teacher at the same university, I sought such explanations for my students... and I have been doing this until now. And tonight, reading the (extremely interesting) discussion here, it occurred to me to explain in this way, through the Wheatstone bridge, how the emitter follower reduces the source resistance many times over. That is how hard ideas mature... and how important it is to have such a creative atmosphere for their appearance...

Answer 4

I know this thread is old and dead but you did want a simple intuitive answer to your inquiry…and so here it is. Imagine applying a source signal directly at the emitter. Initially the source would just see the base impedance (plus emitter resistance re) in parallel with RE…but then all of a sudden…the transistor wants to yank on the source current (because of the current gain) – it wants to draw way more current from the source than it had originally bargained for (reflecting a drop in impedance as seen by the source). Or at least that's how I see it.