Common Source JFET Amplifier

So far we have looked at the bipolar type transistor amplifier and especially the common emitter amplifier, but small signal amplifiers can also be made using Field Effect Transistors or FET’s for short.

These devices have the advantage over bipolar transistors of having an extremely high input impedance along with a low noise output making them ideal for use in amplifier circuits that have very small input signals.

The design of an amplifier circuit based around a junction field effect transistor or “JFET”, (N-channel FET for this tutorial) or even a metal oxide silicon FET or “MOSFET” is exactly the same principle as that for the bipolar transistor circuit used for a Class A amplifier circuit we looked at in the previous tutorial.

Firstly, a suitable quiescent point or “Q-point” needs to be found for the correct biasing of the JFET amplifier circuit with single amplifier configurations of Common-source (CS), Common-drain (CD) or Source-follower (SF) and the Common-gate (CG) available for most FET devices.

These three JFET amplifier configurations correspond to the common-emitter, emitter-follower and the common-base configurations using bipolar transistors. In this tutorial about FET amplifiers we will look at the popular Common Source JFET Amplifier as this is the most widely used JFET amplifier design.

Consider the Common Source JFET Amplifier circuit configuration below.

Common Source JFET Amplifier

The amplifier circuit consists of an N-channel JFET, but the device could also be an equivalent N-channel depletion-mode MOSFET as the circuit diagram would be the same just a change in the FET, connected in a common source configuration. The JFET gate voltage Vg is biased through the potential divider network set up by resistors R1 and R2 and is biased to operate within its saturation region which is equivalent to the active region of the bipolar junction transistor.

Unlike a bipolar transistor circuit, the junction FET takes virtually no input gate current allowing the gate to be treated as an open circuit. Then no input characteristics curves are required. We can compare the JFET to the bipolar junction transistor (BJT) in the following table.

JFET to BJT Comparison

Junction FET	Bipolar Transistor
Gate, ( G )	Base, ( B )
Drain, ( D )	Collector, ( C )
Source, ( S )	Emitter, ( E )
Gate Supply, ( VG )	Base Supply, ( VB )
Drain Supply, ( VDD )	Collector Supply, ( VCC )
Drain Current, ( ID )	Collector Current, ( IC )

Since the N-Channel JFET is a depletion mode device and is normally “ON”, a negative gate voltage with respect to the source is required to modulate or control the drain current. This negative voltage can be provided by biasing from a separate power supply voltage or by a self biasing arrangement as long as a steady current flows through the JFET even when there is no input signal present and Vg maintains a reverse bias of the gate-source pn junction.

In our simple example, the biasing is provided from a potential divider network allowing the input signal to produce a voltage fall at the gate as well as voltage rise at the gate with a sinusoidal signal. Any suitable pair of resistor values in the correct proportions would produce the correct biasing voltage so the DC gate biasing voltage Vg is given as:

Note that this equation only determines the ratio of the resistors R1 and R2, but in order to take advantage of the very high input impedance of the JFET as well as reducing the power dissipation within the circuit, we need to make these resistor values as high as possible, with values in the order of 1 to 10MΩ being common.

The input signal, (Vin) of the common source JFET amplifier is applied between the Gate terminal and the zero volts rail, (0v). With a constant value of gate voltage Vg applied the JFET operates within its “Ohmic region” acting like a linear resistive device. The drain circuit contains the load resistor, Rd. The output voltage, Vout is developed across this load resistance.

The efficiency of the common source JFET amplifier can be improved by the addition of a resistor, Rs included in the source lead with the same drain current flowing through this resistor. Resistor, Rs is also used to set the JFET amplifiers “Q-point”.

When the JFET is switched fully “ON” a voltage drop equal to Rs x Id is developed across this resistor raising the potential of the source terminal above 0v or ground level. This voltage drop across Rs due to the drain current provides the necessary reverse biasing condition across the gate resistor, R2 effectively generating negative feedback.

So in order to keep the gate-source junction reverse biased, the source voltage, Vs needs to be higher than the gate voltage, Vg. This source voltage is therefore given as:

Then the Drain current, Id is also equal to the Source current, Is as “No Current” enters the Gate terminal and this can be given as:

This potential divider biasing circuit improves the stability of the common source JFET amplifier circuit when being fed from a single DC supply compared to that of a fixed voltage biasing circuit. Both resistor, Rs and the source by-pass capacitor, Cs serve basically the same function as the emitter resistor and capacitor in the common emitter bipolar transistor amplifier circuit, namely to provide good stability and prevent a reduction in the loss of the voltage gain. However, the price paid for a stabilized quiescent gate voltage is that more of the supply voltage is dropped across Rs.

The the value in farads of the source by-pass capacitor is generally fairly high above 100uF and will be polarized. This gives the capacitor an impedance value much smaller, less than 10% of the transconductance, gm (the transfer coefficient representing gain) value of the device. At high frequencies the by-pass capacitor acts essentially as a short-circuit and the source will be effectively connected directly to ground.

The basic circuit and characteristics of a Common Source JFET Amplifier are very similar to that of the common emitter amplifier. A DC load line is constructed by joining the two points relating to the drain current, Id and the supply voltage, Vdd remembering that when Id = 0: ( Vdd = Vds ) and when Vds = 0: ( Id = Vdd/RL ). The load line is therefore the intersection of the curves at the Q-point as follows.

Common Source JFET Amplifier Characteristics Curves

As with the common emitter bipolar circuit, the DC load line for the common source JFET amplifier produces a straight line equation whose gradient is given as: -1/(Rd + Rs) and that it crosses the vertical Id axis at point A equal to Vdd/(Rd + Rs). The other end of the load line crosses the horizontal axis at point B which is equal to the supply voltage, Vdd.

The actual position of the Q-point on the DC load line is generally positioned at the mid center point of the load line (for class-A operation) and is determined by the mean value of Vg which is biased negatively as the JFET is a depletion-mode device. Like the bipolar common emitter amplifier the output of the Common Source JFET Amplifier is 180o out of phase with the input signal.

One of the main disadvantages of using Depletion-mode JFET is that they need to be negatively biased. Should this bias fail for any reason the gate-source voltage may rise and become positive causing an increase in drain current resulting in failure of the drain voltage, Vd.

Also the high channel resistance, Rds(on) of the junction FET, coupled with high quiescent steady state drain current makes these devices run hot so additional heatsink is required. However, most of the problems associated with using JFET’s can be greatly reduced by using enhancement-mode MOSFET devices instead.

MOSFETs or Metal Oxide Semiconductor FET’s have much higher input impedance’s and low channel resistances compared to the equivalent JFET. Also the biasing arrangements for MOSFETs are different and unless we bias them positively for N-channel devices and negatively for P-channel devices no drain current will flow, then we have in effect a fail safe transistor.

JFET Amplifier Current and Power Gains

We said previously that the input current, Ig of a common source JFET amplifier is very small because of the extremely high gate impedance, Rg. A common source JFET amplifier therefore has a very good ratio between its input and output impedances and for any amount of output current, Io the JFET amplifier will have very high current gain Ai.

Because of this common source JFET amplifiers are extremely valuable as impedance matching circuits or are used as voltage amplifiers. Likewise, because power = current × voltage, and output voltages are usually several millivolts or even volts, the power gain, Ap is also very high.

In the next tutorial we will look at how the incorrect biasing of the transistor amplifier can cause Distortion to the output signal in the form of amplitude distortion due to clipping and as well as the effect of phase and frequency distortion.

Crossover Distortion in Amplifiers

We have seen that one of the main disadvantages of the Class-A Amplifier configuration is its low full power efficiency rating due to being biased around its central Q-point.

But we also know that we can improve the amplifier and almost double its efficiency simply by changing the output stage of the amplifier to a Class B push-pull type configuration. However, this is great from an efficiency point of view, but most modern Class B amplifiers are transformerless or complementary types with two transistors in their output stage.

This results in one main fundamental problem with push-pull amplifiers in that the two transistors do not combine together fully at the output both halves of the waveform due to their unique zero cut-off biasing arrangement. As this problem occurs when the signal changes or “crosses-over” from one transistor to the other at the zero voltage point it produces an amount of “distortion” to the output wave shape. This results in a condition that is commonly called Crossover Distortion.

Crossover Distortion produces a zero voltage “flat spot” or “deadband” on the output wave shape as it crosses over from one half of the waveform to the other. The reason for this is that the transition period when the transistors are switching over from one to the other, does not stop or start exactly at the zero crossover point thus causing a small delay between the first transistor turning “OFF” and the second transistor turning “ON”. This delay results in both transistors being switched “OFF” at the same instant in time producing an output wave shape as shown below.

Crossover Distortion Waveform

 

In order that there should be no distortion of the output waveform we must assume that each transistor starts conducting when its base to emitter voltage rises just above zero, but we know that this is not true because for silicon bipolar transistors the base voltage must reach at least 0.7v before the transistor starts to conduct thereby producing this flat spot. This crossover distortion effect also reduces the overall peak to peak value of the output waveform causing the maximum power output to be reduced as shown below.

Non-Linear Transfer Characteristics

 

This effect is less pronounced for large input signals as the input voltage is usually quite large but for smaller input signals it can be more severe causing audio distortion to the amplifier.

Pre-biasing the Output

The problem of Crossover Distortion can be reduced considerably by applying a slight forward base bias voltage (same idea as seen in the Transistor tutorial) to the bases of the two transistors via the center-tap of the input transformer, thus the transistors are no longer biased at the zero cut-off point but instead are “Pre-biased” at a level determined by this new biasing voltage.

Push-pull Amplifier with Pre-biasing

 

This type of resistor pre-biasing causes one transistor to turn “ON” exactly at the same time as the other transistor turns “OFF” as both transistors are now biased slightly above their original cut-off point. However, to achieve this the bias voltage must be at least twice that of the normal base to emitter voltage to turn “ON” the transistors. This pre-biasing can also be implemented in transformerless amplifiers that use complementary transistors by simply replacing the two potential divider resistors with Biasing Diodes as shown below.

Pre-biasing with Diodes

 

This pre-biasing voltage either for a transformer or transformerless amplifier circuit, has the effect of moving the amplifiers Q-point past the original cut-off point thus allowing each transistor to operate within its active region for slightly more than half or 180o of each half cycle. In other words 180o + Bias. The amount of diode biasing voltage present at the base terminal of the transistor can be increased in multiples by adding additional diodes in series. This then produces an amplifier circuit commonly called a Class AB Amplifier and its biasing arrangement is given below.

Class AB Output Characteristics

Crossover Distortion Summary

Then to summarise, Crossover Distortion occurs in Class B amplifiers because the amplifier is biased at its cut-off point. This then results in BOTH transistors being switched “OFF” at the same instant in time as the waveform crosses the zero axis. By applying a small base bias voltage either by using a resistive potential divider circuit or diode biasing this crossover distortion can be greatly reduced or even eliminated completely by bringing the transistors to the point of being just switched “ON”.

The application of a biasing voltage produces another type or class of amplifier circuit commonly called a Class AB Amplifier. Then the difference between a pure Class B amplifier and an improved Class AB amplifier is in the biasing level applied to the output transistors. One major advantage of using diodes over resistors is that the PN-junctions compensate for variations in the temperature of the transistors. Therefore, we can say the a Class AB amplifier is a Class B amplifier with “Bias” and we can therefore summarise as:

• Class A Amplifiers – No Crossover Distortion as they are biased in the center of the load line.
• Class B Amplifiers – Large amounts of Crossover Distortion due to biasing at the cut-off point.
• Class AB Amplifiers – Some Crossover Distortion if the biasing level is set too low.

As well as the three amplifier classes above, there are a number of high efficiency Amplifier Classes relating to switching amplifier designs that use different switching techniques to reduce power loss and increase efficiency. Some of these amplifier designs use RLC resonators or multiple power-supply voltages to help reduce power loss and distortion.

How Transistors Work – A Simple Explanation

How transistors work is probably the hardest concept for you to understand as a beginner. At least it was for me. The problem is that almost everyone is trying to teach that a transistor is “…a semiconductor device”. And instead of just telling you what it does, they explain that “…it consists of n-doped and p-doped materials”.

I don’t know about you, but that statement didn’t help me much!

So let me tell you, in a simple way, how transistors work. I even made a video for you, just to make it clearer.

The transistor is like an electronic switch. It can turn a current on and off. A simple way you can think of it is to look at the transistor as a relay without any moving parts. A transistor is similar to a relay in the sense that you can use it to turn something ON and OFF.

Check out the video explanation I made on the transistor:

There are different types of transistors. A very common one is the “bipolar junction transistor” or “BJT”. And it usually looks like this:

It has three pins: Base (b), collector (c) and emitter (e). And it comes in two versions: NPN and PNP. The schematic symbol for the NPN looks like this:

How transistors work

The transistor works because of something called a semiconducting material. A current flowing from the base to the emitter “opens” the flow of current from the collector to the emitter.

In a standard NPN transistor, you need to apply a voltage of about 0.7V between the base and the emitter to get the current flowing from base to emitter. When you apply 0.7V from base to emitter you will turn the transistor ON and allow a current to flow from collector to emitter.

Let’s look at an example:

In the example above you can see how transistors work. A 9V battery connects to an LEDand a resistor. But it connects through the transistor. This means that no current will flow in that part of the circuit until the transistor turns ON.

To turn the transistor ON you need to apply 0.7V from base to emitter of the transistor. Imagine you have a small 0.7V battery. (In a practical circuit you would use resistors to get the correct voltage from whatever voltage source you have)

When you apply the 0.7V battery from base to emitter, the transistor turns ON. This allows current to flow from the collector to the emitter. And thereby turning the LED ON!

More on the transistor

The transistor is also what makes amplifiers work. Instead of having just two states (on or off) it can also be anywhere in between “fully on” and “fully off”.

A small “control current” can then control how big a portion of a bigger “main current” that is going to flow through it. Thereby, the transistor can amplify a signal.

We use transistors in almost all electronics and it’s probably the most important component in electronics.

Do you understand how transistors work? Post your comments and questions below! Then go check out the LDR circuit diagram and see if you can understand it.

Sensorless BLDC Motor Drive IR3230

The IR3230 is a three-phase brushless DC motor controller/driver with many integrated features. They provide large flexibility in adapting the IR3230 to a specific system requirement and simplify the system design.

Purpose of this document is to provide all the information for realizing a sensorless 3-phase DC motor driver using IR3230 and Microchip PIC16F1937.

To rotate the BLDC motor, the stator windings should be energized in a sequence. It is important to know the rotor position in order to understand which winding will be energized following the energizing sequence. Rotor position is sensed using Hall effect sensors embedded into the stator.

Most BLDC motors have three Hall sensors embedded into the stator on the non-driving end of the motor. Whenever the rotor magnetic poles pass near the Hall sensors, they give a high or low signal, indicating the N or S pole is passing near the sensors. Based on the combination of these three Hall sensor signals, the exact sequence of commutation can be determined. Figure shows the typical connection of the IR3230 to a 2-pole BLDC motor with Hall sensors. Based on the physical position of the Hall sensors, there are two versions of output. The Hall sensors may be at 60° or 120° phase shift to each other. This job is referred to a 60° phase shift configuration.

In more and more applications we need to drive a BLDC motor with no Hall sensor embedded into the stator. In this condition we have to find a way to generate the equivalent digital word that the Hall sensors would generate if present. When a BLDC motor rotates, each winding generates a voltage known as back Electromotive Force or back EMF, which opposes the main voltage supplied to the windings according to Lenz's Law. The polarity of this back EMF is in opposite direction of the energized voltage.

Download Application Note AN-1187 (572 Kb)

IR3230 typical connection to a BLDC motor with Hall effect sensors embedded in the stator

Input Impedance of an Amplifier

Input Impedance, Zin or Input Resistance as it is also called, is an important parameter in the design of a transistor amplifier and as such allows amplifiers to be characterized according to their effective input and output impedances as well as their power and current ratings.

An amplifiers impedance value is particularly important for analysis especially when cascading individual amplifier stages together one after another to minimise distortion of the signal.

The input impedance of an amplifier is the input impedance “seen” by the source driving the input of the amplifier. If it is too low, it can have an adverse loading effect on the previous stage and possibly affecting the frequency response and output signal level of that stage. But in most applications, common emitter and common collector amplifier circuits generally have high input impedances.

Some types of amplifier designs, such as the common collector amplifier circuit automatically have high input impedance and low output impedance by the very nature of their design. Amplifiers can have high input impedance, low output impedance, and virtually any arbitrary gain, but were an amplifiers input impedance is lower than desired, the output impedance of the previous stage can be adjusted to compensate or if this is not possible then buffer amplifier stages may be needed.

In addition to voltage amplification ( Av ), an amplifier circuit must also have current amplification ( Ai ). Power amplification ( Ap ) can also be expected from an amplifier circuit. But as well as having these three important characteristics, an amplifier circuit must also have other characteristics like high input impedance ( Zin ), low output impedance ( Zout ) and some degree of bandwidth, ( Bw ). Either way, the “perfect” amplifier will have infinite input impedance and zero output impedance.

Input and Output Impedance

In many ways, an amplifier can be thought of as a type of “black box” which has two input terminals and two output terminals as shown. This idea provides a simple h-parameter model of the transistor that we can use to find the DC set point and operating parameters of an amplifier. In reality one of the terminals is common between the input and output representing ground or zero volts.

When looking from the outside in, these terminals have an input impedance, Zin and an output impedance, Zout. The input and output impedance of an amplifier is the ratio of voltage to current flowing in or out of these terminals. The input impedance may depend upon the source supply feeding the amplifier while the output impedance may also vary according to the load impedance, RL across the output terminals.

The input signals being amplified are usually alternating currents (AC) with the amplifier circuit representing a load Z to the source. The input impedance of an amplifier can be tens of ohms, (Ω’s) to a few thousand ohms, (kΩ’s) for bipolar based transistor circuits up to millions of ohms, (MΩ’s) for FET based transistor circuits.

When a signal source and load are connected to an amplifier, the corresponding electrical properties of the amplifier circuit can be modelled as shown.

Output and Input Impedance Model

Where, VS is the signal voltage, RS is the internal resistance of the signal source, and RLis the load resistance connected across the output. We can expand this idea further by looking at how the amplifier is connected to the source and load.

When an amplifier is connected to a signal source, the source “sees” the input impedance, Zin of the amplifier as a load. Likewise, the input voltage, Vin is what the amplifier sees across the input impedance, Zin. Then the amplifiers input can be modelled as a simple voltage divider circuit as shown.

Amplifier Input Circuit Model

The same idea applies for the output impedance of the amplifier. When a load resistance, RL is connected to the output of the amplifier, the amplifier becomes the source feeding the load. Therefore, the output voltage and impedance automatically becomes the source voltage and source impedance for the load as shown.

Amplifier Output Circuit Model

Then we can see that the input and output characteristics of an amplifier can both be modelled as a simple voltage divider network. The amplifier itself can be connected in Common Emitter (emitter grounded), Common Collector (emitter follower) or in Common Base configurations. In this tutorial we will look at the bipolar transistor connected in a common emitter configuration seen previously.

Common Emitter Amplifier

The so called classic common emitter configuration uses a potential divider network to bias the transistors Base. Power supply Vcc and the biasing resistors set the transistor operating point to conduct in the forward active mode. With no signal current flow into the Base, no Collector current flows, (transistor in cut-off) and the voltage on the Collector is the same as the supply voltage, Vcc. A signal current into the Base causes a current to flow in the Collector resistor, Rc generating a voltage drop across it which causes the Collector voltage to drop.

Then the direction of change of the Collector voltage is opposite to the direction of change on the Base, in other words, the polarity is reversed. Thus the common emitter configuration produces a large voltage amplification and a well defined DC voltage level by taking the output voltage from across the collector as shown with resistor RLrepresenting the load across the output.

Single Stage Common Emitter Amplifier

Hopefully by now we are able to calculate the values of the resistors required for the transistor to operate in the middle of its linear active region, called the quiescent point or Q point, but a quick refresher will help us understand better how the amplifiers values were obtained so that we can use the above circuit to find the input impedance of the amplifier.

Firstly lets start by making a few simple assumptions about the single stage common emitter amplifier circuit above to define the operating point of the transistor. The voltage drop across the the Emitter resistor, VRE = 1.5V, the quiescent current, IQ = 1mA, the current gain (Beta) of the NPN transistor is 100 ( β = 100 ), and the corner or breakpoint frequency of the amplifier is given as: ƒ-3dB = 40Hz.

As the quiescent current with no input signal flows through the Collector and Emitter of the transistor, then we can say that, IC = IE = IQ = 1mA, so by using Ohms Law:

With the transistor switched fully-ON (saturation), the voltage drop across the Collector resistor, Rc will be half of Vcc – VRE to allow for maximum output signal swing from peak-to-peak around the center point without clipping of the output signal.

Note that the DC no signal voltage gain of the amplifier can be found from -RC/RE. Also notice that the gain is negative in value as the output signal is inverted. i.e. 180o out-of-phase with the input signal.

As the NPN transistor is forward biased, the Base-Emitter junction acts like a forward biased diode so the Base will be 0.7 volts more positive than the Emitter voltage ( Ve + 0.7V ), therefore the voltage across the Base resistor R2 will be:

If the two biasing resistors are already given, we can also use the following standard voltage divider formula to find the Base voltage Vb across R2.

The information given stated that the quiescent current is 1mA. Thus the transistor is biased with a Collector current of 1mA across the 12 volt supply, Vcc. This Collector current is proportional to the Base current as Ic = βIb. The DC current gain, Beta ( β ) of the transistor was given as 100, then the Base current flowing into the transistor will be:

The DC bias circuit formed by the voltage divider network of R1 and R2 sets the DC operating point. The Base voltage was previously calculated at 2.2 volts then we need to establish the proper ratio of R1 to R2 to produce this voltage value across the 12 volt supply, Vcc.

Generally, for a standard voltage divider DC biasing network of a common emitter amplifier circuit, the current flowing through the lower resistor, R2 is ten times greater than the DC current flowing into the Base. Then the value of resistor, R2 can be calculated as:

The voltage dropped across resistor R1 will be the supply voltage minus the Base bias voltage. Also if resistor R2 carries 10 times the Base current, upper resistor R1 of the series chain must pass the current of R2 plus the transistors actual Base current, Ib. In other words, 11 times the Base current as shown.

For a common emitter amplifier, the reactance Xc of the Emitter bypass capacitor is usually one tenth (1/10th) the value of the Emitter resistor, RE at the cut-off frequency point. The amplifiers specifications gave a -3dB corner frequency of 40Hz, then the value of capacitor CE is calculated as:

Now we have the values established for our common emitter amplifier circuit above, we can now look at calculating its input and output impedance of amplifier as well as the values of the coupling capacitors C1 and C2.

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Basic Emitter Amplifier Model

The generalised formula for the input impedance of any circuit is ZIN = VIN/IIN. The DC bias circuit sets the DC operating “Q” point of the transistor and as the input capacitor, C1 acts as an open circuit and blocks any DC voltage, at DC (0Hz) the input impedance (ZIN) of the circuit will be extremely high. However when an AC signal is applied to the input, the characteristics of the circuit changes as capacitors act as short circuits at high frequencies and pass AC signals.

The generalised formula for the AC input impedance of an amplifier looking into the Base is given as ZIN = REQ||β(RE+ re). Where REQ is the equivalent resistance to ground (0v) of the biasing network across the Base, and re is the internal signal resistance of the forward biased Emitter layer. Then if we short out the 12 volt power supply, Vcc to ground because Vcc appears as a short to AC signals, we can redraw the common emitter circuit above as follows:

Amplifier Circuit Model

Then we can see that with the supply voltage shorted, there are a number of resistors connected in parallel across the transistor. By taking the input side of the transistor amplifier only and treating capacitor C1 as a short circuit to AC signals, we can redraw the above circuit to define the input impedance of the amplifier as:

Input Impedance of Amplifier

We said in the previous Common Emitter Amplifier tutorial that the internal signal resistance of the Emitter layer was equal to the product of 25mV ÷ Ie with this 25mV value being the internal volt drop and IE = IQ. Then for our amplifier circuit above the equivalent AC resistance value re of the Emitter diode is given as:

Emitter Leg Signal Resistance

Where re represents a small internal resistor in series with the Emitter. Since Ic/Ib = β, then the value of the transistors Base impedance will be equal to β x re. Note that if bypass capacitor CE is not included within the amplifiers design, then the value becomes: β(RE+ re) significantly increasing the input impedance of the amplifier.

In our example bypass capacitor, CE is included, therefore the input impedance, ZIN of the common Emitter amplifier is the input impedance “seen” by the AC source driving the amplifier and is calculated as:

Input Impedance Equation

This 2.2kΩ’s is the input impedance looking into the input terminal of the amplifier. If the impedance value of the source signal is known, and in our simple example above it is given as 1kΩ, then this value can be added or summed with ZIN if required.

But lets assume for one minute that our circuit has no bypass capacitor, CE connected. What would be the input impedance of the amplifier without it. The equation would still be the same except for the addition of RE in the β(RE+ re) part of the equation as the resistor will no longer be shorted at high frequencies. Then the unbypassed input impedance of our amplifier circuit without CE will be:

Input Impedance without Bypass Capacitor

Then we can see that the inclusion of the Emitter leg bypass capacitor makes a huge difference to the input impedance of the circuit as the impedance goes down from 15.8kΩ’s without it to 2.2kΩ’s with it in our example circuit. We will see later that the addition of this bypass capacitor, CE also increases the amplifiers gain.

In our calculations to find the input impedance of the amplifier, we have assumed that the capacitors in the circuit have zero impedance (Xc = 0) for AC signal currents, as well as infinite impedance (Xc = ∞) for DC biasing currents. Now that we know the bypassed input impedance of the amplifier circuit, we can use this value of 2.2kΩ’s to find the value of the input coupling capacitor, C1 required at the specified cut-off frequency point which was given previously as 40Hz. Therefore:

Input Coupling Capacitor Equation

Now that we have a value for the input impedance of our single stage common Emitter amplifier circuit above, we can also obtain an expression for the output impedance of the amplifier in a similar fashion.

Output Impedance of an Amplifier

The Output Impedance of an amplifier can be thought of as being the impedance (or resistance) that the load sees “looking back” into the amplifier when the input is zero. Working on the same principle as we did for the input impedance, the generalised formula for the output impedance can be given as: ZOUT = VCE/IC.

But the signal current flowing in the Collector resistor, RC also flows in the load resistor, RL as the two are connected in series across Vcc. Then again, by taking the output side of the transistor amplifier only and treating the output coupling capacitor C2 as a short circuit to AC signals, we can redraw the above circuit to define the output impedance of the amplifier as:

Output Impedance of Amplifier

Then we can see that the output signal resistance is equal to RC in parallel with RL giving us an output resistance of:

Output Impedance Equation

Note that this value of 833Ω’s results from the fact that the load resistance is connected across the transistor. If RL is omitted, then the output impedance of the amplifier would be equal to the Collector resistor, RC only.

Now that we have a value for the output impedance of our amplifier circuit above, we can calculate the value of the output coupling capacitor, C2 as before at the 40Hz cut-off frequency point.

Output Coupling Capacitor Equation

Again the value of coupling capacitor C2 can be calculated either with or without the inclusion of load resistor RL.

Voltage Gain

The voltage gain of a common emitter circuit is given as Av = ROUT/REMITTER where ROUTrepresents the output impedance as seen in the Collector leg and REMITTER is equal the the equivalent resistance in the Emitter leg either with or without the bypass capacitor connected.

Without the bypass capacitor CE connected, (RE+ re).

and with the bypass capacitor CE connected, (re) only.

Then we can see that the inclusion of the bypass capacitor within the amplifier design makes a dramatic change to the voltage gain, Av of our common emitter circuit from 0.5 to 33. It also shows that the common emitter gain does not go to infinity when the external emitter resistor is shorted by the bypass capacitor at high frequencies but instead the gain goes to the finite value of ROUT/re.

We have also seen that as the gain goes up the input impedance goes down from 15.8kΩ’s without it to 2.2kΩ’s with it. The increase in voltage gain can be considered an advantage in most amplifier circuits at the expense of a lower input impedance.

Input Impedance Summary

In this tutorial we have seen that the input impedance of a common emitter amplifier can be found by shorting out the supply voltage and treating the voltage divider biasing circuit as resistors in parallel. The impedance “seen” looking into the divider network (R1||R2) is generally much less that the impedance looking directly into the transistors Base, β(RE+ re) as the AC input signal changes the bias on the Base of the transistor controlling the current flow through the transistor.

There are many ways to bias the transistor. Thus, there are many practical single transistor amplifier circuits each with their own input impedance equations and values. If you require the input impedance of the whole stage plus source impedance, then you will need to consider Rs in series with the base bias resistors as well, (Rs + R1||R2).

The output impedance of a common emitter stage is just equal to the collector resistor in parallel with the load resistor (RC||RL) if connected otherwise its just RC. The voltage gain, Av of the amplifier is dependant upon RC/RE.

The emitter bypass capacitor, CE can provide an AC ground path for the Emitter, shorting out the emitter resistor, RE leaving only the signal Emitter resistance, re in the Emitter leg. The effect of this is an increase in the gain of the amplifier (from 0.5 to 33) at high frequencies but also a decrease in the amplifiers input impedance value, (from 18.5kΩ to 2.2kΩ).

With this bypass capacitor removed, the amplifiers voltage gain, Av decreases and ZINincreases. One way to maintain a fixed amount of gain and input impedance is to include an additional resistor in series with CE to create what is called a “split-emitter” amplifier circuit that is a trade-off between an unbypassed and a fully bypassed amplifier circuit. Note that the addition or removal of this bypass capacitor has no effect on the amplifiers output impedance.

Then we can see that the input and output impedances of an amplifier can play an important role in defining the transfer characteristics of an amplifier with regards to the relationship between the output current, Ic and the input current, Ib. Knowing an amplifiers input impedance can help to graphically construct a set of output characteristics curves for the amplifier.