Analogue ELEK1289 - Electronic Systems and Practice II - Unit 3 Amplifier Modelling

Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingAnalysis of Amplifier SystemsAmplifier Model We have seen in the previous chapter, that an amplifier is a device that increases the magnitude of a signal for use by a load. Amplifiers are complicated arrangements of transistors, resistors and other components. However a simplified description is all that is necessary when the requirement is to analyse the source and load behaviour. The amplifier can be thought of as an interface between the source and the load.A basic amplifier may be represented as follows:Gain Element Input Output Signal SignalIt is useful to classify amplifiers into four broad categories:The Voltage Amplifier The Transconductance Amplifier Input signal is voltage VIN ,  Input signal is current IIN Output signal is voltage VOUT  Output signal is voltage VOUT VOUT IOUT  The voltage gain AV   The Transconductance gain Gm  VIN VINThe Current Amplifier The Transresistance Amplifier Input signal is current IIN  Input signal is a voltage VIN  Output signal is current IOUT  Output signal is current IOUT IOUT VOUT  The current gain AI   The Transresistance gain Rm  IIN IINIn this chapter we will focus on the analysis of the voltage amplifier and the current amplifier.1 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingNon Ideal Amplifiers Saturation Voltage An ideal amplifier means that the amplifier introduces no noise or distortion to the signal, i.e. the output replicates the input exactly as the input signal increases. However, for real amplifiers the gain will taper off as VIN increases in magnitude. This is known as saturation of the amplifier.Non Ideal CharacteristicVOUT (V)40Saturation voltage = 30V 30200 2 = 10 A0 V (V) 0 1 2 3 IN Saturation in Non-Ideal Voltage AmplifierFrequency Response A non-ideal amplifier will also have a frequency response in that as the frequency increases the gain will fall for a particular magnitude of the input signal.Frequency Response of Non Ideal Amplifier Gain A4030 Ideal20 Practical100 f (Hz) 0 10KHz 100KHz 1MHzFrequency Response of Non-Ideal AmplifierFor an ideal voltage amplifier we assume that the gain remains constant as the frequency increases.2 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingInput impedance For an ideal voltage amplifier we assume that the input impedance is infinite. However, a real voltage amplifier will also have a finite input impedance. RINVIN VOUTInput Impedance3 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingEquivalent Circuit Models for Gain Elements The concept of equivalent circuits, covered in the Electronic Circuits and Devices module to model voltage and current sources, can be applied to more complicated case of an amplifier. By drawing an amplifier as an equivalent circuit, you can simplify equations related to its performance.The Voltage Amplifier A basic voltage amplifier may be represented as follows:VOUT Transfer Charteristic for Various Gains1000 =1 75 0 V 2 A AV = V A V VOUT IN 50 =5 A V250 V 0 5 10 INVOUT AV  or VOUT  AV .VIN VIN Voltage AmplifierThe input signal from a source is applied to the input terminals of the amplifier, and an output is taken from the second set of terminals. In practice the amplifier input and output voltages will be influenced by the characteristics of the source, which supplies the input to the amplifier, and the load, which the amplifier output is required to drive. To analyse this, it is necessary to adopt a model of the interior of the amplifier. R S I IIN OUT A +VOUT RL VIN VS -SOURCE GAIN ELEMENT LOADRS: Source Resistance VS: Source Voltage RL: Load ResistanceSource – Gain Element – Load4 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingThe Voltage Amplifier Equivalent Circuit The amplifier input and output circuits are characterised using Thevenin’s Theorem to produce the model for a voltage amplifier. Gain ElementROUT+ AV VIN VOUT VIN RIN -Voltage Amplifier Equivalent Circuit ModelAs amplifiers are designed to operate from input to output only, VOUT does not influence VIN. The amplifier’s input terminals present an input resistance to the voltage source. Therefore, the input circuit is modelled as an input resistance RIN.  RIN is the Thevenin resistance seen at the input terminals and is called the input resistance of the amplifier. The output circuit of the amplifier can be modelled as Thevenin source, as shown above: ROUT is the Thevenin resistance seen at the output terminals and is called the output resistance of the amplifier AVVIN is the Thevenin voltage seen at the output of the amplifier where AV is the voltage gain of the amplifier. Because the magnitude of the Thevenin source is dependent on the unloaded gain AV and the input voltage VIN., the amplifiers output circuit is said to contain a dependent source. 5 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingAn amplifier does not exist in isolation and in practice the input will be driven by a non-ideal source and the output will be connected to a load. The load is assumed to be a resistance connected across the output terminals as shown below:Source Gain Element LoadRS ROUT+ + AV VIN VOUT VIN RIN RL VS - -Voltage Source Voltage Amplifier Load Equivalent Equivalent Circuit Model Equivalent Circuit CircuitSource/Amplifier/Load Equivalent CircuitsAnalysing the input circuit:Source Gain ElementVRS =IIN RSIINR + S RIN V S VIN -Input Circuit Figure 1: Amplifier Input CircuitThe input resistance RIN affects the input voltage VIN to the amplifier because it forms a voltage divider with the source resistance RS.  RIN  The voltage division rule => VIN  VS  equation 1  RIN  RS  So for example, if the source voltage is 3 V, the source resistance is 500  and the input resistance is 1 k, then the input voltage would be: VIN = 3 ( 1000 / 1500 ) = 2 V6 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingAnalysing the output circuit:Gain Element LoadVROUT = IOUT ROUTIOUT+ VOUT A V RL - V INOutput CircuitThe presence of RL modifies the behaviour of the amplifier. The unloaded voltage gain of the amplifier is called AV VOUT AV  equation 2 IN V RLNote AV is independent of the circuit into which the amplifier is connected RL  Applying the voltage division rule => VOUT  AVVIN  equation 3  ROUT  RL   RIN  But, the input voltage is VIN  VS  .  RIN  RS Therefore substituting VIN into the VOUT equation gives =>:    RIN  RL  VOUT  AVVs      RIN  RS  ROUT  RL The loaded voltage gain of the amplifier is called AVL and is defined as: VOUT AVL  equation 4 VSTherefore dividing the VOUT equation above by VS gives an expression for the loaded voltage gain:    RIN  RL  AVL  AV    equation 5    RIN  RS  ROUT  RL Looking at this equation, it can be seen the loaded gain AVL must be less than or equal to the open circuit gain AV: AVL  AV In practice the loaded voltage gain is less than the unloaded voltage gain.7 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingThe loaded voltage gain will be at its maximum if the amplifier is designed such that:RIN ->  and ROUT -> 0 Thus,    RIN  RL  AVL  AV    => AVL --> AV    RIN  RS  ROUT  RL Ideal voltage amplifier: Real voltage amplifierRIN =  , ROUT = 0  In reality we require: AVL = AV RIN >> RS  VIN  VS=> independent of source and load ROUT << RL  VOUT  AvVSExample Problems Attempt the following example problems. The solutions will be covered during lectures.Exercise 1A non-ideal voltage amplifier has an input resistance of RIN = 2 kW and output resistance of ROUT = 0 W. The gain AV = 20 and the input current is 1 mA. What is VOUT?Exercise 2 A voltage amplifier has the following characteristics:Gain AV = 10, RIN = ¥ W and ROUT = 0 W.Calculate VOUT and IOUT if RL = 100 W, RS = 1 kW, VS = 10 mV.Exercise 3 Repeat example No. 2 for the two cases:(a) RS = 500 W and RIN = 1.5 k.(b) RS = 200 W and RIN = 800 .(c) RS = 200 W and RIN = 800 . and ROUT = 100 .(d) RS = 200 W and RIN = 800 . and ROUT = 1000 . Comment on the results obtainedExercise 4Calculate the power gain AP and the output power in dBW of a voltage amplifier having RIN = 2 kW, AV = 20, VS = 8 V, RS = 200 W and RL = 200 W. Assume ROUT of the amplifier to be ideal.8 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingAnalysis of Amplifier Systems [continued] The Current Amplifier A basic current amplifier may be represented as follows:IOUT (mA) Transfer Charteristic for Various Gains of Current Amplifier40 I IIN OUT 0 1 =5 30 = A AI A2 20 A=100 I (mA) 0 2 4 6 8 10 12 IN IOUT A is the current gain AI  or IOUT = AIIIN IIN Current AmplifierCurrent Amplifier Using Norton’s theorem, the following model can be adopted for a current amplifier:I IIN OUTR R AIIIN ROUT VOUT R IS S IN LCurrent Source Current Amplifier Load Equivalent Equivalent Circuit Model Equivalent Circuit Circuit RIN is the resistance seen at the input terminals and is called the input resistance of the amplifier.  ROUT is the Norton resistance seen at the output terminals and is called the output resistance of the amplifier AIIIN is the Norton current produced at the output of the amplifier where AI is the current gain of the amplifier. 9 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingAnalysing the input circuit:IINRS RIN ISCurrent Source Current Amplifier Equivalent Circuit Input Equivalent CircuitAmplifier Input CircuitThe input resistance RIN affects the input current IIN to the amplifier as follows:  RS  The current division rule => IIN  IS  equation 6  RIN  RS Analysing the output circuit:IOUTA I V I IN ROUT OUT RLCurrent Amplifier Load Output Equivalent Circuit Equivalent CircuitThe zero-loaded current gain of the amplifier is called AI IOUT AI  equation 7 IN I RL 0Note AI is independent of the circuit into which the amplifier is connected ROUT  Applying the current division rule => IOUT  AIIIN  equation 8  ROUT  RL 10 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling RS  But, the input current is IIN  IS   RIN  RS Therefore substituting IIN into the IOUT equation gives =>:    RS  ROUT  IOUT  AIIs      RIN  RS  ROUT  RL The loaded current gain of the amplifier is called AIL and is defined as: IOUT AIL  equation 9 IS Therefore dividing the IOUT equation above by IS gives an expression for the loaded current gain:    RS  ROUT  AIL  AI    equation 10    RIN  RS  ROUT  RL Looking at this equation, it can be seen the loaded gain AIL must be less than or equal to the open circuit gain AI: Ideal current amplifier: Real current amplifier Ideally: In reality we require:RIN = 0 , ROUT =   RIN << RS => AIL =AI => independent of source and load  IIN  IS resistance ROUT >> RL IOUT  AIIS VOUT  AIISRL Example Problems Exercise 5A non-ideal current amplifier has an input resistance of RIN = 100 W, output resistance of ROUT =  W. The gain AI = 10 and the input voltage is 2 V. What is IOUT if the load resistance of RL = 10 kW.?Exercise 6 A current amplifier has the following characteristics:Gain AI = 5, RIN =0 W and ROUT = ¥ W.Calculate VOUT and IOUT if RL = 100 W, RS = 2 kW, IS = 10 mA.Exercise 7 Repeat previous example for the two cases:(a) RS = 2 k W and RIN = 2 k.(b) RS =2 k W and RIN = 100 .(c) RS =2 k W, RIN = 100  and ROUT =100 W.(d) RS =2 k W, RIN = 100  and ROUT =10 k W. Comment on the results obtained11 Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier ModellingAmplifiers in Cascade We can see in the previous examples that the amplifier equivalent circuit models are useful for analysis of source and load effects. These models are also useful to analysis the internal loading when two or more stages are cascaded to form a single amplifier. Consider the two- stage cascaded voltage amplifier shown below. The overall gain is affected by loading effects from each of the three loops. The loops are simple series circuits, so the voltages can easily be calculated with the voltage division rule.Source Two-Stage Cascaded Amplifier Load Stage 1 Stage 2 RS ROUT1 ROUT+ + +RL VS VIN1 RIN1 AV1 VIN1 VIN2 RIN2 AV2 VIN2 V - - - OUT2. Two-Stage Cascaded Voltage AmplifierConsider the two-stage cascaded current amplifier shown below. Again, the current can easily be calculated with the current division rule.Two-Stage Cascaded Current AmplifierStage 1 Stage 2 IIN1 IOUT2A I AI2IIN2 R RS R I1 IN1 ROUT1 R OUT2 IS IN1 IN2 RLTwo-Stage Cascaded Current AmplifierSample problems covered in class12

Analogue ELEK1289 - Electronic Systems and Practice II - Unit 3 Amplifier Modelling

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