Small Signal Analysis Tutorial

Amplifiers are fundamental components in various electronic devices, from radios to stereo systems. While there are many different kinds of transistor amplifier circuits, this tutorial focuses on the analysis and characteristics of these amplifiers, particularly their small-signal behavior.

Amplifiers: An Overview

Generally, amplifiers are used in audio systems like radios and CD players. The following sections will explore the different types of amplifiers, their properties, and how to analyze their performance.

Types of Amplifiers

• Small Signal Amplifiers (Voltage Amplifiers): These amplifiers are characterized by their input resistance, output resistance, and gain.
• Large Signal Amplifiers (Power Amplifiers): These are categorized into different classes, such as Class A, Class B, and Class AB amplifiers.

In this amplifiers summary section we looked at the amplifier circuit based on a single bipolar transistor as shown below, but there are several different kinds of transistor amplifier circuits that we could use.

Typical Single Stage Amplifier Circuit

Small Signal Amplifiers (Voltage Amplifiers)

Small Signal Amplifiers are also known as Voltage Amplifiers. Voltage Amplifiers have 3 main properties, Input Resistance, Output Resistance and Gain.

Gain

The Gain of a small signal amplifier is the amount by which the amplifier “Amplifies” the input signal. Gain is a ratio of output divided by input, therefore it has no units but is given the symbol (A) with the most common types of transistor gain being, Voltage Gain (Av), Current Gain (Ai) and Power Gain (Ap). The power Gain of the amplifier can also be expressed in Decibels or simply dB.

DC Base Biasing

In order to amplify all of the input signal distortion free in a Class A type amplifier, DC Base Biasing is required. DC Bias sets the Q-point of the amplifier half way along the load line. This DC Base biasing means that the amplifier consumes power even if there is no input signal present.

The transistor amplifier is non-linear and an incorrect bias setting will produce large amounts of distortion to the output waveform. Too large an input signal will produce large amounts of distortion due to clipping, which is also a form of amplitude distortion. Incorrect positioning of the Q-point on the load line will produce either Saturation Clipping or Cut-off Clipping.

Common Amplifier Configurations

The Common Emitter Amplifier configuration is the most common form of all the general purpose voltage amplifier circuit using a Bipolar Junction Transistor. The Common Source Amplifier configuration is the most common form of all the general purpose voltage amplifier circuit using a Junction Field Effect Transistor.

BJT Amplifier to JFET Amplifier Comparison

Here's a comparison between Common Emitter (BJT) and Common Source (JFET) amplifiers:

BJT vs JFET Amplifiers: A Comparative Analysis

Parameter	Common Emitter Amplifier	Common Source Amplifier
Voltage Gain, (AV)	Medium/High	Medium/High
Current Gain, (Ai)	High	Very High
Power Gain, (AP)	High	Very High
Input Resistance, (Rin)	Medium	Very High
Output Resistance, (Rout)	Medium/High	Medium/High
Phase Shift	180o	180o

Large Signal Amplifiers (Power Amplifiers)

Large Signal Amplifiers are also known as Power Amplifiers. Power Amplifiers can be sub-divided into different Classes, for example:

• Class A Amplifiers - where the output device conducts for all of the input cycle.
• Class B Amplifiers - where the output device conducts for only 50% of the input cycle.
• Class AB Amplifiers - where the output device conducts for more than 50% but less than 100% of the input cycle.

An ideal Power Amplifier would deliver 100% of the available DC power to the load. Class A amplifiers are the most common form of power amplifier but only have an efficiency rating of less than 40%. Class B amplifiers are more efficient than Class A amplifiers at around 70% but produce high amounts of distortion. Class B amplifiers consume very little power when there is no input signal present.

By using the “Push-pull” output stage configuration, distortion can be greatly reduced. However, simple push-pull Class B Power amplifiers can produce high levels of Crossover Distortion due to their cut-off point biasing. Pre-biasing resistors or diodes will help eliminate this crossover distortion. Class B Power Amplifiers can be made using Transformers or Complementary Transistors in its output stage.

Small-Signal Model of MOSFET

This is the second article in our analog circuit design series. After understanding the structure of the MOSFET, we will now delve into its characteristics to construct a small signal model, which will facilitate circuit analysis later on.

When a MOSFET operates in the saturation region, it essentially functions as a voltage-controlled current source. Let’s examine the relationship between V_GS and I_D.

Figure 1: I_D vs. V_GS Relationship

This graph shows how I_D varies with V_GS. Our primary interest is how I_D responds to changes in V_GS. We define a parameter, g_m, to represent the slope of this relationship. g_m is called transconductance.

Equation 1: g_m is defined as the partial derivative of I_D with respect to V_GS

In practice, the I_D vs. V_GS relationship shown in Figure 1 is non-linear. Therefore, we analyze the behavior for small signals around a specific operating point Q. When the signal amplitude is very small, this change in I_D relative to V_GS can be approximated as linear (meaning g_m is treated as a constant).

In the small-signal model, we use lowercase i_d and v_gs to represent these “small changes” (AC signals). Thus, the relationship from Equation 1 can be expressed in the small-signal model as:

i_d = gm * v_gs

This is precisely why this analysis is called the “small-signal model.”

Channel Length Modulation (CLM)

In addition to g_m, we must also discuss the impact of Channel Length Modulation (CLM). In the previous article on MOS Principles and Structure, we mentioned that once V_DS exceeds a certain value, the current ideally stops increasing with V_DS. Without considering CLM, the ideal saturation current equation is:

Equation 2: Ideal Saturation Current Formula

As Equation 2 shows, the current depends only on V_GS and not V_DS. However, in a practical device, the actual relationship is as follows:

Figure 2: The Practical Case with Channel Length Modulation

As seen in Figure 2, the current does slightly increase as V_DS rises. The reason for this is a phenomenon called Channel Length Modulation. When V_DS increases after the channel has already pinched off, it causes the effective channel length to shorten, as shown in the diagram below.

Figure 3: Channel Length Modulation

As seen in Figure 3, the new effective channel length, L’, is shorter than the original length L. If we substitute this L’ back into Equation 2, we can see that a shorter L’ results in a larger current.

Equation 3: With L’ < L, substituting L’ for L in Equation 2 causes I_D to increase.

As Equation 3 shows, because L’ < L, I_D increases. To simplify this and incorporate the effect into our model, we modify the formula using the channel length modulation parameter, lambda:

Equation 4: Current Formula Including Channel Length Modulation

The new term (1 + lambda V_DS) in Equation 4 is what causes the “upward slope” in the saturation region. We define the reciprocal of this slope (from the I_D-V_DS curve in Figure 2) as the output resistance, r_o:

Equation 5: Output Resistance r_o

With this, we can now draw the complete small-signal equivalent circuit for an NMOS. In future analyses, we can simply replace the NMOS symbol with this model to analyze its behavior using standard circuit theory.

Figure 4: Small-Signal Model of an Enhancement-Type NMOS

As mentioned in the MOS Structure article, there is an insulating layer under the Gate (G). Therefore, the gate is an open circuit to the drain (D) and source (S); no current flows into the gate.

Current only flows between the drain and source. The magnitude of this (small-signal) current is determined by the transconductance, g_m, multiplied by v_gs (the small-signal gate-source voltage). We also include the output resistance r_o in parallel.

Steps for Small-Signal Analysis

1. Find the quiescent point (DC operating point) of the circuit.
2. Linearize the non-linear circuit components at the DC operating point. The most basic example of this is a diode, whose current is approximately proportional to the square of the voltage across it. But as long as the signal is small (hence the name, small-signal analysis!) compared to the DC operating point of the diode, you can approximate this to a linear relationship, i.e. ΔV=ΔIR.
3. Find the small-signal solution: Now the DC sources are removed. Voltage sources are replaced with 0Ω resistors (short circuit), current sources are replaced with ∞Ω resistors (open circuit).

Resistor divider, before replacing Vin.

Resistor divider, Vin replaced with short.

Circuit re-arranged to highlight both resistors in parallel.

Equivalent resistance of the two resistors in parallel.

How to Measure the Small Signal Characteristics of a BJT

A bipolar junction transistor, or BJT, is a type of transistor. They are commonly found in electronic amplifier circuits such as those used to transmit data wirelessly, and in radios.

Schematic of a BJT

C is the Collector, B is the Base, and E is the Emitter. The internal characteristics of a BJT can vary with temperature, voltage, and current. Because of this, before putting a transistor into your project or device you may want to measure the internal characteristics of the transistor at the voltage and current you will have it operating.

This Instructable shows you how to calculate the small signal characteristics of a BJT using what is called the hybrid-pi model, and by taking measurements of the currents going into the BJT. These characteristics are called the transconductance, current gain, and internal resistance.

To measure these values, you need to take the following steps:

1. Create a circuit to test your transistor in.
2. Measure the voltage or current going into the transistor.
3. Calculate the Beta value (current gain at low frequencies) and the transconductance.
4. Using the values calculated for current gain and transconductance, find R-π, the internal resistance.

An important safety note: If the current through the transistor is too high, it may melt or explode. For most small transistors, anything close to an amp of current is high.

Step 1: Set Up a Test Circuit

To test the collector and base currents you will first need to set up a circuit to test them in. Because BJTs usually operate at very low currents, we attach 1000 ohm resistors on each node of the transistor to ensure that the transistors are not damaged. The above image shows the diagram for the a test circuit. It uses a 2n3904 BJT (labeled Q1 in the diagram), using 1000 ohm resistors (R1, R2, R3) on each terminal, with 15 volts on the collector terminal and 2 volts on the base terminal.

Step 2: Measure the Voltage Across the Resistors

Because BJTs operate at such low current, instrument noise can make it difficult to accurately measure the current. Instrument noise is small, random variations in the measured values. In most multimeters, this is a very small amount, but it is similar in magnitude to the currents in the test circuit. In order to compensate for this issue, you may instead want to measure the voltage across the resistors and then use Ohm's law to calculate the current.

Ohm's law states the the voltage across a component equals the current through it multiplied by its resistance: V = I * R. To find the current through the resistors divide the voltage by the resistance: I = V / R.

Test Circuit: The pictures shown above show the values you would measure if you constructed the test circuit. The voltage on the collector resistor is 1.359 volts, and the voltage on the base resistor is 8.107 millivolts.

Because the resistors in the test circuit are all 1000 ohms, dividing by the resistance simply changes the prefix to the one below it on the SI scale: 1.359 volts on the collector becomes 1.359 milliamps through the collector, and 8.107 millivolts on the base becomes 8.107 microamps through the base.

Step 3: Calculate the Internal Characteristics Based on Input Currents

Now that you've calculated the internal currents you can calculate the small signal values of the BJT.

These are as follows:

• β0: Current Gain This value represents the ratio between the current into the collector and the current into the base. The particular value is specific to each transistor, and can also vary with temperature.
• gm: Transconductance This value represents the ratio between the current out of the transistor and the voltage into it. V is thermal voltage, which is a constant equal to 26 millivolts at room temperature.
• rπ: Internal Resistance This value is the resistance between the base and emitter terminals of the transistor. Like current gain, it is unique to each transistor.

Step 4: Calculate Current Gain

To calculate current gain, divide the value of the current going into the base terminal by the value of the current going into the collector terminal.

This value represents how much the transistor amplifies the current going into the collector terminal. It is just a number, with no units. However, to make it easier to understand what it is doing, you may find it useful to write this value as amps per amps.

Test Circuit: For the test circuit the collector current is 1.359 milliamps, and the base current is 8.107 microamps. Dividing the collector current by the base current gives a current gain of 167.6 amps per amp.

Step 5: Calculate Transconductance

Transconductance is an intimidating word, but it is just the ratio between the voltage into the transistor and current out of the transistor. To calculate it, divide the current into the collector terminal by VT, thermal voltage. Doing so gives you a value in inverse ohms, written as mhos.

Test Circuit: For the test circuit the collector current is 1.359 milliamps. The thermal voltage is constant at 26 millivolts. This results in a value of 52.27 millimhos.

Step 6: Calculate Internal Resistance

The last step is to calculate the internal resistance of the BJT. This value represents how much of the voltage placed on the collector terminal makes it to the emitter terminal. To calculate it, divide the value of current gain by the transconductance, both calculated in previous steps. This calculation gives you a result in ohms.

Test Circuit: For the test circuit the current gain is 167.6 amps per amp, and the transconductance is 52.27 milliohms. Performing the calculation gives a result of 3207 ohms.

After having gone through these steps, you will have calculated the main internal characteristics of whatever BJT you were testing. Keep in mind that these values are only valid for lower frequencies, and will change at high frequencies. One thing you may have noticed after going through these steps is that you did not calculate Ro. This value is the output resistance due to the Early effect, and changes depending on the voltage and is unique to each transistor.