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Chapter 6Chapter 6BJT AmplifiersBJT Amplifiers

ObjectivesObjectives

Understand the concept of amplifiers

Identify and apply internal transistor parameters

Understand and analyze common-emitter, common-base, and common-collector amplifiers

Discuss multistage amplifiers

Troubleshoot amplifier circuits.

IntroductionIntroduction

One of the primary uses of a transistor is to amplify ac signals. This could be an audio signal or perhaps some high frequency radio signal. It has to be able to do this without distorting the original input.

Amplifier OperationAmplifier Operation

Recall from the previous chapter that the purpose of dc biasing was to establish the Q-point for operation. The collector curves and load lines help us to relate the Q-point and it’s proximity to cutoff and saturation. The Q-point is best established where the signal variations do not cause the transistor to go into saturation or cutoff.

What we are most interested in is the ac signal itself. Since the dc part of the overall signal is filtered out in most cases, we can view a transistor circuit in terms of just its ac component.

Amplifier OperationAmplifier OperationFor the analysis of transistor circuits from both dc and ac perspectives the ac subscripts are lower case and italicized. Instantaneous values use both italicized lower case letters and subscripts.

Review of previous classReview of previous class

Amplifier OperationAmplifier OperationThe boundary between cutoff and saturation is called the linear region. A transistor which operates in the linear region is called a linear amplifier. Note that only the ac component reaches the load because of the capacitive coupling and that the output is 180º out of phase with input.

Review of Output Characteristic Review of Output Characteristic

Example 1Example 1

Example 1

Let Q as the selected operating point(quiscent point).

If Ib varies about 10microamps, find Vce and Ic variations

Transistor Equivalent CircuitsTransistor Equivalent Circuits

• We represent a transistor with a network of several elements to simplify the analysis

Transistor Equivalent CircuitsTransistor Equivalent CircuitsWe can view transistor circuits by use of resistance or r parameters for better understanding. Since the base resistance, rb is small it is normally is not considered and since the collector resistance, re is fairly high we consider it as an open. The emitter resistance, re is the main parameter that is viewed.

You can determine rc from this simplified equation.

re = 25 mV/IE

Example 2Example 2

• A current measurement device gave a reading of a direct current value for current flowing at the Emitter of a transistor as 2mA. Find resistance internal to the transistor between the emitter and base terminal…

• It is re’ and is given by 25mV/IE=12.5ohm• Note: we can use this equation to find IE for

given re’ value

Transistor Equivalent CircuitsTransistor Equivalent CircuitsThe two graphs best illustrate the difference between βDC and βac. The two only differ slightly.

Transistor Equivalent CircuitsTransistor Equivalent CircuitsSince r parameters are used throughout the rest of the textbook we will not go into deep discussion about hparameters. However, since some data sheets include or exclusively provide h parameters these formulas can be used to convert them to r parameters.

r’e = hre/hoe r’c = (hre + 1)/hoe r’b = hie - (1+ hfe) hre/hoe

Direct conversion: αac = hfb and βac = hfe

h parameters are labeled in this manner: hxy

x: i (input impedance), r (voltage feedback ratio),f (forward current gain), o (output admittance)

y : e (common emitter), b (common base), c (common collector)

The CommonThe Common--Emitter AmplifierEmitter Amplifier

The common-emitter amplifier exhibits high voltage and current gain. The output signal is 180º out of phase with the input. Now lets use our dc and ac analysis methods to view this type of transistor circuit.

Classroom exercise: draw signals internal to the amplifier circuit and explain why…

The Common Emitter AmplifierThe Common Emitter AmplifierDC Analysis DC Analysis

The dc component of the circuit “sees” only the part of the circuit that is within the boundaries of C1, C2, and C3 as the dc will not pass through these components. The equivalent circuit for dc analysis is shown.

The methods for dc analysis are just are the same as dealing with a _____?_____ circuit. Classroom exercise!

Circuit Redraw method for ACCircuit Redraw method for AC Common Emitter AmplifierCommon Emitter AmplifierAC Equivalent CircuitAC Equivalent Circuit

The ac equivalent circuit basically replaces the capacitors with shorts, being that ac passes through easily through them. The power supplies are also effectively shorts to ground for ac analysis.

Common Emitter AmplifierCommon Emitter AmplifierAC Equivalent CircuitAC Equivalent Circuit

The ac equivalent circuit basically replaces the capacitors with shorts, being that ac passes through easily through them. The power supplies are also effectively shorts to ground for ac analysis.

Classroom exercise: Prove that Rin(base)=βacre’


We can look at the input voltage in terms of the equivalent base circuit (ignore the other components from the previous diagram). Note the use of simple series-parallel analysis skills for determining Vin.

Example : For the circuit in slide 17, an ac input with Vs=10mV, source resistance of 300Ω and IE of 3.8mA, find the signal voltage at base.


Example : For the circuit in slide 17, an ac input with Vs=10mV, source resistance of 300Ω and IE of 3.8mA, find the signal voltage at base.

re’ = 25mV/3.8mA = 6.58Ω

Rin(base) = βacre’ = 160*6.58

= 1.05kΩ

Rin(tot) = R1//R2//Rin(base)

= 873

Vb = Vs*Rin(tot)/(Rin(tot)+Rs)

= 7.44mV

Common Emitter Amplifier Common Emitter Amplifier AC Equivalent CircuitAC Equivalent Circuit

The input resistance as seen by the input voltage can be illustrated by the r parameter equivalent circuit. The simplified formula below is used.

Rin(base) = βacr’e

The output resistance is for all practical purposes the value of RC.


Voltage gain can be easily determined by dividing the ac output voltage by the ac input voltage.

Av = Vout/Vin = Vc/Vb

Voltage gain can also be determined by the simplified formula below.

(Please do the derivation)

Av = RC/r’e

AttenuationAttenuation

Taking the attenuation from the ac supply internal resistance and input resistance must be determined since it affects the overall gain.

Attenuation = (Vb/Vs) = Rin(total) / (Rs + Rin(total))


Taking the attenuation from the ac supply internal resistance and input resistance must be determined since it affects the overall gain.

Attenuation = (Vb/Vs) = Rin(total) / (Rs + Rin(total))


Taking the attenuation from the ac supply internal resistance and input resistance into consideration is included in the overall gain.

A’v = (Vb/Vs)Av


Example: Suppose we have a source of 10mV. Due to resistances, the voltage felt at base(amplifier input) is 5mV.

Then, attenuation factor is 5mV/10mV = 0.5

Let say the gain from input(base) to output is 20, then the output voltage is 20*5mV = 100mV.

The overall gain is therefore 100mV/10mV = 10.

This is equivalent to multiplying attenuation factor and amplifier gain:

Which is 0.5*20 = 10

The CommonThe Common--Emitter AmplifierEmitter Amplifier

The emitter bypass capacitor helps increase the gain by allowing the ac signal to pass more easily.

The XC(bypass) should be about ten times less than REat the minimum frequency where the circuit will operate! Write the simple formula based on this statement. Homework: Calculate the minimum frequency for the circuit below!

EXAMPLE 6-4Select a minimum value for a emitter bypass capasitor, C2, in figure 6-16 if the amplifier must operate over a frequency range from 2 k Hz to 10 kHz.FIGURE 6-16

Solution Since RE=560Ω, Xc of the bypass capasitor, C2, should be no greater than

10Xc= REXc= RE/10 = 560Ω/10 = 56Ω

The capacitance value is determined at the minimum frequency of 2 kHz as follows:

C2=1/(2fXc) = 1/(2(2 kHz)( 56Ω)) = 1.42µFThis is a minimum value for the bypass capasitor for this circuit.You can use a large value, although cost and physical size usually impose limitations.Related Problem If the minimum frequency is reduced to 1 kHz, what value of bypass capasitor must you use?

Voltage Gain w/o Bypass CapacitorVoltage Gain w/o Bypass CapacitorThe emitter is no longer at AC Ground, the RE is now seen by the signal and adds to internal re’ Av = Rc/(re’+RE). Gain is somewhat reduced!

Voltage Gain w/o Bypass CapacitorVoltage Gain w/o Bypass CapacitorEXAMPLE 6-5

Calculate the base-to-collector voltage gain of the amplifier in Figure 6-16 both without and with an emitter bypass capasitor if there is ni load resistor.

Solution From Example 6-3,r’e=6.58Ω from this same amplifier. Without C2 the gain is

Av=Rc/(r’e+RE) = 1.0 kΩ/6.58Ω=152

With C2, the gain isAv=Rc/r’e=1.0kΩ/6.58Ω= 152

As you can see, the bypass capasitor makes quite a difference.Related Problem Determind the base-to-collector voltage gain in figure 6-16 with RE bypassed, for the following circuit values: Rc=1.8kΩ, RE=1.0kΩ,R1 = 33kΩ, and R2=6.8kΩ.

Effect of Inserting LoadEffect of Inserting LoadWhen load RL is added at the output capacitor end, the effective resistance seen by the signal is Rc in parallel with RL.

Rc(tot) = RC//RL

Effect of Inserting LoadEffect of Inserting LoadEXAMPLE 6-6

Calculate the base-to-collector voltage gain of the amplifier in Figure 6-16 when a load resistance of 5 kΩ in connected to the output. The emitter is effectively bypassed and r’e=6.58Ω.Solution The ac collector resistance is

Rc=RcRl/(Rc+Rl) = ((1.0 kΩ)(5 kΩ))/6 kΩ= 833 Ω

Therefore,Av=Rc/r’e = 833 Ω/6.58 Ω = 127

The unloaded gain was found to be 152 in Example 6-5Related Problem Determine the base-to-collector voltage gain in Figure 6-16 when a 10 kΩ load resistence is connected from collector to ground. Change the resistence values as follows: Rc=1.8 kΩ, R1=33 kΩ, and R2=6.8 kΩ. The emitter resistor is effectively bypassed and r’e=10 Ω.

Voltage Gain StabilityVoltage Gain Stability

• Bypassing RE with capacitor causes the gain to be more dependent on re’- reduced stability.

• Without bypassing the gain is Av=RC/(re’+RE) and if RE is big enough. Av=RC/RE…. Which is less dependent on re’ or IE.

• We need to intorduce “swamping techniques”• RE2 is introduced, RE1>10re’ is not bypassed but RE2 is

bypassed making Av= Av=RC/(re’+RE1) or…..?

The CommonThe Common--Emitter AmplifierEmitter AmplifierThe bypass capacitor makes the gain unstable since transistor amplifier becomes more dependant on IE. This effect can be swamped or somewhat alleviated by adding another emitter resistor(RE1).

The CommonThe Common--Collector AmplifierCollector AmplifierThe common-collector amplifier is usually referred to as the emitter follower because there is no phase inversion or voltage gain. The output is taken from the emitter. The common-collector amplifier’s main advantages are it’s high current gain and high input resistance.

The CommonThe Common--Collector AmplifierCollector Amplifier

Because of it’s high input resistance the common-collector amplifier used as a buffer to reduce the loading effect of low impedance loads. The input resistance can be determined by the simplified formula below.

Rin(base) ≅ βac(r’e + Re)

The CommonThe Common--Collector AmplifierCollector Amplifier

The output resistance is very low. This makes it useful for driving low impedance loads.

The current gain(Ai) is approximately βac.

The power gain is approximately equal to the current gain(Ai).

The voltage gain is approximately 1.

The CommonThe Common--Collector AmplifierCollector AmplifierThe darlington pair is used to boost the input impedance to reduce loading of high output impedance circuits. The collectors are joined together and the emitter of the input transistor is connected to the base of the output transistor. The input impedance can be determined the formula below.

Rin = βac1βac2Re

The CommonThe Common--Base AmplifierBase AmplifierThe common-base amplifier has high voltage gain with a current gain no higher than 1. It has a low input resistance making it ideal for low impedance input sources. The ac signal is applied to the emitter and the output is taken from the collector.

Theodore f Bogart: Electronic Devices and Circuits

The CommonThe Common--Base AmplifierBase Amplifier

The common-base voltage gain(Av) is approximately equal to Rc/r’eThe current gain is approximately 1.

The power gain is approximately equal to the voltage gain.

The input resistance is approximately equal to r’e.

The output resistance is approximately equal to RC.

The CommonThe Common--Base AmplifierBase AmplifierYou may see circuits biased this way….. How do we analyse?

IE=(VEE-VBE)/RE

VC=VCC-ICRC

Av? With and w/o load?

Multistage AmplifiersMultistage Amplifiers

Two or more amplifiers can be connected to increase the gain of an ac signal. The overall gain can be calculated by simply multiplying each gain together.

A’v = Av1Av2Av3 ……


Gain can be expressed in decibels(dB). The formula below can be used to express gain in decibels.

A v(dB) = 20logAv

Each stage’s gain can now can be simply added together for the total.


The capacitive coupling keeps dc bias voltages separate but allows the ac to pass through to the next stage.


The output of stage 1 is loaded by input of stage 2. This lowers the gain of stage 1. This ac equivalent circuit helps give a better understanding how loading can effect gain.

Multistage AmplifiersMultistage AmplifiersDirect coupling between stage improves low frequency gain. The disadvantage is that small changes in dc bias from temperature changes or supply variations becomes more pronounced.

TroubleshootingTroubleshooting

Troubleshooting techniques for transistor amplifiers is similar to techniques covered in Chapter 2. Usage of knowledge of how an amplifier works, symptoms, and signal tracing are all valuable parts of troubleshooting. Needless to say experience is an excellent teacher but having a clear understanding of how these circuits work makes the troubleshooting process more efficient and understandable.


The following slide is a diagram for a two stage common-emitter amplifier with correct voltages at various points.

Utilize your knowledge of transistor amplifiers and troubleshooting techniques and imagine what the effects would be with various faulty components, for example, open resistors, shorted transistor junctions or capacitors. More importantly how would the output be effected by these faults. In troubleshooting it is most important to understand the operation of a circuit

What faults could cause low or no output?

What faults could cause a distorted output signal?


SummarySummary

Transistor circuits can be view in terms of it’s ac equivalent for better understanding.

The common-emitter amplifier has high voltage and current gain.

The common-collector has a high current gain and voltage gain of 1. It has a high input impedance and low output impedance.

Most transistors amplifiers are designed to operate in the linear region.

SummarySummary

Multi-stage amplifiers are amplifier circuits cascaded to increased gain. We can express gain in decibels(dB).

Troubleshooting techniques used for individual transistor circuits can be applied to multistage amplifiers as well.

The common-base has a high voltage gain and a current gain of 1. It has a low input impedance and high output impedance

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