For AC equivalent circuits, it is written that, in the case of small input signals the emitter diode does not rectify, instead it offers a resistance called AC resistance. Why is this so? Why doesn't it rectify?


instead it offers a resistance called ac resistance. why it is so ? why dont it rectify ?

Let the voltage across the junction be of the following form

$$v_{BE} = V_{BE} + v_{be}$$

where \$V_{BE}\$ is positive and constant while \$v_{be}\$ is AC and is, in some sense, small (I'll later clarify what that means).

Now, recall the equation for the base current:

$$i_B = \frac{I_S}{\beta}e^{\frac{v_{BE}}{V_T}}$$

Substituting the form for \$v_{BE}\$ given earlier yields

$$i_B = \frac{I_S}{\beta}e^{\frac{V_{BE} + v_{be}}{V_T}}=\frac{I_S}{\beta}e^{\frac{V_{BE}}{V_T}}e^{\frac{v_{be}}{V_T}}$$

But, the quiescent (no signal) base current is just

$$I_B = \frac{I_S}{\beta}e^{\frac{V_{BE}}{V_T}}$$

Thus, the total base current is

$$i_B = I_Be^{\frac{v_{be}}{V_T}}$$

The crucial next step is to assume that \$v_{be}\$ is small enough \$^1\$ such that we can approximate the exponential term with

$$e^{\frac{v_{be}}{V_T}} \approx \left(1 + \frac{v_{be}}{V_T}\right)$$

In that case, we can write the base current in the following form:

$$i_B = I_B + i_b $$


$$i_b = \frac{I_B}{V_T}v_{be} = \frac{v_{be}}{r_{\pi}}$$

We call \$r_{\pi} = \frac{V_T}{I_B}\$ the small signal base-emitter resistance but it's not a true resistance, it's really just the (inverse) slope of the base-emitter IV curve at the point \$I_B, V_{BE}\$.

So, as long as the junction voltage is approximately the constant \$V_{BE}\$ with only a small variation \$v_{be}\$ added, the current is approximately the constant \$I_B\$ with a small variation \$i_b\$ added that is linearly proportional to \$v_{be}\$.

\$^1\$ e.g., \$v_{be}\lt \lt 4V_T \approx 100\mathrm{mV}\$


The full operation of the circuit is analyzed in two parts: the small-signal model and the bias circuit.
In the bias circuit, one considers the nature of the rectifier base-emitter base-collector junctions to establish the operating point (point Q), together with external components needed.

In the small signal model, considering all the factors that alter the AC signal to be amplified. In this case, it is considered that the base-emitter diode, polarized, can be viewed as a resistance.
The value of this resistance is determined by the ratio VI for that diode. This "linearization" is valid only when the AC signal is small compared to the levels of DC. If the value of AC is greater, you might consider not only the resistance of the diode.

The full operation of the circuit, is established by the superposition of two models.


Because if you look closely enough at a small enough portion of a nonlinear curve it looks pretty much like a straight line.

There's a detailed mathematical treatment here, but I'll also reproduce the Wiki graph to illustrate intuitively (I hope) what is going on:

enter image description here

In the small signal model we are looking at very small changes from the bias (operating point) Id and Vd on the above graph.

If you look closely near the operating point, you can see that a straight line is a pretty good approximation of the behavior near the operating point.

enter image description here

You need large signal analysis (such as the Shockley equation) to determine the operating point. For small deviations around the operating point, it's often advantageous to assume linear behavior.

It's a mathematical model- and models will inevitably deviate from reality in one way or another. For example, some nonlinearity remains so that if you put a tiny and pure sine wave voltage across a biased diode, the current will not be quite a pure sine wave.

Incidentally, this is of practical concern when RFI (say millivolts or tens of millivolts) is present. The slight nonlinearity can cause rectification and a resulting DC offset of many microvolts- which can be very significant in precision instrumentation applications.

  • \$\begingroup\$ I'll swap you an answer expansion for one of mine... \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 19 '14 at 15:00
  • \$\begingroup\$ You want to see some math? \$\endgroup\$ – Spehro Pefhany Aug 19 '14 at 15:05
  • \$\begingroup\$ Well (I think) the model also assumes the transistor is turned on. Vbe is always positive, so no rectification. \$\endgroup\$ – George Herold Aug 19 '14 at 15:17
  • \$\begingroup\$ Maybe the question just went a little over my head. I don't claim to know, well, much. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 19 '14 at 15:25
  • 2
    \$\begingroup\$ @sherrellbc If there is nonlinearity that is not exactly symmetrical about the zero point of the input (in this case, the operating point), then a sine wave input (mean value exactly zero) will not have a mean value of zero on the output. Like a rectifier. In the inertial sensor world we sometimes call this "vibration rectification". Even though subsequent stages may filter out the unwanted sine wave itself, you're left with a DC offset. \$\endgroup\$ – Spehro Pefhany Aug 19 '14 at 19:56

It is rather a property of the small signal model than a property of the diode you are applying the model to.

It is the very purpose of the small signal model to look at any non linear device as if it was a linear device, i.e. a resistor. Of course it can only be applied with restrictions, e.g. that the function is differentiable at the operation point.

Here is another useful link.

  • \$\begingroup\$ If I recall correctly, one of these major restrictions is that the "small-signal" be no larger than 5-10mV peak or so. \$\endgroup\$ – sherrellbc Aug 19 '14 at 15:37
  • 2
    \$\begingroup\$ @sherrellbc, the restriction is that the errors in the approximation are no larger than what is important for your application. 5-10 mV might be a reasonable value in lots of circuits, but others might be able to take a 1 V signal as small-signal and for others the small-signal approximation might be inaccurate for a 1 uV signal. \$\endgroup\$ – The Photon Aug 19 '14 at 15:58
  • \$\begingroup\$ @sherrellbc: yes, of course it's also a restriction that the signal must be small (but I think that is obvious because the model is called small signal model) \$\endgroup\$ – Curd Aug 19 '14 at 16:53

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