3.5.5: Small Signal Models

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In order to create the linear model, we need to introduce the concept of bias, and large signal and small signal device behavior. Consider the following circuit, shown in Figure \(\PageIndex{1}\). We are applying the sum of two voltages to the diode, \(V_{B}\), the bias voltage (which is assumed to be a DC voltage), and \(v_{s}\), the signal voltage (which is assumed to be AC, or sinusoidal). By definition, we will assume that \(\left| v_{s} \right|\) is much less than \(\left| V_{B} \right|\). As a result of these voltages, there will be a current \(I_{D}\) flowing through the diode. This will consist of two currents: \(I_{B}\), the so-called bias current, and \(i_{s}\), which will be the signal current. Again, we assume that \(i_{s}\) is much smaller than \(I_{B}\).

A circuit contains a voltage source V_B, a sinusoidal voltage source v_s, and a diode in series. The current flowing through the diode from anode to cathode is I_D, which is the sum of I_B and i_s. — Figure \(\PageIndex{1}\): Putting together a large signal bias, and a small signal AC excitation

What we would like to do is to see if we can find a linear relationship between \(v_{s}\) and \(i_{s}\), which we could use in our signal analysis. There are two ways we can attack the problem: a graphical approach, and a purely mathematical approach. Let's try the graphical approach first, as it is more intuitive, and then we will confirm what we find out with a mathematical method.

Let's remind ourselves about the \(I \text{-} V\) characteristics of a diode (Figure \(\PageIndex{2}\)). In the present situation, \(V_{D}\) is the sum of two voltages: a DC bias voltage \(V_{B}\) and an AC signal, \(v_{s}\). Let's plot \(V_{D} (t)\) on the \(V_{D}\) axis as shown in Figure \(\PageIndex{3}\).

Graph of diode current on I_D vs V_D axes. Graph equation is given by e (the natural exponent) to the power of q V_D over kT, minus 1, the whole multiplied by I_sat. — Figure \(\PageIndex{2}\): Diode \(I \text{-} V\) behavior

Standard first-quadrant coordinate plane with V_D on the x-axis and I_D on the y-axis, showing the graph of V_B as an exponential growth function starting from the origin. A third vertical axis of time extends downwards from a point on the V_D axis, with the direction of increasing time being to the bottom. A sine function of v_s is shown on the V_D vs t axes. — Figure \(\PageIndex{3}\): Bias and signal excitation of a diode \(I \text{-} V\) curve

How are we going to figure out what the current is? What we need to do is to project the voltage up onto the characteristic I-V curve, and then project over to the vertical current axis. We do this in Figure \(\PageIndex{4}\).

The graph from Figure 3 above has one vertical dotted line each extending upward from the t axis, the minimum of the v_s function, and the maximum of the v_s function. The vertical lines extend upwards until they intersect with the V_B graph, and then extend horizontally to the left. A sinusoidal function bounded by the top and bottom horizontal lines extends to the left, beginning on the center dotted line where it intersects the I_D axis at the value I_B. It rises to the left, reaching a maximum value of i_s, and then falls to a minimum value that is closer to the I_B dotted line than the i_s line is. — Figure \(\PageIndex{4}\): Copy and Paste Caption here. (Copyright; author via source)

Note that the output current signal is somewhat distorted, which means we do not have linear behavior yet. Let's reduce the amplitude of the signal voltage, as shown in Figure \(\PageIndex{5}\). Now we see two things: a) the output is much less distorted, so we must getting a more linear behavior, and b) we could get the amplitude of the output signal \(i_{s}\) simply by multiplying the input signal \(v_{s}\) by the slope of the I-V curve at the point where the device is biased. We have replaced the non-linear I-V curve of the diode by a linear one, which is applicable over the range of the signal voltage.

\[i_{s} = \left. \frac{\partial}{\partial V_{D}} \left(V_{D}\right) \right| _{I_{D} = I_{B}} \nonumber \]

Figure 4 above is repeated with a v_s function of smaller magnitude, leading to the horizontal dotted lines extending out from it to be more equally spaced. The value of i_s, the topmost horizontal line, is the equal to the partial derivative of the I_D-vs-V_D curve evaluated where the value of V_D is V_B. — Figure \(\PageIndex{5}\): With a smaller signal, the response is more linear

To get the slope, we need a few simple equations: \[I_{D} = I_{\text{sat}} \left( e^{\frac{q V_{D}}{kT}} - 1 \right) \simeq I_{\text{sat}} e^{\frac{q V_{D}}{kT}} \nonumber \]

\[ \frac{\partial}{\partial V_{D}} \left(I_{D}\right) = \frac{q}{kT} I_{\text{sat}} e^{\frac{q V_{D}}{kT}} \nonumber \]

When we evaluate the partial derivative at voltage \(V_{D}\), we note that \[I_{\text{sat}} e^{\frac{q V_{D}}{kT}} = I_{B} \nonumber \]

and hence, the slope of the curve is just \(\frac{q}{kT} I_{B}\) or \(40 I_{B}\), since \(\frac{q}{kT}\) just has a value of \(40 \mathrm{~V}^{-1}\) at room temperatures. Note that current divided by voltage is just conductance, (which is just the inverse of resistance) and so we have found the small signal linear conductance for the diode.

As far as the AC signal generator is concerned, we could replace the diode with a resistor whose value is the inverse of the conductance, or \(r = \frac{1}{40} I_{B}\), where \(I_{B}\) is the DC bias current through the diode.

Students are sometimes confused about how we can replace a diode, which only conducts in one direction, with a resistor, which conducts both ways. The answer is to look carefully at Figure \(\PageIndex{5}\). As the AC signal voltage rises and falls, the AC output current also increases and decreases in the same manner. Over the limited range of the AC signal parameters, the diode is indeed a linear signal element, not a rectifying one, as it is for large signal applications.

Now let's get the same answer from a purely mathematical approach. \[I_{D} = I_{B} + i_{s} = I_{\text{sat}} \left( e^{\frac{q V_{D}}{kT}} - 1 \right) \simeq e^{\frac{q \left( V_{B}+v_{s} \right)}{kT}} \nonumber \]

In the last expression, we dropped the \(-1\) as it is very small compared to the exponential term and can be neglected.

Now we note that: \[e^{\frac{q \left( V_{B}+v_{s} \right)}{kT}} = e^{\frac{q V_{B}}{kT}} e^{\frac{q v_{s}}{kT}} \nonumber \]

And, for the second exponential, if \(q V_{B}\) is much less than \(kT\), \[e^{\frac{q V_{s}}{kT}} \simeq 1 + \frac{q v_{s}}{kT} + \ldots \nonumber \]

where we have used the power series expansion for the exponential, but have only taken the first two terms. Thus \[I_{B} + i_{s} \simeq I_{\text{sat}} e^{\frac{q V_{B}}{kT}} \left( 1 + \frac{q v_{s}}{kT} \right) \nonumber \]

Obviously \[I_{B} = I_{\text{sat}} e^{\frac{q V_{B}}{kT}} \nonumber \]

and \[\begin{array}{l} i_{s} &= I_{\text{sat}} e^{\frac{q V_{B}}{kT}} \left( \frac{q}{kT} v_{s} \right) \\ &= \frac{q}{kT} I_{B} v_{s} \end{array} \nonumber \]

which gives us the same result as before: \[\frac{i_{s}}{v_{s}} = \frac{q}{kT} I_{B} \nonumber \]