Im having trouble to understand the meaning of output impedance of an active circuit and in this case the emitter follower. I have read several information but yet failed to get the meaning. Im looking for an easy but correct definition.

If we call the output impedance of an emitter follower Zout. This is what I understand about the meaning of Zout: If we couple a variable load R and vary it, the output impedance Zout is then the change in Vce relative to the change in I R as I have drawn below?:

enter image description here

Is that the correct meaning of Zout in layman’s term? Definitions containing “Looking into” makes thing more complicated at the moment. If mine wrong could you provide such explanation similar to mine? Im completely confused and this is my maybe tenth time I struggle to understand.

The definition might be the inverse slope of Vce Ic curves but I need a more concrete definition showing also how it is obtained?

  • 1
    \$\begingroup\$ Yes, this is the correct meaning of what is called "output impedance". To be more precise, it is "differential output impedance" in contrast to the "static output impedance" but this is implied. Note it is shown with a slight backward slope in the figure (like negative resistance) but do not pay attention to it. \$\endgroup\$ Dec 2, 2019 at 22:19
  • \$\begingroup\$ It's better stated as: \$Z_\text{OUT}=\frac{\text{d}\,V_\text{E}}{\text{d}\,I_\text{R}}\$. Note that this is the tilt or slope located at some point along some curve. As you can probably guess, this isn't a straight line that has only one slope to it. But actually is a curve where the slope is ever-changing. So the value only has good meaning nearby some point on that curve. You need to know where that point is in order to compute the value. So it's kind of "mathy" in that sense. Abstractly, it's just the tangent you might draw with a ruler if pointed to a spot on a curve drawn on paper. \$\endgroup\$
    – jonk
    Dec 2, 2019 at 22:55
  • \$\begingroup\$ I'll write something. Hopefully, it's more along the lines of "a more concrete definition showing also how it is obtained." \$\endgroup\$
    – jonk
    Dec 3, 2019 at 1:48

3 Answers 3


... I need a more concrete definition showing also how it is obtained

Since the base of the BJT is nailed down hard (zero impedance voltage source), the dynamic output impedance is (you can find the starting equation at this Wiki page on the BJT and the Ebers-Moll model):

$$\begin{align*} \operatorname{D}\,I_\text{E}&=\operatorname{D}\left[I_\text{ES}\left(e^{^\left[\frac{V_{_\text{BE}}}{\eta\,V_T}\right]}-1\right)\right]\\\\ &=I_\text{ES}\,\operatorname{D}\left[e^{^\left[\frac{V_{_\text{BE}}}{\eta\,V_T}\right]}-1\right]\\\\ &=I_\text{ES}\,\:e^{^\left[\frac{V_{_\text{BE}}}{\eta\,V_T}\right]}\operatorname{D}\left[\frac{V_{_\text{BE}}}{\eta\,V_T}\right]\\\\ &=\frac{I_\text{ES}\,\:e^{^\left[\frac{V_{_\text{BE}}}{\eta\,V_T}\right]}}{\eta\,V_T}\:\:\operatorname{D}\,V_{_\text{BE}}\\\\ &\approx \frac{I_\text{E}}{\eta\,V_T}\:\:\operatorname{D}\,V_{_\text{BE}}\\\\&\therefore\\\\ r_e=\frac{\text{d}\,V_{_\text{BE}}}{\text{d}\,I_\text{E}} &= \frac{\eta\,V_T}{I_\text{E}} \end{align*}$$

(\$\eta\$ is the emission co-efficient and is often just taken as \$\eta=1\$.)

There is also some Ohmic base resistance, \$r_b^{'}\$, and Ohmic emitter resistance, \$r_e^{'}\$, to account for. (For small signal BJTs, \$5\:\Omega \le r_b^{'}\le 20\:\Omega\$ and \$50\:\text{m}\Omega \le r_e^{'}\le 400\:\text{m}\Omega\$.)

Roughly speaking, this Ohmic portion adds another \$r_e^{'}+\frac{r_b^{'}}{\beta+1}\$. So the total, including Ohmic and dynamic resistances, is:


(If the voltage source at the BJT's base has some source resistance, then just treat it similarly to how \$r_b^{'}\$ was treated, above.)

The above only accounts for the simplified BJT portion which doesn't include, for example, the Early Effect. It also assumes that the temperature is dead-stable and doesn't move. (The saturation current, \$I_\text{ES}\$, is highly temperature-dependent -- on the order of the 3rd power of the absolute temperature, proportionally. So these equations become seriously bogged down if you want to start taking into account changes in temperature due to changes in the collector current, for example.)

Finally, it doesn't account for \$R_\text{E}\$, which will appear to be "in parallel" with the above formula for \$r_e\$. The value of \$R_\text{E}\$ can be selected so that it is near the expected load current (higher or lower) in order to stabilize the net apparent output impedance (if that is needed for some reason.) However, \$R_\text{E}\$ may be there to provide a very low, minimum load for the circuit, with the output impedance now guaranteed to be no higher than \$R_\text{E}\$.

Because the dynamic resistance portion often dominates, the total value may also change rapidly with variations in the emitter current.

Let's test the above idea using a Spice program to see if the above simplified, theoretical treatment is supported by the vastly more sophisticated calculations used by Spice. I'll avoid the complexities of using the .MEAS statement to automatically compute this. Instead, I'll do it manually and in plain view.

Here's the circuit in LTspice:

enter image description here

From the BJT information, together with an estimated emitter current of \$I_\text{E}\approx \frac{6\:\text{V}-700\:\text{mV}}{1.0\:\text{k}\Omega}\approx 5.3\:\text{mA}\$, we find that \$r_e\approx \frac{26\:\text{mV}}{5.3\:\text{mA}}+200\:\text{m}\Omega+\frac{10\:\Omega}{201}\approx 5.2\:\Omega\$, with most of that coming from the first term. Technically, we'd need to put that in parallel with the \$1\:\text{k}\Omega\$ resistor, dropping it to about \$5.17\:\Omega\$. But I already rounded the above value to the nearest tenth, so this means we'll stick with \$r_e\approx 5.2\:\Omega\$ for a theoretical estimate.

(The .temp card on the above schematic is there so that \$V_T=26\:\text{mV}\$.)

Now let's see what LTspice tells us:

enter image description here

Just by eye, I can read off the following two voltages from above: \$5.303677(6)\:\text{V}\$ and \$5.303682(8)\:\text{V}\$. We know that the injected current is \$1\:\mu\text{A}\$. So we compute, \$r_e=\frac{5.3036828\:\text{V}-5.3036776\:\text{V}}{1\:\mu\text{A}}=5.2\:\Omega\$!!!

Which is remarkably good, as I didn't even try this out before writing the above text.

An important note about the above process is that I didn't inject \$10\:\text{mA}\$. This would have substantially moved the point along that curve I talked about earlier and therefore the computation would be a very different secant instead of an exact tangent. I chose an injection current that was less than a thousandth of the current in \$R_1\$ to test the idea.

That doesn't mean it isn't useful to explore how \$r_e\$ varies with different loads. It's just that if you want to find out the exact tangent value with Spice, you need to keep the change tiny. Otherwise, you get conflated results and you can't use that to verify the earlier theory that I developed.

Just a note.

  • \$\begingroup\$ Thanks a lot for this for both the explanation and example. This really clears up the confusion I had. Just one more question if you don't mind. Otherwise let me know I can open it as another question. You convert the voltage source impedance or ohmic base resistance by simply dividing it to (β+1). I have seen this conversion before and I don't know whether it comes from basic circuit theory or not; but could you also show what is the reason behind this 1/(β+1) factor when we translate the input/source impedance as a part of the output impedance? \$\endgroup\$
    – user1999
    Dec 3, 2019 at 10:51
  • \$\begingroup\$ @user1999 Suppose the emitter current changes by a milliamp. All of that must change the voltage drop through \$r_e^{'}\$. But because the base current only changes by an amount that is \$\beta+1\$ smaller, the change in the voltage drop across \$r_b^{'}\$ is similarly less. In effect, the base Ohmic resistance "looks smaller" when "seen, looking backwards" from the emitter, subjected to emitter current changes. \$\endgroup\$
    – jonk
    Dec 3, 2019 at 12:23
  • \$\begingroup\$ @user1999 Similarly, if instead you were thinking only about base current changes (not likely for this circuit, but it does matter when thinking about CE amplifier input impedance), then base current changes are multiplied before reaching the emitter. So the Ohmic emitter resistance looks larger from the perspective of changes in the base current. That's a good thing for CE amplifier stages as it reduces the emitter's loading effect as seen by the base. \$\endgroup\$
    – jonk
    Dec 3, 2019 at 12:46
  • \$\begingroup\$ @jonk Why did you use the collector current instead of emitter current? \$\endgroup\$
    – G36
    Dec 3, 2019 at 16:00
  • \$\begingroup\$ @G36 I think I'll adjust the text a little and refer to a Wiki page for the equation I'll use. This will slightly change the nomenclature. But it won't harm the clarity of what I wrote. Thanks for making me take note. If you see a mistake here, please let me know. Thanks, again! \$\endgroup\$
    – jonk
    Dec 3, 2019 at 19:17

For the purposes of this qualitative explanation, you can replace the emitter resistor Re with the varying load R(L). Here is my intuitive explanation:

In the circuit of the emitter follower, the input base-emitter voltage of the transistor is formed as a difference between two single-ended voltages - the input voltage VIN and the output voltage VOUT. The task of the transistor is to keep this difference (almost) equal to zero. For this purpose, it “observes” the difference and changes its output voltage in the right direction while zeroing it. As a result of this “game” called negative feedback, the output voltage follows the input voltage; hence the name “follower”. The simpler case is when the input voltage is constant; then the circuit acts as a voltage stabilizer.

To see the meaning of this configuration, let’s disturb the output of this voltage stabilizer. We can do it in various ways but the question is about the case when the load resistance varies. I have illustrated my intuitive explanation of the circuit operation with the picture below:

enter image description here

The IV curves of the transistor and load are drawn in the same coordinate system. Their intersection point represents the operating point (the present current through and voltage across the load).

When, for example, the load resistance increases, its IV curve begins rotating clockwise.. and if this resistor was connected in the collector, the transistor would react in a different manner. Its output IV curve would be immovable… and the operating point would move horizontally along it… i.e., the load voltage would significantly change.

But the negative feedback has totally changed the transistor behavior… and it begins moving down its IV curve simultaneously with the R curve rotation. As a result, the operating point totally changes its trajectory and begin moving down along a new (almost) vertical line. This line represents the extremely low output differential resistance of the emitter follower. It is low since, when the load resistance varies, the current varies... but the voltage does not vary.

So what is the meaning of the low differential output resistance? It is not the ordinary static (ohmic) resistance; it is varying resistance. You can see in this configuration a dynamic voltage divider consisting of two "resistors" - the collector-emitter output part of the transistor and the load. When the load resistance R increases/decreases, the transistor increases/decreases its "resistance" RCE so that to keep the divider ratio K = R/(R + RCE) constant... and accordingly, the output voltage stays constant.

"Looking into" here means that the load R "sees" it is driven by a very good voltage source.


Rout = Re//re

Where re = the intrinsic emitter resistance of the transistor = 25mV/Ie

(This simplification ignores any out output impedance of the battery, which will be small).

The output impedance,Rout forms a potential divider with your variable resistor,R. This potential divider is between the emitter voltage (VB-0.7V) and ground.


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