Lecture 1 Fundamentals of Electric Circuits PDF

Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential and Operational Amplifier Lecture No. 3: Analysis of Transistor Amplifier Circuits UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering...

Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential and Operational Amplifier Lecture No. 3: Analysis of Transistor Amplifier Circuits UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals Before we proceed to differential amplifier circuits, we need to first learn the difference of Differential and Single-Ended Signals. Consider the arrangement of the voltage sources shown. Voltage readings can be done in two ways Voltage of a node with respect to the ground Voltage across two nodes UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals Signal reading from a single node to the ground is called the single-ended signal. Example: The voltage signal from node 1 (or 2) to the ground. We will call these signals as V1 and V2. Signal reading from a single node to another node is called the differential signal. Example: The voltage signal from node 1 to node 2. We will call this signal VD VCM is the common mode voltage. VCM will appear in both V1 and V2 readings, but not in VD. VCM can be any signal (e.g., common mode noise), but most of the time this is some DC voltage to bias our amplifiers! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals So, can we convert two single ended signals into a differential signal? Yes! We can do some voltage readings! Voltage at node 1: 𝑉1 = 𝑉𝑋 + 𝑉𝐶𝑀 Voltage at node 2: 𝑉2 = −𝑉𝑋 + 𝑉𝐶𝑀 The differential voltage, 𝑉𝐷 , is the voltage between node 1 and node 2 𝑽 𝑫 = 𝑽𝟏 − 𝑽𝟐 𝑉𝐷 = 𝑉𝑋 + 𝑉𝐶𝑀 − −𝑉𝑋 + 𝑉𝐶𝑀 𝑉 𝑉𝑋 = 𝐷 2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals We can say that 𝑉𝐶𝑀 is the average of voltages V1 and V2: 𝑉𝑋 +𝑉𝐶𝑀 −𝑉𝑋 +𝑉𝐶𝑀 𝑉𝐶𝑀 = 2 𝑽𝟏 +𝑽𝟐 𝑽𝑪𝑴 = 𝟐 If we know voltages 𝑉1 and 𝑉2 , we will also know 𝑉𝐷 and 𝑉𝐶𝑀 ! ^_^ UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals We can treat the 𝑉𝐷 and 𝑉𝐶𝑀 equations as a system of linear equations with 2 equation and 2 unknowns (V1 and V2) Solving this system of linear equations, 𝑽𝑫 𝑽𝟏 = + 𝑽𝑪𝑴 𝟐 𝑽𝑫 𝑽𝟐 = − + 𝑽𝑪𝑴 𝟐 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential vs. Single-Ended Signals UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Differential Pair The circuit shown is the typical configuration of a differential amplifier The emitter of each transistor is tied together and connected to a constant current source, 𝐼𝑇𝐴𝐼𝐿. These pair of transistors is called the differential pair. The current sharing is controlled by the difference of 𝑉𝑖1 and 𝑉𝑖2. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Differential Pair The circuit shown is the typical configuration of a differential amplifier The emitter of each transistor is tied together and connected to a constant current source, 𝐼𝑇𝐴𝐼𝐿. These pair of transistors is called the differential pair. The current sharing is controlled by the difference of 𝑉𝑖1 and 𝑉𝑖2. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Differential Pair Recalling Module 1,𝑣 𝑏𝑒 𝐼𝐶 ∝ 𝑒 𝑉𝑇 Therefore, 𝑣 𝑉𝑖1 −𝑉𝐸 𝑉𝑖1 𝑏𝑒1 𝑉𝐸 − 𝐼𝐶1 = 𝑘𝑒 𝑉𝑇 = 𝑘𝑒 𝑉𝑇 = 𝑘𝑒 𝑉𝑇 𝑒 𝑉𝑇 𝑣𝑏𝑒2 𝑉𝑖2 −𝑉𝐸 𝑉𝐸 𝑉𝑖2 − 𝐼𝐶2 = = 𝑘𝑒 𝑉𝑇 = 𝑘𝑒 𝑉𝑇 𝑘𝑒 𝑉𝑇 𝑒 𝑉𝑇 However, 𝐼𝐸1 ≅ 𝐼𝐶1 ; 𝐼𝐸2 ≅ 𝐼𝐶2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Differential Pair Using KCL: 𝐼𝑇𝐴𝐼𝐿 𝑉= 𝐼𝑉𝐸1 + 𝐼𝐸2 𝑉𝐸 𝑉𝑖2 − 𝐸 𝑖1 − 𝐼𝑇𝐴𝐼𝐿 = 𝑘𝑒 𝑉𝑇 𝑉𝑇 𝑒 + 𝑘𝑒 𝑒 𝑉𝑇 𝑉𝑇 𝑉𝐸 𝑉𝑖1 𝑉𝑖2 − 𝐼𝑇𝐴𝐼𝐿 = 𝑉 𝑘𝑒 𝑉𝑇 𝑉𝑇 𝑒 +𝑒 𝑉𝑇 − 𝐸 1 𝑘𝑒 𝑉𝑇 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖1 𝑉𝑖2 𝑒 𝑉𝑇 +𝑒 𝑉𝑇 Substitute this to 𝐼𝐶1 and 𝐼𝐶2 equations! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Differential Pair 𝑉𝑖1 𝑒 𝑉𝑇 𝐼𝐶1 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖1 𝑉𝑖2 𝑒 𝑉𝑇 + 𝑒 𝑉𝑇 𝑉𝑖2 𝑒 𝑉𝑇 𝐼𝐶2 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖1 𝑉𝑖2 𝑒 𝑉𝑇 + 𝑒 𝑉𝑇 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Differential Pair Simplifying these equations, 1 𝐼𝐶1 = 𝐼𝑇𝐴𝐼𝐿 ⋅ − 𝑉𝑖1 −𝑉𝑖2 1+𝑒 𝑉𝑇 1 𝐼𝐶2 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖1 −𝑉𝑖2 1+𝑒 𝑉𝑇 OR 1 𝐼𝐶1 = 𝐼𝑇𝐴𝐼𝐿 ⋅ −𝑉𝑖𝑑 1+𝑒 𝑉𝑇 1 𝐼𝐶2 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖𝑑 1+𝑒 𝑉𝑇 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering  Lecture No. 3: Analysis of Transistor Amplifier Circuits The Differential Pair 1 1 I 𝐼𝐶1 = 𝐼𝑇𝐴𝐼𝐿 ⋅ − 𝑉𝑖1 −𝑉𝑖2 0.9 C1 I C2 1+𝑒 𝑉𝑇 0.8 1 For 𝑉𝑖2 ≫ 𝑉𝑖1 For 𝑉𝑖1 ≫ 𝑉𝑖2 𝐼𝐶2 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖1 −𝑉𝑖2 0.7 𝐼𝐶1 ≅ 0 𝐼𝐶1 ≅ 𝐼𝑇𝐴𝐼𝐿 𝐼𝐶2 ≅ 𝐼𝑇𝐴𝐼𝐿 𝐼𝐶2 ≅ 0 1+𝑒 𝑉𝑇 0.6 I (mA) OR 0.5 1 0.4 𝐼𝐶1 = 𝐼𝑇𝐴𝐼𝐿 ⋅ −𝑉𝑖𝑑 0.3 1+𝑒 𝑉𝑇 1 0.2 𝐼𝐶2 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖𝑑 0.1 1+𝑒 𝑉𝑇 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 V =V (V) id i1 i2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load The circuit shown is the differential amplifier with resistive load (differential output, single supply). In which region do the transistors operate? This is an amplifier! The BJT should operate in the active region! ^_^ If the BJT is operating in the active region, we should have proper biasing voltages! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load Since we are using single supply in this case, we should have a biasing voltage in the base terminal of each BJT! During quiescent operation, currents 𝐼𝐶1 and 𝐼𝐶2 are equal! 𝐼𝑇𝐴𝐼𝐿 𝐼𝐶1 = 𝐼𝐶2 ≅ 2 For this condition to happen in a symmetrical differential amplifier, the base voltages of each BJT should be equal! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load 𝑉𝐵𝐵1 = 𝑉𝐵𝐵2 → 𝑉𝑖1 − 𝑉𝑖2 ቚ =0 𝐷𝐶 The biasing voltages do not appear in the differential input voltage! This indicates that the biasing voltage is a common mode voltage! We will call this biasing voltage as the input common mode voltage, 𝑉𝑖𝐶𝑀. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load What should be the value of 𝑉𝑖𝐶𝑀 ? We need to set 𝑉𝑖𝐶𝑀 such that both transistors will operate in the active region. Conditions: 𝑉𝐵𝐸1 = 𝑉𝐵𝐸2 = ~0.7𝑉 𝑉𝐶𝐸1 , 𝑉𝐶𝐸2 > 𝑉𝐶𝐸 𝑠𝑎𝑡 We need to use our favorite circuit theorem! KVL! ^_^ We can consider only one circuit since the circuit due to Q1 is identical to the circuit due to Q2. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load KVL at the output loop: 𝑉𝐶𝐶 = 𝐼𝐶 𝑅𝐿 + 𝑉𝐶𝐸 + 𝑉𝐸 + 𝐼𝐶1 𝑅𝐿 KVL at the input loop: − 𝑉𝑖𝐶𝑀 = 𝑉𝐵𝐸 + 𝑉𝐸 𝑉𝑖𝐶𝑀 + From the input loop equation, we 𝑉𝐶𝐸1 observed that 𝑉𝐸 is input bias dependent! + − 𝑉𝐸 = 𝑉𝑖𝐶𝑀 − 𝑉𝐵𝐸 𝑉𝐵𝐸1 𝑉𝐸 − Substituting 𝑉𝐸 to the output loop equation: 𝑉𝐶𝐶 = 𝐼𝐶 𝑅𝐿 + 𝑉𝐶𝐸 + 𝑉𝑖𝐶𝑀 − 𝑉𝐵𝐸 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load Recalling the conditions to operate in the active region: + 𝑉𝐵𝐸1 = 𝑉𝐵𝐸2 = ~0.7𝑉 𝐼𝐶1 𝑅𝐿 − 𝑉𝐶𝐸1 , 𝑉𝐶𝐸2 > 𝑉𝐶𝐸 𝑠𝑎𝑡 Therefore, 𝑉𝑖𝐶𝑀 + 𝑉𝐶𝐶 = 𝐼𝐶 𝑅𝐿 + 𝑉𝐶𝐸(𝑠𝑎𝑡) + 𝑉𝑖𝐶𝑀,𝑚𝑎𝑥 − 𝑉𝐵𝐸 𝑉𝐶𝐸1 − 𝑉𝑖𝐶𝑀,𝑚𝑎𝑥 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐿 − 𝑉𝐶𝐸(𝑠𝑎𝑡) + 𝑉𝐵𝐸 + 𝑉𝐵𝐸1 𝑉𝐸 In terms of 𝐼𝑇𝐴𝐼𝐿 − 𝐼𝑇𝐴𝐼𝐿 𝑅𝐿 𝑉𝑖𝐶𝑀,𝑚𝑎𝑥 = 𝑉𝐶𝐶 − − 𝑉𝐶𝐸(𝑠𝑎𝑡) + 𝑉𝐵𝐸 2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load To determine the minimum 𝑉𝑖𝐶𝑀 , we need to consider the minimum voltage required for the constant current source to operate properly. 𝑉𝑖𝐶𝑀,𝑚𝑖𝑛 = 𝑉𝐵𝐸 + 𝑉𝐸,𝑚𝑖𝑛 𝑉𝑖𝐶𝑀 The input common mode voltage range will be + 𝑉𝑖𝐶𝑀,𝑚𝑖𝑛 < 𝑉𝑖𝐶𝑀 < 𝑉𝑖𝐶𝑀,𝑚𝑎𝑥 𝑉𝐵𝐸1 𝑉𝐸 − 𝑉𝐵𝐸 + 𝑉𝐸,𝑚𝑖𝑛 < 𝑉𝑖𝐶𝑀 < 𝐼𝑇𝐴𝐼𝐿 𝑅𝐿 𝑉𝐶𝐶 − − 𝑉𝐶𝐸(𝑠𝑎𝑡) + 𝑉𝐵𝐸 2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem Determine the input common mode voltage range of the differential amplifier circuit shown. Assume 𝑉𝐶𝐶 = 10𝑉, 𝑉𝐶𝐸(𝑠𝑎𝑡) = 0.2𝑉 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load The input common mode voltage can be removed by using dual supply. What will provide the proper biasing? Removing 𝑉𝑖𝐶𝑀 indicates that there will be no DC voltage connected to the base! During quiescent operation, the base terminals are grounded! Biasing will now be provided by −𝑉𝐸𝐸 Input Loop Equation: 𝑉𝐸𝐸 = 𝑉𝐵𝐸 + 𝑉𝑇𝐴𝐼𝐿 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem Convert the dual supply differential amplifier shown into a single supply one. Assume 𝑉𝐵𝐸 = 0.79𝑉 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering  Lecture No. 3: Analysis of Transistor Amplifier Circuits The Differential Pair 1 1 I 𝐼𝐶1 = 𝐼𝑇𝐴𝐼𝐿 ⋅ − 𝑉𝑖1 −𝑉𝑖2 0.9 C1 I C2 1+𝑒 𝑉𝑇 0.8 1 For 𝑉𝑖2 ≫ 𝑉𝑖1 For 𝑉𝑖1 ≫ 𝑉𝑖2 𝐼𝐶2 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖1 −𝑉𝑖2 0.7 𝐼𝐶1 ≅ 0 𝐼𝐶1 ≅ 𝐼𝑇𝐴𝐼𝐿 𝐼𝐶2 ≅ 𝐼𝑇𝐴𝐼𝐿 𝐼𝐶2 ≅ 0 1+𝑒 𝑉𝑇 0.6 I (mA) OR 0.5 1 0.4 𝐼𝐶1 = 𝐼𝑇𝐴𝐼𝐿 ⋅ −𝑉𝑖𝑑 0.3 1+𝑒 𝑉𝑇 1 0.2 𝐼𝐶2 = 𝐼𝑇𝐴𝐼𝐿 ⋅ 𝑉𝑖𝑑 0.1 1+𝑒 𝑉𝑇 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 V =V (V) id i1 i2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load (AC) AC analysis will be easier if we will divide the circuit into two identical common emitter amplifier. ^_^ We will only analyze the half circuit! Analysis using half circuit = Analysis similar to M1! However, we need to convert first the differential signals into single-ended signals Find an expression of single-ended signal in terms of the differential and common mode signals UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-Ended Signals Transformation UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Output to Single-Ended Output Conversion /2 /2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Input to Single-Ended Input Conversion /2 /2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load (AC) The common mode voltages in the transformed signals are the quiescent voltage levels (DC voltages) These common mode voltages will disappear during AC analysis! ^_^ 𝑉𝑖𝑑 𝑉𝑖1 = + 𝑉𝑖𝐶𝑀 2 𝑉𝑖𝑑 Biasing Small signal 𝑉𝑖2 = − + 𝑉𝑖𝐶𝑀 and 2 quiescent AC 𝑉𝑜𝑑 voltages 𝑉𝑜1 = + 𝑉𝑜𝐶𝑀 voltages 2 𝑉𝑜𝑑 𝑉𝑜2 = − + 𝑉𝑜𝐶𝑀 2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits AC Equivalent Circuit of a Differential Amplifier UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits AC Equivalent Circuit of a Differential Amplifier 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits AC Equivalent Circuit of a Differential Amplifier We know that 𝐼𝑇𝐴𝐼𝐿 = 𝐼𝐸1 + 𝐼𝐸2 Therefore, ↑ 𝐼𝐸1 will lead to ↓ 𝐼𝐸2 ↓ 𝐼𝐸1 will lead to ↑ 𝐼𝐸2 We can say that 𝑖𝑒1 = −𝑖𝑒2 Therefore, 𝑖 𝑇𝐴𝐼𝐿 = 0 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits AC Equivalent Circuit of a Differential Amplifier The small signal current 𝑖 𝑇𝐴𝐼𝐿 will only flow through 𝑅𝑂𝑈𝑇. If 𝑖 𝑇𝐴𝐼𝐿 = 0 Then 𝑉𝑅𝑂𝑈𝑇 = 0𝑉 This will make the emitter node an AC ground. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits AC Equivalent Circuit of a Differential Amplifier We can now redraw the circuit! As we can observe, the differential amplifier AC circuit is composed of two half circuits. Each half circuit is a Common Emitter AC circuit (Fixed Bias) UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits AC Equivalent Circuit of a Differential Amplifier HC#1: 𝑉𝑜1 𝑟𝑜 ||𝑅𝐿 = − HC#1 HC#2 𝑉𝑖1 𝑟𝑒 HC#2: 𝑉𝑜2 𝑟𝑜 ||𝑅𝐿 = − 𝑉𝑖2 𝑟𝑒 We know that: 𝑉𝑜𝑑 𝑉𝑜1 = ; 2 𝛽𝑖𝐵 𝛽𝑖𝐵 𝑉𝑜𝑑 𝛽𝑟𝑒 𝛽𝑟𝑒 𝑉𝑜2 = − ; 2 𝑉𝑖𝑑 𝑉𝑖1 = ; 2 𝑉𝑖𝑑 𝑉𝑖2 = − ; 2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits AC Equivalent Circuit of a Differential Amplifier HC#1: 𝑉𝑜1 𝑟𝑜 ||𝑅𝐿 = − HC#1 HC#2 𝑉𝑖1 𝑟𝑒 HC#2: 𝑉𝑜2 𝑟𝑜 ||𝑅𝐿 = − 𝑉𝑖2 𝑟𝑒 Therefore 𝑉𝑜𝑑 𝑟𝑜 ||𝑅𝐿 𝐴𝑉𝑑 = = − 𝛽𝑖𝐵 𝛽𝑖𝐵 𝑉𝑖𝑑 𝑟𝑒 𝛽𝑟𝑒 𝛽𝑟𝑒 Since the output and input are both differential signals, we will call this ratio as the differential voltage gain, 𝑨𝑽𝒅. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits AC Equivalent Circuit of a Differential Amplifier HC#1: 𝑉𝑜1 𝑟𝑜 ||𝑅𝐿 = − HC#1 HC#2 𝑉𝑖1 𝑟𝑒 HC#2: 𝑉𝑜2 𝑟𝑜 ||𝑅𝐿 = − 𝑉𝑖2 𝑟𝑒 Therefore 𝑉𝑜𝑑 𝑟𝑜 ||𝑅𝐿 𝐴𝑉𝑑 = = − 𝛽𝑖𝐵 𝛽𝑖𝐵 𝑉𝑖𝑑 𝑟𝑒 𝛽𝑟𝑒 𝛽𝑟𝑒 Since the output and input are both differential signals, we will call this ratio as the differential voltage gain, 𝑨𝑽𝒅. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Noise As mentioned before, the common mode voltage can be either AC or DC. The DC common mode voltage 𝑉𝑜 = is for biasing! 2𝑚𝑉𝑝𝑘−𝑝𝑘 Of course, we need that! ^_^ The AC common mode voltage is typically a type of noise. 𝑉𝑖𝐶𝑀 = 1𝑚𝑉𝑝𝑘−𝑝𝑘 We don’t want noise! >.< Some differential amplifiers can not suppress this common mode noise. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Noise The common-mode rejection ratio (CMRR) is a figure of merit the effectiveness of a differential amplifier to amplify 𝑉𝑜 differential signals and rejecting = common-mode signals. 2𝑚𝑉𝑝𝑘−𝑝𝑘 Mathematically, it is defined as 𝐴𝑉𝑑 𝐶𝑀𝑅𝑅 = 𝐴𝑉𝑐 𝑉𝑖𝐶𝑀 = 1𝑚𝑉𝑝𝑘−𝑝𝑘 In dB, 𝐴𝑉𝑑 𝐶𝑀𝑅𝑅𝑑𝐵 = 20 log 𝐴𝑉𝑐 Higher value is better! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain For the differential amplifier circuit shown, the common mode voltage gain is defined as 𝑉𝑂𝑑 𝐴𝑉𝑐 = 𝑉𝑖𝑐𝑚 Where 𝑉𝑜𝑑 = 𝑉𝑜1 − 𝑉𝑜2 Since 𝑉𝑖1 and 𝑉𝑖2 are in phase, ↑ 𝐼𝐸1 will lead to↑ 𝐼𝐸2 ↓ 𝐼𝐸1 will lead to↓ 𝐼𝐸2 Therefore, 𝑖𝑒1 + 𝑖𝑒2 = 𝑖 𝑇𝐴𝐼𝐿 ≠ 0 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain We can not assume an AC ground in the emitter! We need to consider the effect of 𝑅𝑂𝑈𝑇 ! 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 How can we divide the circuit into two half circuits? We can remove the DC current source. We can represent 𝑅𝑂𝑈𝑇 as two resistors in parallel. Each will have a value of 2𝑅𝑂𝑈𝑇. Each resistor will now have a current of 𝑖 𝑇𝐴𝐼𝐿 /2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain HC#1 HC#2 Current in this wire is zero! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain HC#1 HC#2 We can observe that the differential amplifier’s AC circuit is composed of two half circuits. Each half circuit is a 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 Common Emitter AC circuit (Emitter Stabilized with 𝑅𝐸 = 2 ⋅ 𝑅𝑂𝑈𝑇 ) UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain HC#1: HC#1 HC#2 𝑉𝑜1 𝛽𝑅𝐿 = − 𝑉𝑖𝑐𝑚 𝛽𝑟𝑒 +2 𝛽+1 𝑅𝑂𝑈𝑇 HC#2: 𝑉𝑜2 𝛽𝑅𝐿 = − 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 𝑉𝑖𝑐𝑚 𝛽𝑟𝑒 +2 𝛽+1 𝑅𝑂𝑈𝑇 If 𝛽 is high enough, 𝑉𝑜1 𝑅𝐿 =− 𝑉𝑖𝑐𝑚 𝑟𝑒 +2𝑅𝑂𝑈𝑇 𝑉𝑜2 𝑅𝐿 = − 𝑉𝑖𝑐𝑚 𝑟𝑒 +2𝑅𝑂𝑈𝑇 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain HC#1 HC#2 The common mode gain is 𝑉𝑜𝑑 𝑉𝑜1 −𝑉𝑜2 𝐴𝑉𝑠𝑐 = = 𝑉𝑖𝑐𝑚 𝑉𝑖𝑐𝑚 We can say that 𝐴𝑉𝑐 = 𝐴𝑉𝐻𝐶1 − 𝐴𝑉𝐻𝐶2 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 Since HC#1 is identical with HC#2, 𝐴𝑉𝑐 = 0 Therefore, CMRR will be very high (approaching infinity) UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain HC#1 HC#2 “If the components in the first half circuit are perfectly identical with the second half 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 circuit’s components, the common-mode voltage gain will be zero” UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain HC#1 HC#2 Practically, no components will be perfectly identical! Try finding a resistor with a resistance of exactly 1k. Good luck! Haha! Therefore, we can not 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 achieve 𝐴𝑉𝑐 = 0 IRL. If the BJTs are perfectly identical, but 𝑅𝐿 are not, Δ𝑅𝐿 𝐴𝑉𝑐 = 𝑟𝑒+2𝑅𝑂𝑈𝑇 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Common-Mode Voltage Gain If BJTs and 𝑅𝐿 are both not HC#1 HC#2 identical to the other half circuit, 𝐴𝑉𝑐 = 𝐴𝑉𝐻𝐶1 − 𝐴𝑉𝐻𝐶2 Therefore, 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 𝑅𝐿1 𝑅𝐿2 𝐴𝑉𝑐 = − + 𝑟𝑒1 +2𝑅𝑂𝑈𝑇 𝑟𝑒2 +2𝑅𝑂𝑈𝑇 CMRR is dependent on device mismatch. ☺ UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load Summary: Inputs and Outputs Differential Input Differential Output ∴ Fully Differential Differential Gain 𝑟𝑜 ||𝑅𝐿 𝑅𝐿 𝐴𝑉𝑑 = − =− 𝑟𝑒 𝑟𝑒 Common-mode Gain 𝐴𝑉𝑐 ≅ 0 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-ended Output Conversion The circuit we analyzed before is a fully differential differential amplifier. It is because the input and output are both differential! However, most circuits after the differential amplifier are single- ended input! We need to convert the differential output into single- ended output! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-ended Output Conversion Instead using two outputs to have a differential output, we can use one output to have a single-ended signal! Instead of using 2 half circuits, we will only consider 1 half circuit! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-ended Output Conversion The output is in the HC#2, therefore, use HC#2! HC#1 HC#2 HC#2: 𝑉𝑜2 𝑟𝑜 ||𝑅𝐿 =− 𝑉𝑖2 𝑟𝑒 We know that 𝑉𝑜2 = 𝑉𝑜 𝑉𝑖𝑑 𝑉𝑖2 = − 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 2 Therefore, 𝑉𝑜 𝑟𝑜 ||𝑅𝐿 𝐴𝑉 𝑠 = = 𝑉𝑖𝑑 2𝑟𝑒 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-ended Output Conversion When considering the HC#1 HC#2 common mode gain, again, we will just consider one of the half circuits! Considering HC#2, 𝛽𝑟𝑒 𝛽𝑖𝐵 𝛽𝑖𝐵 𝛽𝑟𝑒 𝛽𝑅𝐿 𝐴𝑉𝑐 = − 𝛽𝑟𝑒 +2(𝛽+1)𝑅𝑂𝑈𝑇 If 𝛽 is high enough, 𝑅𝐿 𝐴𝑉𝑐 = − 𝑟𝑒 +2𝑅𝑂𝑈𝑇 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Resistive Load (S.E.) Summary: Inputs and Outputs Differential Input Single-ended Output Differential Gain 𝑟𝑜 ||𝑅𝐿 𝑅𝐿 𝐴𝑉𝑠 = = 2𝑟𝑒 2𝑟𝑒 Common-mode Gain 𝑅𝐿2 𝐴𝑉𝑐 = − 𝑟𝑒2 +2𝑅𝑂𝑈𝑇 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-ended Output Conversion Observations: The voltage gain was reduced! 𝑅𝐿 𝑅𝐿 From 𝐴𝑉𝑑 = to 𝐴𝑉𝑠 = 𝑟𝑒 2𝑟𝑒 The common-mode gain was increased! 𝑅𝐿 From 𝐴𝑉𝑐 = 0 to 𝐴𝑉𝑐 = 𝑟𝑒 +2𝑅𝑂𝑈𝑇 Therefore, max value of CMRR was reduced! 𝑟𝑒 +2𝑅𝑂𝑈𝑇 From 𝐶𝑀𝑅𝑅 = ∞ to 𝐶𝑀𝑅𝑅 = 2𝑟𝑒 Main drawback of single ended output  UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-ended Output Conversion The differential amplifier’s performance was degraded! So, is there a way to convert a differential output into a single- ended one without degrading the circuit’s performance? Yes! By using active load! Obviously, we will not use resistors anymore. ^_^ UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering  Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-ended Output Conversion The two resistors were replaced by a current mirror circuit! The gain is effectively increased! This is due to higher output impedance! It was increased from 𝑟𝑜𝑁 ||𝑅𝐿 to 𝑟𝑜𝑃 ! The voltage gain* will be 𝑉𝑜 𝑟𝑜𝑃 𝐴𝑉𝑠 = = 𝑉𝑖𝑑 𝑟𝑒𝑁 *See the derivation in the class notebook, we can not use half circuit analysis due to the non-symmetrical circuit  UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering  Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential to Single-ended Output Conversion The equivalent circuit for computing the common mode gain is shown. 𝛽𝑖𝑏4 𝛽𝑖𝑏3 𝑖𝑏3 𝛽𝑟𝑒3 𝑖𝑏4 𝛽𝑟𝑒4 The resulting common mode voltage gain* will be 𝛽 𝑟𝑒3 𝛽 𝑉𝑋 𝐴𝑉𝑐 = ⋅ − ⋅ 𝑟𝑜𝑃 𝛽𝑟𝑒1 + 2 𝛽 + 1 𝑅𝑂𝑈𝑇 𝑟𝑒4 𝛽𝑟𝑒2 + 2 𝛽 + 1 𝑅𝑂𝑈𝑇 𝛽𝑖𝑏1 𝛽𝑖𝑏2 𝛽𝑟𝑒1 𝛽𝑟𝑒2 If 𝑟𝑒1 = 𝑟𝑒3 = 𝑟𝑒3 = 𝑟𝑒4 , 𝐴𝑉𝑐 will be zero. However, based on module 3, 𝐼𝐶1 ≠ 𝐼𝐶2 due to the BJT current mirror! Therefore, 𝑟𝑒1 ≠ 𝑟𝑒2 and 𝐴𝑉𝑐 ≠ 0!  *See the derivation in the class notebook, we can not use half circuit analysis due to the non-symmetrical circuit  UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Differential Amplifier with Active Load (S.E.) Summary: Inputs and Outputs Differential Input Single-ended Output Differential Gain 𝑟𝑜𝑃 𝐴𝑉𝑠 = 𝑟𝑒 Common-mode Gain 𝛽 𝑟𝑒3 𝛽 𝐴𝑉𝑐 = ⋅ − ⋅ 𝑟𝑜𝑃 𝛽𝑟𝑒1 +2 𝛽+1 𝑅𝑂𝑈𝑇 𝑟𝑒4 𝛽𝑟𝑒2 +2 𝛽+1 𝑅𝑂𝑈𝑇 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The 5-Transistor Operational Transconductance Amplifier 5T-OTA = Differential Amplifier w/ Active Load The current source was replaced by a BJT operating in the active region! Differential Gain 𝑟𝑜𝑃 𝐴𝑉𝑠 = 𝑟𝑒 Common-mode Gain 𝛽 𝑟𝑒3 𝛽 𝐴𝑉𝑐 = ⋅ − ⋅ 𝑟𝑜𝑃 𝛽𝑟𝑒1 +2 𝛽+1 𝑅𝑂𝑈𝑇 𝑟𝑒4 𝛽𝑟𝑒2 +2 𝛽+1 𝑅𝑂𝑈𝑇 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Operational Amplifier An Operational Amplifier behaves quite similar to a differential amplifier. It amplifies the voltage difference between its two inputs! The main difference of an OP-AMP to the differential amplifier is its very high open loop gain! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits How to make an OP-AMP? An OP-AMP is typically comprised of three stages Differential Amplifier Stage Additional amplifier stages for more gain Output stage (Voltage Follower) 𝑉+ Differential Second Stage 𝑉𝑂𝑈𝑇 Output Stage Amplifier Amplifier 𝑉− UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits How to make an OP-AMP? UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits How to make an OP-AMP? LM2902 OP-AMP LM741 OP-AMP UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP as a Black Box An operational amplifier can be abstracted as “black box” having two inputs and one output. The op amp symbol distinguishes between the two inputs by the plus and minus sign. Vin1 and Vin2 are called the “noninverting” and “inverting” inputs, respectively. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP as a Black Box We view the op amp as a circuit that amplifies the difference between the two inputs, arriving at the equivalent circuit depicted in the figure below The voltage gain is denoted by A0: 𝑉𝑜𝑢𝑡 = 𝐴0 (𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 ) *𝐴0 → Open Loop Gain UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits The Ideal OP-AMP Ideal LM2902 (TI) LM741 (TI) Voltage Gain, ∞ 100,000 200,000 A0 = 𝑉𝑜𝑢𝑡 /(𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 ) Input Impedance, ∞ Several MΩ 2 MΩ 𝑍𝑖𝑛 Output Impedance, < 100 Ω 0 < 100 Ω 𝑍𝑜𝑢𝑡 Speed/Bandwidth, ∞ 1.2MHz 1.5MHz 𝐵 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP Circuit Analysis The very high gain of the op amp leads to an important observation. Since realistic circuits produce finite output swings, e.g., 2 V, the difference between Vin1 and Vin2 will be 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 = 𝐴0 As we can observe in this equation, if 𝑉𝑜𝑢𝑡 is a finite value, the difference between 𝑉𝑖𝑛1 and 𝑉𝑖𝑛2 will be quite small. For an ideal OP-AMP: 𝐼𝑓 𝐴0 → ∞, 𝑡ℎ𝑒𝑛 𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 → 0 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Ideal OP-AMP with Supply Voltages Practically, as we observed in the its internal circuit, an OP-AMP requires some voltage sources to power it up. This can be represented similar to the figure shown. The output voltage swing of an ideal OP-AMP will be 𝑉𝐸𝐸 ≤ 𝑉𝑜𝑢𝑡 ≤ 𝑉𝐶𝐶 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: The Comparator A comparator compares the voltage levels between the two inputs. An OP-AMP–based comparator is essentially an OP-AMP operating in open- loop. Therefore, 𝑉𝑜𝑢𝑡 = 𝐴0 𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: The Comparator 𝑉𝑜𝑢𝑡 = 𝐴0 𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 If 𝑉𝑖𝑛1 > 𝑉𝑖𝑛2 , and 𝐴0 = ∞ 𝑉𝑜𝑢𝑡 = +∞ If 𝑉𝑖𝑛1 < 𝑉𝑖𝑛2 , and 𝐴0 = ∞ 𝑉𝑜𝑢𝑡 = −∞ However, we learned that for an ideal OP-AMP with supply voltages, 𝑉𝐸𝐸 ≤ 𝑉𝑜𝑢𝑡 ≤ 𝑉𝐶𝐶 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: The Comparator Therefore, the output voltage of the OP-AMP–based comparator is ~𝑉𝐶𝐶 ; for Vin1 > Vin2 𝑉𝑜𝑢𝑡 =ቊ ~𝑉𝐸𝐸 ; for Vin1 < Vin2 In the example circuit shown, +5𝑉; for Vin1 > Vin2 𝑉𝑜𝑢𝑡 = ቊ −5𝑉; for Vin1 < Vin2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: Negative Feedback As we observed in the open-loop operation of an ideal OP-AMP, the output voltage will be either +∞ or – ∞ (if there’s no supply) We need to somehow reduce the voltage gain to make the the output finite. One way to reduce the circuit’s gain is to use negative feedback. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: Negative Feedback The figure below is the typical block diagram of a negative amplifier The circuit voltage gain will be 𝐴0 𝐴𝑉 = 1+𝐴0 𝛽 𝐴0 → Open-loop Voltage Gain; 𝛽 → Feedback Factor + 𝑉𝐼𝑁 𝐴0 𝑉𝑂𝑈𝑇 − 𝛽 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: Negative Feedback If 𝐴0 is high enough, 1 lim 𝐴𝑉 = 𝐴0 →∞ 𝛽 The voltage gain will now be dependent on the feedback factor, 𝛽. 𝛽 is typically dependent on the external components like resistors. ^_^ + 𝑉𝐼𝑁 𝐴0 𝑉𝑂𝑈𝑇 − 𝛽 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: Unity-Gain Amplifier The circuit shown is the unity gain amplifier Ideally, it has a voltage gain of 1. Voltage Gain: 𝐴0 𝐴𝑉 = 1 + 𝐴0 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: Unity-Gain Amplifier The circuit shown is the unity gain amplifier Ideally, it has a voltage gain of 1. Voltage Gain: 𝐴0 𝐴𝑉 = 1 + 𝐴0 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: Non-Inverting Amplifier The circuit shown is the Non-Inverting Amplifier The output voltage is in phase with the input voltage. Voltage Gain: 𝐴0 𝑅1 + 𝑅2 𝐴𝑉 = 𝑅1 + 𝐴0 + 1 𝑅2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: Inverting Amplifier The circuit shown is the Inverting Amplifier Ideally, the output voltage is 180° out of phase with the input voltage. Voltage Gain: 𝐴0 𝑅1 𝐴𝑉 = − 𝑅1 + 𝐴0 + 1 𝑅2 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits OP-AMP–Based Circuits: Alternative Analysis If the OP-AMP has high enough voltage gain, high enough input impedance, and low enough output impedance we can treat the OP-AMP as ideal to simplify the analysis. For an ideal OP-AMP and finite 𝑉𝑜𝑢𝑡 , we can assume the following: 𝑉𝑖𝑛1 = 𝑉𝑖𝑛2 𝑖𝑖𝑛1 = 𝑖𝑖𝑛2 = 0 𝑖𝑖𝑛2 𝑉𝑖𝑛2 𝑖𝑖𝑛1 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛1 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem Derive the voltage gain of the circuits shown below. Assume all OP-AMPs are ideal. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem Show that the circuit below is an inverting summing amplifier. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem Show that the circuit below is a difference amplifier. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem Determine the output to input voltage relationship of the following circuits shown below UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Analog Filter Circuits Lecture No. 3: Analysis of Transistor Amplifier Circuits UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits General Considerations To define the performance parameters of filters, we first take a brief look at some applications. Suppose a cellphone receives a desired signal, 𝑋(𝑓), with a bandwidth of 200 kHz at a center frequency of 900 MHz. Now, let us assume that, in addition to 𝑋(𝑓), the cellphone receives a large interferer centered at 900 MHz + 200 kHz. Since the information is in the signal 𝑋(𝑓), we need to “reject” the interferer by means of a filter. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Filter Characteristics The frequency response of every filter is divided into three bands: Passband; Transition Band; and Stopband Some characteristics of a filter should be considered: ▪ The filter must not affect the desired signal. It must provide a “flat” frequency response across the bandwidth of 𝑋(𝑓). ▪ The filter must attenuate the interferer signal. It must exhibit a “sharp” transition band. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem In a wireless application, the interferer in the adjacent channel may be 25 dB higher than the desired signal. Determine the required stopband attenuation of the filter in the figure below if the signal power must exceed the interferer power by 15 dB for proper detection. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Classification of Filters The figure below summarizes the types of filters. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Classification of Filters: Summary UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Filter Transfer Function Every filter is characterized mathematically by its transfer function. 𝑠 + 𝑧1 𝑠 + 𝑧2 … (𝑠 + 𝑧𝑚 ) Π𝑖 (𝑠 + 𝑧𝑖 ) 𝐻 𝑠 = 𝛼 =𝛼 𝑠 + 𝑝1 𝑠 + 𝑝2 … (𝑠 + 𝑝𝑛 ) Π𝑗 (𝑠 + 𝑝𝑗 ) where 𝑧𝑘 and 𝑝𝑘 denote the zero and pole frequencies, respectively. It is common to express 𝑧𝑘 and 𝑝𝑘 as 𝜎 + 𝑗𝜔, where 𝜎 represents the real part and 𝜔 the imaginary part. For this course we will only deal with systems that has zeros and poles that are not complex numbers (only real numbers). UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Passive Low Pass Filter The circuit shown is a first order LPF. Its transfer function, 𝐻(𝑠), is derived as 𝑉𝑜 𝑠 1 1 =𝐻 𝑠 = ∙ 𝑉𝑖 (𝑠) 𝑅𝐶 𝑠 + 1 𝑅𝐶 The frequency response can be plotted by substituting 𝑠 = 𝑗𝜔 1 1 𝐻 𝑗𝜔 = ∙ 𝑅𝐶 𝑗𝜔 + 1 𝑅𝐶 This will result to a complex plot that is composed of a Magnitude Plot and a Phase plot. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Passive Low Pass Filter The frequency response shown in the previous slides is the magnitude response. To get the equation of the magnitude response, we can take the magnitude of the transfer function, 𝐻(𝑗𝜔). 1 1 𝐻(𝑗𝜔) = ∙ 𝑅𝐶 𝑗𝜔 + 1 𝟏 𝟏 𝑅𝐶 𝑯(𝒋𝝎) = ∙ 𝑹𝑪 𝟏 𝟐 𝝎𝟐 + 𝑹𝑪 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Passive Low Pass Filter 𝟏 𝟏 𝑯(𝒋𝝎) = ∙ 𝑹𝑪 𝟏 𝟐 The cut − off frequency is the transfer function’s pole! 𝝎𝟐 + 𝑹𝑪 𝟏 𝝎𝒄 = 𝒑 = 𝑹𝑪 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Passive High Pass Filter The circuit shown is a first order HPF. Its transfer function, 𝐻(𝑠), is derived as 𝑉𝑜 𝑠 𝑠 =𝐻 𝑠 = 𝑉𝑖 (𝑠) 1 𝑠+ 𝑅𝐶 The frequency response can be plotted by substituting 𝑠 = 𝑗𝜔 𝑗𝜔 𝐻 𝑗𝜔 = 1 𝑗𝜔 + 𝑅𝐶 This will result to a complex plot that is composed of a Magnitude Plot and a Phase plot. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Passive High Pass Filter The frequency response shown in the previous slides is the magnitude response. To get the equation of the magnitude response, we can take the magnitude of the transfer function, 𝐻(𝑗𝜔). 𝑗𝜔 𝐻 𝑗𝜔 = 1 𝑗𝜔 + 𝑅𝐶 𝝎 𝑯(𝒋𝝎) = 𝟐 𝟏 𝝎𝟐 + 𝑹𝑪 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Passive High Pass Filter 𝝎 𝑯(𝒋𝝎) = 𝟐 𝟏 The cut − off frequency is the transfer function’s pole! 𝝎𝟐 + 𝑹𝑪 𝟏 𝝎𝒄 = 𝒑 = 𝑹𝑪 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Active Filters The simplest way to construct an active filter circuit is to cascade a first stage passive filter into a second stage active circuit such as a non-inverting amplifier. The purpose of the amplifier is to increase the system gain! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Active Filters 𝑹𝟏 𝟏 𝟏 𝑹𝟏 𝒔 𝑯 𝒔 = 𝟏+ 𝑯 𝒔 = 𝟏+ 𝑹𝟐 𝑹𝑪 𝟏 𝑹𝟐 𝟏 𝒔+ 𝒔+ 𝑹𝑪 𝑹𝑪 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits First Order Filters: Active Filters Another way to construct an active filter is to use frequency dependent components in the feedback network of an amplifier. 𝑹𝟏 𝟏 𝑹𝟏 𝟏 𝒔+ 𝟏+ 𝟏+ 𝒔+ 𝑹𝟐 𝑹𝟏 𝑪 𝑹𝟐 𝑹𝟐 𝑪 𝑯 𝒔 = 𝑯 𝒔 = 𝟏 𝟏 𝒔+ 𝒔+ 𝑹𝟏 𝑪 𝑹𝟐 𝑪 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem Determine the type of filter, transfer function, and cut-off frequency of the filter circuits shown in the figure. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Second Order Filters The general transfer function of a second order filter is defined as 𝛼𝑠 2 + 𝛽𝑠 + 𝛾 𝐻 𝑠 = 𝜔𝑛 2 𝑠 + 𝑠 + 𝜔𝑛2 𝑄 𝜔𝑛 → Natural Frequency ▪ Take note that natural frequency is not always the same as the -3dB frequency ^_^ ▪ For second order filter circuits, we will set the natural frequency as the cutoff frequency 𝑄→ Quality Factor We don’t care on the values of 𝛼, 𝛽, 𝑎𝑛𝑑 𝛾 in finding the cut-off frequency UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Second Order Filters: Passive Filters Second order filter circuits are typically composed of 2 frequency dependent components. For passive filters, it is typically composed of R, L, and C. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Second Order Filters: Active Filters Typical active second order filters are realized using the Sallen-Key Filter. ▪ Two capacitors are used for the two frequency dependent components. ▪ The circuit shown below is an Active Low Pass Filter. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Second Order Filters: Active Filters The circuit shown below is an Active High Pass Filter. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Band Pass Filters The simplest way to implement a Band Pass Filter is to cascade a Low Pass Filter stage and a High Pass Filter Stage. However, we need to ensure that 𝑓𝑐𝐻𝑃𝐹 < 𝑓𝑐𝐿𝑃𝐹. This kind of BPF is only suitable for wideband applications. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Narrowband BPF Bandpass filters for narrowband applications has a transfer function of 𝐾𝑠 𝐻 𝑠 = 𝜔𝑛 2 𝑠 + 𝑠 + 𝜔𝑛2 𝑄 𝜔𝑛 → Natural Frequency In this case, it is the same as the center frequency, 𝜔0. 𝑄 → 𝑄𝑢𝑎𝑙𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 It is the ratio of the center frequency, 𝜔0, to the bandwidth, 𝐵𝑊. Higher Q factor → The filter is more selective. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Passive Narrowband BPF Narrowband BPF can be implemented using passive elements such as resistors, inductors, and capacitors as shown in the figure. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Active Narrowband BPF An active narrowband BPF is implemented using a Multiple Feedback Filter which is shown in the figure. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Output Stages and Power Amplifiers Lecture No. 3: Analysis of Transistor Amplifier Circuits UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits General Considerations Imagine an audio amplifier with an 8-Ω speaker load … If the amplifier is a common emitter or common source amplifier, the output voltage is derived as 𝑉𝑂𝑈𝑇 = 𝐼𝑂𝑈𝑇 (𝑍𝑂𝑈𝑇 ||𝑅𝐿 ) The output voltage will decrease with the load resistance! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits General Considerations What if we want to deliver a power of 1W to the 8-Ω speaker load? The load power for sinusoid signals will be 𝑉𝑃 2 1 𝑃𝐿𝑂𝐴𝐷 = ⋅ 2 𝑅𝐿 If the load power is 1W, the output peak voltage should be 𝑉𝑃 = 4𝑉 The peak current through the load will be 𝑉𝑃 𝐼𝑃 = = 0.5 𝐴. 𝑅𝐿 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits General Considerations Observations: The resistance that must be driven by the amplifier is much lower than the typical values (from hundreds to thousands of ohms) The current levels involved in this example are much greater than the typical currents (in milliamperes) The voltage swings delivered by the amplifier can hardly be viewed as “small” signals The power drawn from the supply voltage, at least 1 W, is much higher than our typical values A transistor carrying such high currents and sustaining several volts (e.g., between collector and emitter) dissipates a high power and, as a result, heats up. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits General Considerations Based on the observations, the parameters of interest in power amplifier stages will be 1. Distortion 2. Power Efficiency 3. Voltage Rating Distortion It is the non-linearity resulting from large signal operation. A high-quality audio amplifier must achieve a very low distortion to reproduce music with high fidelity. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits General Considerations Power Efficiency It is defined as Power Delivered to Load 𝜂= Power Drawn from Supply Measure of how much power from the supply goes to the load. In the previous modules, this parameter was not considered because the absolute value of the power consumption was quite small (a few milliwatts) Voltage Rating Higher output power levels → Higher output voltage swings. The transistor should not breakdown under these conditions. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Emitter Follower as a Power Amplifier Considering the emitter follower circuit, The small signal voltage gain is 𝑟𝑜 ||𝑅𝐿 𝐴𝑉 = 𝑟𝑒 +𝑟𝑜 ||𝑅𝐿 The input impedance is 𝑍𝑖 = 𝛽𝑟𝑒 + 𝛽 + 1 𝑟𝑜 The output impedance is 𝑍𝑜 = 𝑟𝑒 ||𝑟𝑜 ≅ 𝑟𝑒 With its relatively low 𝑍𝑜 , the emitter follower may be considered a good candidate for driving “heavy” loads, i.e., low impedances. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Emitter Follower as a Power Amplifier If the load resistance, 𝑅𝐿 , is 8Ω, a near unity gain will be achieved if 𝑟𝑒 ≪ 𝑅𝐿. For example, for 𝑟𝑒 = 0.8Ω, the voltage gain will be 8Ω 𝐴𝑉 = = 0.9091 0.8Ω+8Ω → 𝑉𝑜𝑢𝑡 will be 0.9091 of 𝑉𝑖𝑛 If 𝑟𝑒 is 0.8Ω, the quiescent emitter current will be If the thermal voltage, 𝑉𝑇 , is 26mV (reference book) 𝐼𝐸𝑄 = 32.5 mA UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Emitter Follower as a Power Amplifier The voltage gain presented in the previous slide is only true for small signals. However, during heavy loads, the circuit’s output swing is large; Small signal analysis will not be valid! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Emitter Follower as a Power Amplifier To this end, consider the the emitter follower shown, in which 𝐼1 is the biasing current source To simplify the analysis, consider the assumptions: The circuit operates rom negative and positive power supplies 𝑉𝑖𝑛𝑄 = 0.8 𝑉; 𝑉𝑜𝑢𝑡𝑄 = 0 𝑉 Ideally, 𝑉𝑖𝑛 is always greater than 𝑉𝑜𝑢𝑡 by 0.8 V UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Emitter Follower as a Power Amplifier For 𝑉𝑖𝑛 = 0.8 𝑉, 𝑉𝑜𝑢𝑡 will be zero! Therefore, 𝐼𝐸 ≅ 𝐼𝐶 = 32.5 mA If 𝑉𝑖𝑛 rises from 0.8 𝑉 to 4.8 𝑉, the emitter voltage follows the base voltage with a relatively constant difference of 0.8 𝑉. This will produce a 4-V swing in the output! 𝑉𝑜𝑢𝑡 𝐼𝐸 = + 32.5 mA 𝑅𝐿 𝑉𝑜𝑢𝑡 = 𝑅𝐿 𝐼𝐸 − 32.5 mA UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Emitter Follower as a Power Amplifier Now, suppose 𝑉𝑖𝑛 begins from 0.8 V and gradually goes down. Since, ideally, the voltage difference between the input and output is 0.8V, if 𝑉𝑖𝑛 becomes 0.7V, then 𝑉𝑜𝑢𝑡 becomes -0.1V. The output is still following the input. But until when the output follows the input? The output voltage swing of the circuit shown is −𝐼1 𝑅𝐿 ≤ 𝑉𝑜𝑢𝑡 ≤ 𝑉𝐶𝐶 − 𝑉𝐶𝐸,𝑠𝑎𝑡 During this range, the output ideally follows 𝐼𝐸 = 𝑉𝑜𝑢𝑡 + 32.5 mA 𝑅𝐿 the input. 𝑉𝑜𝑢𝑡 = 𝑅𝐿 𝐼𝐸 − 32.5 mA UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Emitter Follower as a Power Amplifier With large signal sinusoid input, the output waveform will be: The output voltage swing is limited.  → To increase swing, we need to increase the biasing current The input-to-output characteristic curve will be: → This will increase the power dissipation of the BJT! UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Push-Pull Stage As we observed in the previous section, to improve the output swing, the biasing current 𝐼1 needs to increase. This will lead to a higher power dissipation; Less efficiency. We want to increase 𝑰𝟏 only when 𝑉𝑖𝑛 becomes more negative. We need to replace the current source with another device! The constant current source is replaced by a PNP transistor. When the NPN transistor is OFF, the PNP transistor 𝑉𝑜𝑢𝑡 starts to “kick” in and allows 𝑉𝑜𝑢𝑡 to track 𝑉𝑖𝑛. 𝐼𝐸 = + 32.5 mA 𝑅𝐿 𝑉𝑜𝑢𝑡 = 𝑅𝐿 𝐼𝐸 − 𝐼1 UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Push-Pull Stage UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Push-Pull Stage Why is it called “Push-Pull”?! When 𝑉𝑖𝑛 is sufficiently positive, Q1 operates as an emitter follower and Q2 remains OFF; Q1 pushes current into 𝑅𝐿. When 𝑉𝑖𝑛 is sufficiently negative, Q2 operates as an emitter follower and Q1 remains OFF; Q2 pulls current from 𝑅𝐿. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Sample Problem Sketch the input/output characteristic of the push-pull stage for very positive or very negative inputs. UNIVERSITY Department OF SANTO TOMAS of Electronics – DEPARTMENT OF ELECTRONICS ENGINEERING Engineering Lecture No. 3: Analysis of Transistor Amplifier Circuits Push-Pull Stage What will happen if 𝑉𝑖𝑛 is approaching zero? When 𝑉𝑖𝑛 is zero, we can observe that both transistors are turned OFF. As long as 𝑉𝑖𝑛 is not high enough to turn on one of the transistors, the output will remain at