Circuit analysis is a fundamental subject in the field of electrical engineering. Knowledge of circuit analysis is esssntial, since it is applied in several areas of electrical engineering. Electronic devices are used in a huge number of applications. The field of electronics can be divided into areas such as analogue electronics, digital electronics, communication electronics, and power electronics. Knowledge of circuit analysis is essential to analyze electronic circuits, including a digital circuit. Hence this page provides an introduction to electronic devices. The field of devices is vast, and a detailed description of the devices is outside of the scope of this text. The treatment of electronic devices in this page is from the viewpoint of circuit analysis. Since it is not possible to model a device without some knowledge of how the device functions, a brief description of the physics of devices is provided, where it is necessary.
In this page, we study the basics of an operational amplifier, a bipolar junction transistor, a junction field-effect transistor, and a diode. An operational amplifier is the most important integrated circuit in the area of analogue electronics. It is used in the linear mode in many applications. In this page, an operational amplifier is treated more or less as a black box. Its internal circuitry is far too complex, to be introduced at this stage. In addition, there are many families of operational amplifier, and the internal circuitry differs from family to family. However, they have a common input-output characteristics, and it is appropriate to use the black box approach, with regard to an operational amplifier.
A diode is the most basic semiconductor device. It is essentially a non-linear device, and the necessity to include description of a diode in a course on linear circuits can be questioned. However it is felt that it is necessary to know about a diode and its applications, since the scope of this text need not be restrcted only to linear circuits. Some transgressions are welcome, especially where these transgressions are used to show how the knowledge of circuit analysis can be extended to analyze and improve understanding of non-linear circuits.
It is necessary to know how a bipolar junction transistor and a field-effect transistor operate, before we find out how they can be modelled. Hence the physics of these devices is presented. These devices are used in both linear and switching circuits.
The first device taken up for study is the operational amplifier. It turns out that it is easy to analyse a circuit with operational amplifier, because it can be treated as ideal voltage amplifier. An operational amplifier is referred to as the op-amp. Hence this term is used more often in this page. In this section, some of the properties of an ideal op amp are presented. The remaining properties of an ideal op amp will be presented later in another page.
The circuit symbol of an ideal op amp is shown in Fig. 5.1. This symbol shows the op amp as a three terminal device, and this symbol does not include the details of external power supplies are connected to an op amp. Normally, a dual power supply is needed for an op amp.
The three terminals of an op amp, shown in its symbol are:
The voltages at both the inputs and the output are expressed with reference to the ground. As shown in Fig. 5.1, the input terminal with the negative sign is known as the inverting input terminal, and the input terminal with the positive sign is known as the inverting input terminal.The third terminal is the output terminal.
An op amp is a voltage amplifier, since both its input and output are voltages. The input voltage of is defined, as shown in Fig. 5.2.
As expressed by equation (5.1) The input voltage, denoted as vi, is the difference between the potential v-, at the inverting input terminal, and the potential v+, at the non-inverting input terminal. When the input voltage is defined in this manner, the output is defined, as expressed by equation (5.2).
It can be seen that the output voltage is the product of the input voltage vi, and the open-loop gain - A of the op amp.
When the open-loop gain of the op amp tends to infinity, the input voltage vi, tends to zero for any finite output voltage, as expressed by equations (5.3) and (5.4).
In other words, both the input voltages, v- and v+, are equal to each other, as shown by equation (5.6). It is preferable to remember equation (5.6), for it is more helpful in analyzing circuits containing ideal op amps.
At this some of the properties of an ideal op amp are presented and described. They are as follows:
These properties of an ideal op amp are explained with the help the diagram in Fig. 5.3.
When the open-loop gain is infinite, both the inputs are at the same potential. In addition, input currents i- and i+, are also zero. The significance of zero input currents is that it implies that an ideal op amp has infinite input resistance. An ideal op amp has zero output resistance too. The concept of infinite input resistance and zero output output resistance can be understood better from the circuit model of a voltage amplifier, presented in Fig. 5.4.
When the input current is zero, it means that the input resistance Rin, is infinite. When the output resistance Ro, is zero, it means that the drop across it is zero, and the output voltage equals the voltage dependent voltage source. In the case of an ideal op amp, the gain of the voltage amplifier is also infinite.
Four basic applications of an op amp are presented here, to show how the properties of an ideal op amp can be used to analyze these circuits. These applications are as follows:
The circuit of an inverting amplifier is presented in Fig. 5.5.
Given that the op amp is ideal, its input currents are zero, and the voltages at the inverting and non-inverting terminals are equal to each other. Since the open-loop gain of an ideal op amp is infinite, the input voltage vi, is also zero. Hence the inverting input terminal is at zero potential. Hence in applications where the imverting input terminal is at zero potential, it is referred to asthe virtual ground. It is preferable not to refer to the inverting terminal as the virtual ground, since this concept may be inadvertently applied to circuits, where the inverting input terminal is not at ground potential. On the other hand, both the input voltages, v- and v+, are always equal to each other, whatever the circuit be, so long as the op amp is ideal, and it is operating in the linear region. The concept of what a linear range is explained later in this sub-section.
When the input current i- , is zero, currents flowing through resistors R1 and R2, are equal to each other, as shown in Fig. 5.6. Then an expression for current I1, can be obtained, and it is expressed by equation (5.7).
Since the voltage at the inverting terminal is zero, current I1, is equal to the ratio of source voltage to the value of resistor R1, as expressed by equation (5.8).
An expression for current I2, can be obtained, as shown in the sketch in Fig. 5.7. Equating the expressions for currents I1 and I2, we can get the relation between the output voltage and the source voltage of the inverting amplifier circuit. Equation (5.9) expresses the gain of an inverting amplifier circuit.
It can be seen from Fig. 5.7 that resistor R2, is connected between the output terminal and the inverting input terminal, and hence this resistor is referred to as the feedback resistor. As shown by equation (5.9), the gain of the inverting amplifier is simply the negative of the ratio of two resistors. It is hence to design an amplifier with the desired gain by choosing proper values for the two resistors. Since the polarity of the output voltage and the polarity of the source voltage are the opposites of each other, the circuit is called the inverting amplifier.
We can find out how the gain of the inverting amplifier is affected by the finite gain of an op amp. The aim here is to find out whether it is justified to analyze an op amp circuit based on the assumption that the op amp is ideal.
As shown in Fig. 5.8, the input voltage vi, is not zero, when the open-loop gain of the op amp is not finite. So long as current i-, is zero, currents I1 and I2, are equal, and the expressions for these currents can be equated with each other, as shown in Fig. 5.8. We can replace voltage vi, by its corresponding expression to get the gain of the circuit, and obtain equation (5.10).
Equation (5.11) is derived from equation (5.10). It can be seen from equation (5.11) that when the open-loop gain becomes infinite, the gain of the circuit reduces to the negative of the ratio of two resistors, as expressed by equation (5.9).
The effect of open-loop gain on the gain of the inverter amplifier can be illustrated with a plot. Let the gain of the circuit with an ideal op amp be - 10. As the value of A changes from 100 to 10000, the plot of the absolute value of the gain of the circuit against A, in log scale, can be obtained using a Matlab script shownb below.
% Matlab Script to obtain plot of inverting amplifier
for n= 1:26,
A(n)= 10000/(1.2023^(n-1));
G(n) = -10/(1+11/A(n));
end;
semilogx(A,abs(G))
grid
title('Effect of Open-loop Gain')
xlabel('Open-loop Gain')
ylabel('Absolute Value of Circuit gain')
axis([100 10000 9 10])
As the open-loop gain of the op amp varies from - 100 to -10000, the gain inverting amplifier varies from - 9, to - 10. It can be seen that when the open-loop gain is between - 8000 and - 10000, the open-loop gain is very close to - 10. A practical op amp has an open-loop gain of the order of - 10000, and hence it is justified to analyse an op amp circuit, based on the assumption that the op amp is ideal. When the open-loop gain of the op amp is -10000, the gain of the inverting amplifier is - 9.989, which is quite close to - 10.
The gain of the inverting amplifier can be obtained using the circuit model of voltage amplifier. When an op amp has a finite gain, infinite input resistance, and zero output resistance, its model contains only the voltage dependent source, as shown in the circuit in Fig. 5.9. The gain of this circuit can be obtained as follows.
For the equivalent circuit of the inverting amplifier shown on the right side of Fig 5.9, the input voltage vi, is represented by equation (5.12). Applying Kirchoffs current law to the node representing the input terminal, we get the equation (5.13). It can be re-presented by the equation (5.14). It can be seen that equations (5.10) and (5.14) are one and the same.
A practical op amp circuit need external power supplies. Due to power supply and the nature of internal circuitry, the output voltage of an op amp gets clamped at a high level, as well as a low level. For example, if an op amp has a dual power supply of + 15 V and - 15 V connected to it, its output voltage cannot exceed + 15 V, and cannot dip below - 15 V. In practice, the output voltage may be clamped at a voltage lower than + 15 V at the high end, and at a voltage greater than - 15 V at the low end. When the output voltage of an op amp is not clamped, it is acting in the linear region, and in such a case, its input voltage vi, is zero. On the other hand, when the output voltage is clamped, its input voltage vi, is no longer zero.
The effect of limits on output voltage can be understood from the circuit in Fig. 10. When the op amp is connected as shown, the output voltage is at its highest, if the source voltage is negative. The output voltage is at its lowest, if the source voltage is positive. The output voltage changes abruptly from one level to the other when the polarity of source voltage changes. It can be seen that the gain of circuit is infinite, when the source voltage is zero. We can see how the open-loop characteristic is changed, when the gain of the circuit becomes finite due to the inverting amplifier configuration.
The inverting amplifier circuit is shown in Fig. 11. When resistor R2, is infinite, the input output characteristic of the circuit remains the same as that shown for the circuit in Fig. 10. The output voltage changes abruptly as the polarity of source voltage changes. Let us see what happens when resistor R2, has a finite value.
When the output voltage is between the two levels, it is equal to the product of the source voltage and the gain of the circuit. When the source voltage is higher than a certain value, the output voltage is at its lowest value, as shown in Fig. 12. Remember that both Vomin and G, are negative values and hence the ratio of Vomin to G is a positive value, as shown by the plot in Fig. 12. The output voltage cannot sink any lower, due to the power supply constraint and the internal circuitry of the op amp. Similarly, when the source voltage is below a certain value, the output voltage is clamped at its highest level. It cannot rise any further. When the output voltage is clamped, the input voltage, vi, is no longer zero.
An example is presented now to explain the input-output characteristic displayed above. Let the gain of the inverting amplifier be - 6. Let the output voltage clamping levels be + 12 V and -12 V. In this case, the output voltage equals the product of source voltage and the gain, when the source voltage is within the range of -2 and + 2 V. When the source voltage is less than - 2 V, the output is clamped at + 12 V. If the source voltage is greater than 2 V, then the output is clamped at - 12 V. When the output voltage is clamped, the voltage at the inverting input terminal is not zero.
An expression for the input voltage vi , can be obtained from equation (5.13). It is presented as equation (5.15), and we can obtain equation (5.16) from equation (5.15). It can be seen that when the output of the op amp is in the linear region, input voltage vi is zero, as expressed by equation (5.17). Let the source voltage vary from - 6 V to + 6 V. The plots of output voltage, Vo, and the input voltage vi , can be obtained using Matlab. The script used is shown below.
% Matlab Script to obtain plots of output voltage and input
voltage
% of Inverting Amplifier with limits on the output voltage
G = - 6; % Gain of the circuit
Vomax = 12; % Maximum output voltage
Vomin = - 12; % Minimum output voltage
R1 = 1; % Value of Resistor R1
R2 = 6; % Value of Resistor R2
for n= 1:121,
Vs(n) = - 6 + 0.1*(n-1);
Vo(n) = G*Vs(n);
if (Vo(n) > 12) Vo(n) = 12; end;
if (Vo(n) < -12) Vo(n) = -12; end;
Vi(n) = (R2*Vs(n) + Vo(n)*R1)/(R1+R2);
end;
subplot(211)
plot(Vs,Vo)
title('Output Voltage')
ylabel('Volts')
axis([-6 6 -15 15])
grid
subplot(212)
plot(Vs,Vi)
title('Input Voltage')
ylabel('Volts')
xlabel('Source Voltage')
axis([-6 6 -4 4])
grid
In this sub-section, we determine the gain of a non-inverting amplifier circuit, containing an ideal op amp. The circuit is presented in Fig. 13, and it can be seen that the voltage source is connected to the non-inverting input terminal. As in the case of the inverting amplifier circuit, there is a feedback resistor, R2, connected from the output terminal to the inverting input terminal. Resistor R1, is connected from the inverting input terminal to the ground. Resistor R3, connected between the source and the non-inverting input terminal, does not carry any current, when the op amp is considered to be ideal. In the case of a non-inverting amplifier circuit, both the source voltage, and the output voltage have the same polarity. On the other hand, the output voltage and the source voltage have opposite polarities, in the case of an inverting amplifier.
A non-inverting amplifier is the preferred amplifier, when the source resistance is large. Resistor R3, shown in Figure 13, can be considered to be the source resistance. Since an ideal op amp draws no current from the source, an error due to loading effect can be avoided. On the other hand, the inverting amplifier configuration is more suitable, for an application such as a summing amplifier. This circuit will be presented in the next section.
The gain of the non-inverting amplifier circuit, shown in Fig. 14, can be obtained as follows. When the op amp is ideal, the input voltage vi , is zero, and the voltage at the inverting input terminal is equal to the voltage at the non-inverting input terminal. Since current i+, is zero, the voltage at the non-inverting terminal is equal to the source voltage. Hence the voltage at the inverting input terminal is equal to the source voltage. Since the voltage at the inverting input terminal is known, the current through resistor R1, can be expressed as shown. It is equal to the negative of the source voltage over resistance R1. The negative sign for the current is due to the direction, assigned to current I1. If it is marked as flowing into the ground, then the sign for current I1, will be positive. Since the op amp is ideal, current i-, is zero, and current I2, equals current I1, as marked in Fig. 14. Then the output voltage is equal to the voltage at the inverting input terminal, minus the voltage drop across resistor, R2. The voltage at the inverting input terminal is equal to the source voltage. We know the expression for current I1. Since current I1, is equal to current I2, we can find the drop across resistor R2. Hence we can express the output voltage, as a function of source voltage, and resistor values, R1 and R2, as shown by the expression in red colour in Fig. 14. It can be seen that the gain is a positive value, meaning that both the source voltage and the output voltage have the same polarity.
The gain of the non-inverting amplifier is slightly different when the open-loop gain of the op amp is finite. When the open-loop gain of the op amp is finite, its input voltage vi , is not zero, as shown in Fig. 15. It is equal to the expression shown in Fig. 15. Then the voltage at the inverting input terminal is equal to the sum of source voltage, and the input voltage vi , as shown in Fig. 15. Once the voltage at the inverting input terminal is known, an expression for current I1, can be obtained, as shown in Fig. 15.
As expressed by equation (5.19), the output voltage is the voltage at the inverting input terminal minus the drop across resistor R2. Since current I1, is equal to current I2, we can find the drop across resistor R2, and the output voltage can be expressed, as shown by equation (5.19). This equation can be simplified, as shown below.
We can group terms containing the output voltage and equate this group with the group of terms containing the source voltage, as shown by equation (5.20). The gain of circuit obtained is expressed by equation (5.21).
The circuit of a summing amplifier with inputs, V1 and V2, is shown in Fig. 16. When the op amp is considered to be ideal, the voltage at inverting input terminal, and current i-, are both zero. Hence the sum of currents, I1 and I2, is equal to current If, flowing through the feedback resistor Rf. Expressions for currents I1 and I2, can be obtained from the circuit in Fig. 16, and they are displayed in red colour.
When the op amp has infinite gain, current If, can be expressed by equation (5.22). Since the output voltage is the negative of the drop across resistor Rf. we get equation (5.23). Equation (5.23) expresses as a function of the resistor values, and the voltages of the two sources. When all the three resistors have the same value, the output voltage is the negative of the sum of voltages, V1 and V2. It can be seen why this circuit is called the summing amplifier. This circuit can be extended to have inputs from additional sources, and an expression for the output voltage can be derived in a similar manner.
When the gain of the op amp, the output voltage can be expressed as shown by equation (5.25). It is known that the sum of currents, I1 and I2, is equal to current If. The expressions for currents I1 and I2, are shown in Fig. 16. The input voltage vi, can be expressed in terms of the output voltage and the open-loop gain. Hence, we get equation (5.26) from equation (5.25). When the open-loop gain if the op amp is large, the effect of open-loop gain on output voltage is negligible. This raises the question of relevance of obtaining equations such as equation (26). The point here is to show that the knowledge of circuit analysis is needed to analyze circuits with some complexity.
An op amp can be used to build the circuit of a
difference amplifier. The circuit in Fig. 17 shows how a difference amplifier
can be realized. The analysis here is based on the assumption that the op amp
is ideal. The difference amplifier has two external sources connected to it.
Let us start with the external source, V2. Due to this
source, the voltage at the non-inverting input terminal can be obtained, and
the expression for this voltage is shown inside the blue box. This expression
is obtained by using the voltage division rule, which in turn is derived from
the Kirchoff's voltage law. This law states that the algebraic sum of
voltages in a loop is zero.
When an op amp is ideal, the voltage at the inverting input terminal is the same as that at the non-inverting input terminal. Hence we get equation (5.27). Since the voltages at the two ends of resistor R1, are known, an expression for current I1, can be obtained, and equation (5.28) presents the expression for I1.
Since current I2, equals current I1, the expression for current I2, is known. Then the output voltage can be expressed as the voltage at the inverting input terminal minus the drop across the resistor R2. Replace the voltage at the inverting input terminal, and current I2, by the corresponding expressions. We can then express the output voltage as a function of voltage V1, voltage V2, and the resistor values, as shown by equation (5.29). If all the four resistors have the same value, the output voltage is equal to the difference between the voltages V2, and V1. It can be seen why this circuit is known as the difference amplifier.
The Effect of Finite gain
The circuit of a difference amplifier, with an op amp that has a finite gain, is shown in Fig. 18. When the open-loop gain of an op amp is finite, its input voltage vi, is not zero. Hence the voltage at the inverting input terminal is the sum of the voltage at the non-inverting terminal and its input voltage. The voltage at the non-inverting terminal has been expressed in Fig. 18.
When the op amp has a finite open-loop gain, the input voltage vi, is equal to the ratio of the output voltage to the open-loop gain. Since the voltages at the two ends of resistor R1, are known, an expression for current I1, can be obtained, and equation (5.31) presents the expression for I1. Since current I2, equals current I1, the expression for current I2, is known. Then the output voltage can be expressed as the voltage at the inverting input terminal minus the drop across the resistor R2. Replace the voltage at the inverting input terminal, and current I2, by the corresponding expressions, as shown by equation (5.32). We can then express the output voltage as a function of voltage V1, voltage V2, the resistor values, and the open-loop gain, as shown by equation (5.33). The purpose of obtaining equation (5.33) is to highlight the usefulness of knowledge of circuit analysis for analyzing electronic circuits.
Among the semiconductor devices, the diode is the basic device. The circuit symbol of a diode is shown in Fig. 19. As shown in Fig. 19, a diode has two terminals, called the anode and the cathode. When the potential at the anode is positive with respect to the potential at its cathode, the diode conducts, and the direction of conventional current flow is as shown in Fig. 19. Current iD, enters the diode through its anode, and leaves the diode via is cathode. The voltage vD, is positive, and is of the order of 0.7 V for low power diodes. When the potential at the anode is positive with respect to that at the cathode, the diode is forward-biased. In other words, a diode conducts when it is forward-biased. When the potential at the anode is negative with respect to that at the cathode, the diode is reverse-biased, and a diode blocks conduction when it is reverse-biased. The reverse current through is almost zero. The volt-ampere characteristic of a semiconductor diode is shown in Fig. 19.
As shown in Fig. 20, a pn junction is formed by doping a wafer intrinsic semiconductor with acceptors on one side, and donors on the other side. The two sides of the pn junction are known as the p-side andn-side. The p-side contains holes and an equal number of acceptor ions, leading to a charge free region. The n-side contains free electrons and an equal number of donor ions, leading to a charge free region. In the case of a diode with no external connections, a depletion region is formed on either side of the junction, with the diffusion of holes and electrons to the other side. This leads to presence of acceptor ions within the depletion region on the p-side, and donor ions within the depletion region on the n-side. Hence a potential barrier is formed, which prevents further diffusion of holes and electrons to the other side.
In order to make the diode conduct, it is necessary to forward-bias the device. As the forward-bias increases, the depletion region shrinks, and the current through the diode increases. For low power diodes, the depletion region disappears when the forward-bias reaches about 0.7 V. In a practical circuit, the forward current through a diode is restricted or controlled by the resistance in series with the diode and the power supply. This aspect is clarified later. When the diode is reverse-biased, the depletion region widens, and the reverse current through the diode is almost zero.
The current through the diode can be expressed by equation (5.34), called the Shockley equation. The parameter IS, known as the saturation or scale current, is a function of the donor and acceptor densities in the diode, as well as the diode temperature, the area of the junction, and other constants. The value of IS varies from 10 nA to 0.01 pA for discrete diodes. The constant n, is known as the emission coefficient, and it lies in the range 1 to 2, depending on the properties of the diode. The voltage VT, is known as the thermal voltage and it equals approximately 26 mV at room temperature of 300o K. When the diode is forward-biased, the exponential term is far greater than 1, and it is much smaller than 1 when the diode is reverse-biased. Hence the current can be expressed as follows:
If n = 1, and VT = 26 mV, the diode current double for every 18 mV rise in vD. The exponential relationship is used to find the change in forward-bias for a specified change in iD. Equations (5.38) and (5.39) illustrate how the change in forward-bias can be calculated for for a specified change in iD. If the diode current increase ten times, then the change in forward-bias is as expressed by equation (5.40).
The modelling of a diode is described later. Next we take up the study of a bipolar junction transistor.
A bipolar junction transistor is a 3-layer device. There are two types of bipolar junction transistor. They are:
The npn transistor is introduced first. Since the operation of a pnp transistor is similar to that of an npn transistor, description of a pnp transistor is omitted.
The symbol and the simplified structure of an npn transistor is shown in Fig. 22. Both the n-regions have a higher doping density than the base region. The base region is also thinner in width than the n-regions.
The operation of the npn transistor can be explained with the help of the diagram in Fig. 23. In order to make the BJT conduct, the base is forward-biased with respect to the emitter. Hence the depletion layer close to the emitter-base junction is almost non-existent. Because of the forward-bias, holes migrate from the base region to the emitter region, and electrons migrate from the emitter region to the base region. Since the doping density of donors is much higher than the doping density of acceptors, the number of electrons migrating to the base is far greater than the number of holes and a tiny fraction of the electrons forms part of the base current. Majority of electrons from the emitter cross over to the collector region, due to the electric field across the collector-base junction. The depletion layer at the collector-base junction is wider in the base region than in the collector region, due to the difference in doping densities.
The operation of the pnp transistor is similar.
Given that an npn transistor is biased as shown in Fig. 23, the currents in the device can be expressed as follows:
The parameter IS is known as the transistor saturation current. It is dependent on the device parameters and can have a wide range values, as shown by equation (5.41). Equation (5.42) expresses the base current, and parameter βF is known as the common-emitter current gain. It varies over a range, as specified in equation (5.42). The emitter current is the sum of the base current and the collector current.
The collector current varies as the base current varies. In addition, as the voltage from collector to emitter increases, there is a slight increase in collector current, given that the base current remains steady.
Since the emitter current is the sum of the base current and the collector current, the emitter current can be expressed as shown by equation (5.44). Then the ratio of collector current to emitter current can be expressed as shown by equation (5.45).
A small-signal model of the BJT can be obtained from its DC operating point.
Let the voltage across the base-emitter junction be VBE, and let the corresponding collecting current be IC. Equation (5.46) shows how VBE and IC are related. For a small increment in base-to-emitter voltage, the collector current can be expressed, as shown by equation (5.47). Given that vbe is much smaller than VT, we get equation (5.48). Hence the corresponding increment in collector current can be expressed as shown by equation (5.49). Here gm is transconductance of the BJT.
The corresponding increment in base current can be expressed as shown by equation (5.50). For a small-signal around the operating point, the base-emitter junction can be modelled as a resistor, called as rπ. The small-signal model of the BJT, called the hybrid-π model, is represented by the circuit in Fig. 25. In Fig. 25, ro represents the output resistance of the BJT. This resistance accounts for the variation in collector current as the collector-to-emitter voltage varies.
There are many types of field-effect transistors. In this section, a brief introduction to the enhancement-type, n-channel MOSFET is provided. The circuit symbol and a circuit to describe the operation of this type of MOSFET are provided in Fig. 26. A MOSFET is essentially a 3-terminal device. The terminals are the gate, the drain and the source. The body is usually connected to the source.
The current through the drain can be defined as follows:
When VGS is less than the threshold voltage VTR, the MOSFET is cut-off. For an enhancement n-channel MOSFET, the threshold voltage is a positive value. Given that VGS is greater than the threshold voltage VTR, and the drain-to-source voltage is less than (VGS - VTR), the MOSFET operates in the triode region. This region is referred to as the ohmic region too. Equation (5.52) defines how the drain current varies in the triode region. Given that VGS is greater than the threshold voltage VTR, and the drain-to-source voltage is greater than (VGS - VTR), the MOSFET operates in the saturation region. When the MOSFET is used as a linear amplifier, it operates in the saturation region. The value of K is dependent on the device parameters, and it is usually specified.
The operating point is defined by a set of values of VGS and ID, as stated by equation (5.54). The transconductance of the device is defined when vGS varies near about VGS, and it is obtained as shown by equation (5.56). The small-signal model of the MOSFET is displayed by the circuit in Fig. 28. The increase in drain current with increase in drain-to-source voltage is accounted for by the output resistance rd. The value of output resistance is computed as the ratio VA to ID. The parameter VA is present in equation (5.53), to allow for increase in drain current. It is a large value, of the order of 100 Volts.
This page has described briefly the devices used in linear electronic circuits. Examples of analogue electronic circuits are presented, as part of the topics described later.
