This might sound a stupid question but I'm confused with the following excerpt:

enter image description here

If Vbe is constant 0.7V which means ΔVbe is zero, does that mean the dynamic resistance is zero?

I think I have problem with interpretation of the graph above. If there is no change in the Vbe signal and Vbe is kept always constant at 0.7V, can we still talk about any kind of resistance here?

  • \$\begingroup\$ For a semi-ideal diode, sure! Ideal part being: \$V_{be}\$ is constant 0.7 V. - Non-ideal part being: \$V_{be}>0\$. - For a real, non-ideal diode, then the ideal part will be removed. The "constant" \$V_{be}\$ is not in reality constant. \$\endgroup\$ – Harry Svensson Oct 30 '17 at 4:00

If Vbe is constant 0.7V which means ΔVbe is zero, does that mean the dynamic resistance is zero?

All that means is that you can't experimentally measure it, since you are refusing to vary \$V_{BE}\$. On the other hand, you can compute it from theory (with or without the Early Effect -- which I see in your chart.)

For a BJT, the following equation is broadly true over quite a range of collector currents (often 4 or 5 or even 6 orders of magnitude):

$$\begin{align*} I_C&=I_{SAT}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\\\\ I_B&=\frac{I_{SAT}}{\beta}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right) \end{align*}$$

Now, there is also an Early Effect that can be approximated this way:

$$\begin{align*} I_C=\frac{I_{SAT}}{1+\frac{V_{BC}}{V_A}}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\\\\ I_B=\frac{I_{SAT}}{\beta}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right) \end{align*}$$

But even if you don't actually vary \$V_{BE}\$, it can be computed from theory.

Dynamic resistance is the local slope of a curve where the y-axis is voltage at the x-axis is current; or else it is the multiplicative inverse of the slope where the y-axis is current and the x-axis is voltage.

[Slope calculations are the meat and potatoes of 1st year calculus, starting with the limit theorem definition of the derivative of a function (grounded rigorously in mathematics by Dedekind and Weierstrass in the mid-late 1800s and impressively complemented by Abraham Robinson's "Non-Standard Analysis" work in the early to mid 1960's.)]

Just take the derivative and you are done. Using the version without the Early Effect and applying the D() operator (see Heaviside):

$$\begin{align*} \operatorname{D}\left(I_C\right)&=\operatorname{D}\left(I_{SAT}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\right)\\\\ &=I_{SAT}\operatorname{D}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\\\\ &=I_{SAT}\operatorname{D}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}\right)\\\\ &=I_{SAT}\:e^{\left[\frac{V_{BE}}{n\:V_T}\right]}\operatorname{D}\left(\frac{V_{BE}}{n\:V_T}\right)\\\\ &=\frac{I_{SAT}\:e^{\left[\frac{V_{BE}}{n\:V_T}\right]}}{n\:V_T}\operatorname{D}\left(V_{BE}\right) \end{align*}$$

At this point, it's pretty easy to see that for almost all useful, realistic values in life:

$$I_C=I_{SAT}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}-1\right)\approx I_{SAT}\left(e^{\left[\frac{V_{BE}}{n\:V_T}\right]}\right)$$


$$\begin{align*} \operatorname{D}\left(I_C\right)&=\frac{I_{SAT}\:e^{\left[\frac{V_{BE}}{n\:V_T}\right]}}{n\:V_T}\operatorname{D}\left(V_{BE}\right)\\\\ \operatorname{D}\left(I_C\right)&=\frac{I_C}{n\:V_T}\operatorname{D}\left(V_{BE}\right)\\\\ &\therefore\\\\ \frac{\operatorname{d}\left(V_{BE}\right)}{\operatorname{d}\left(I_C\right)}&=\frac{n\:V_T}{I_C} \end{align*}$$

Which is the slope of the curve at any given value for \$I_C\$. (Note that \$V_T=\frac{k\: T}{q}\$, which is approximately \$26\:\textrm{mV}\$ at room temperatures. Also note that \$n\$ is the emission coefficient and, for small signal BJTs, is almost always very close to 1. [It is never less than 1.] It is almost always more than 1 when discussing diodes and LEDs, but can also be more than 1 for BJTs. But usually just take it as 1, unless there are reasons to say otherwise.)

This value is assigned to the emitter and not the collector, so the value you want to compute is:

$$\begin{align*} r_i&=\frac{n\:V_T}{I_E} \end{align*}$$

At \$I_C=0.5\:\textrm{mA}\$ (and \$n=1\$) you'd expect \$r_i\approx 52\:\Omega\$ (as \$I_E\approx I_C\$.)

Note that this is in the range that they said. No surprise. It's based on the physics going on. So it pretty much has to be that way. Do note, though, that their range is mentioned in a narrow context. But the above equation applies over a very, very wide context they do not discuss. So their range is not a limitation. You should use the equation and not use their range. The equation tells you the value. Their range does not. Their range is was given because of the context of their discussion.

Take careful note that \$r_i\$ is heavily dependent on the collector/emitter current and that changing the quiescent point changes this dynamic resistance. Also take careful note about the fact that it is also dependent on temperature. (Finally, take careful note that if your emitter resistor is small-valued, then \$r_i\$ in the common base mode matters more and is temperature dependent and this leads to a circuit whose behavior is too easily varied as ambient temperatures vary.)

There are adjustments made due to the Early Effect (modeled with \$V_A\$ parameter.) But I think we can avoid this for now, as it is very simple to include in the above calculations, if needed.

If there is no change in the Vbe signal and Vbe is kept always constant at 0.7V, can we still talk about any kind of resistance here?

There is always a slope to a curve, even if you aren't changing the point on the curve that you are sitting at and can't yet tell what the slope might be. If you are sitting on the road up a steep hill and not moving, you might not notice the slope. But the slope is still there and if you start moving up or down the road you will soon find out about it, too. Kind of like that, I suppose.

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  • \$\begingroup\$ Very thorough informative answer, thanks. Just one more question about these. At the end of the day ri is always between 20 to 100 Ohm. Which means its value is small. Is it important to quantify it in any application? In other words, does the value of ri have significant effects on the other parameters? Let's say we increase the biasing and quiescent current and this changes the ri. ri becomes smaller. Do we have to quantify it or use this change to update the circuit? Or we just say ri decreases and thats it? \$\endgroup\$ – user1245 Oct 30 '17 at 10:08
  • \$\begingroup\$ @user134429 Sorry, I should have added details about n. It's called the "emission coefficient" and is almost always very close to 1 for small signal BJTs. It is more important for diodes. Just think of its value is close to 1 here. \$\endgroup\$ – jonk Oct 30 '17 at 10:25
  • \$\begingroup\$ @user134429 Also, no. \$r_i\$ can be quite a bit less or more -- as it is determined by the collector current. You can think of it as there being a small battery caused by the thermal voltage \$V_T\$ (temperature causes atomic jostling yielding a small voltage) and that the emitter current through this battery implies a certain impedance. This impedance also applies in the common emitter mode and pretty much anywhere a BJT is used, at all. \$\endgroup\$ – jonk Oct 30 '17 at 10:26
  • \$\begingroup\$ @user134429 In the case of the common emitter arrangement, \$r_i\$ is actually part of the input so here called \$r_i\$. But in the common emitter arrangement, since the base is where the input is arriving, it is usually called \$r_e\$. But it is the same value and is there for the same reasons. It's just now in the "common emitter" part of the circuit, instead. \$\endgroup\$ – jonk Oct 30 '17 at 10:28
  • \$\begingroup\$ @user134429 There are Ohmic resistances (lead resistance, for example, or resistance due to the actual doped silicon bulk resistance.) And \$r_i\$ is independent of those and adds to them. The base of a small signal BJT might, because it is lightly doped, have a bulk resistance of 20 Ohms, for example. But since the emitter is the most heavily doped, it's bulk resistance is usually a lot smaller -- less than an Ohm, often. But \$r_i\$ is due to the differential equation I derived for you and exists over many, many orders of magnitude of collector current and regardless of how a BJT is used. \$\endgroup\$ – jonk Oct 30 '17 at 10:31

I believe your interpretation is slightly off. The dynamic input resistance is essentially just the slope of the V-I graph, assuming it is linear over some small range. This slope changes depending on the actual voltage, because of course the graph is not linear. Delta Veb in this case is not implying the signal is actually changing, just that we are looking at a small, linear portion of the graph.

That is what the excerpt means when it says "the value of ri varies with the point of measurement", depending on what value Veb is, the resistance will be different, but it does not need to be changing in time.

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    \$\begingroup\$ So at a "constant" 0.7V Vbe there is dynamic resistance. So when the ac input very small we assume ri as constant? Is that the idea? \$\endgroup\$ – user1245 Oct 30 '17 at 2:28
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    \$\begingroup\$ Yep, it's often called small-signal analysis. \$\endgroup\$ – jramsay42 Oct 30 '17 at 3:20

Dynamic resistance is the ratio of change in voltage to corresponding change in voltage at some DC bias. Usually this change is much smaller than the DC bias. And this resistance is also called as small signal ac resistance.

Now if you are not changing the input, output also won’t change. Basically you can not measure the dynamic response by giving static input. Doesn’t mean that dynamic resistance is zero.

The inverse of slope of IV curve at any bias gives the dynamic resistance at that point.

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