ECD-II
Unit 1 Multistage Amplifier Q 1) Explain the need for cascading?A 1)Cascading amplifiers are used to increase signal strength in Television receiver. In cascading amplifier output of first stage is connected to input of second stage. Although the gain of amplifier depends on device parameters and circuit components, there exists upper limit for gain to be obtained from single stage amplifier. To overcome this problem, we need to cascade two or more stage of amplifier to increase overall voltage gain of amplifier. When more than one stages used in succession it is known as multi-stage amplifier. The disadvantage is bandwidth decrease as number of stages increases. Q 2) Derive the equations for voltage gain?A 2)Voltage gain:In the small signal equivalent circuit model. Remove RL and RS
Figure. Small signal equivalent circuit model Therefore, we get
Figure. Equivalent Small signal equivalent model vout = -gm vt (ro || RD)Then unloaded voltage gain Avo = vout / vt = -gm (ro||RD) Q3) Explain RC coupled Amplifier?A 3)The individual stages of the amplifier are connected using a resistor-capacitor combination hence named as RC Coupled. R1 and R2 form the biasing network while the emitter resistor RE forms the stabilization network. CE is called bypass capacitor which passes AC and restricts DC. The coupling capacitor CC also increases the stability of the network as it blocks the DC while offers a low resistance path to the AC signals, thereby preventing the DC bias conditions of one stage affecting the other.
Figure . RC coupled Amplifier The AC component of this signal is coupled to second stage of the RC coupled amplifier through coupling capacitor CC and thus appears as an input at the base of the second transistor Q2.This is amplified further and passed-on as output of the second stage which is available at the collector terminal of Q2 after getting shift by 180o in its phase.This means that the output of the second stage will be 360o out-of-phase with respect to the input, which in turn indicates that the phase of the input signal and the phase of the output signal obtained at stage II will be identical.The frequency response of a RC coupled amplifier shown by Figure 2, indicates that the gain of the amplifier is constant over a wide range of mid-frequencies while it decreases considerably both at low and high frequencies.
Figure . Frequency Response of RC Coupled Amplifier Q 4) Explain Transformer Coupled Amplifier?A 4)Transformer Coupled
Figure . Transformer Coupled In the circuit diagram, one stage output is connected as an input to the second stage amplifier through a coupling transformer. In the RC coupling amplifier, cascading of the first & second stage amplifier is done through coupling capacitor. The coupling transformer T1 & it is primary and secondary windings are P1 and P2. Similarly, secondary transformer T2 having the primary windings p1 and secondary windings are indicated by p2.R1 & R2 resistors provide the biasing and stabilization for the circuit. Cin isolates DC and allows only AC components from the input signal to the circuit. The emitter capacitor provides a low reactance path to the signal and offers stability to the circuit. The first stage of output is connected as an input to the second stage through secondary windings (p2) of the primary transformer. Here, the input signal is applied to the base of the first transistor. If the input signal has any DC signal, then the components can be eliminated by the input capacitor Cin. When the signal is applied to the transistor it amplifies & forwards to collector terminal. Hence the amplified output is connected as an input to the second stage of the transformer-coupled amplifier through secondary windings (p2) of the coupling transformer.This amplified voltage is applied to the base terminal of the second transistor of the secondary stage of the transformer-coupled amplifier Q 5) Explain its frequency response?A 5) The frequency response of an amplifier allows us to analyze the output gain and phase response for a particular frequency over a wide range of frequencies.
Figure 8. Frequency Reponse The transformer-coupled amplifier offers constant gain over a small range of frequencies. At low frequencies, due to the reactance of the primary transformer p1, the gain is decreased. At higher frequencies, the capacitance between the turns of the transformer will act as a condenser and this reduces the output voltage, and this leads to decrement in gain. Q 6) What is Direct Coupling?A 6)As shown in the figure the direct coupled amplifier (DC ) consists of two transistors Q1 and Q2 voltage divider bias network (R1,R2) which is provided on the transistor Q1 base two collector resistors Rc1,Rc2 transistor Q2 is self-biased and we use emitter bypass resistors RE1 and Re2 . The direct coupled amplifier is operated without using frequency sensitive component like capaciotr, inductor and transformer etc. The direct coupled amplifier amplify the AC signal with frequency as low a fraction of Hertz.
Figure . Direct Coupled Amplifier When +ve half cycle is applied at the input Q1 transistor biased through the divider bias network. The positive half cycle forward biases the transistor Q1 which starts conduction and provides an inverted and amplified output at the collector.We know that VCE = VCC – Ic Rc The amplified negative signal is provided as base to transistor Q2 which is self bias. The base of Q2 transistor is reverse and does not conduct , the output of Q2 is the amplified signal when Q2 does not conduct and the voltage drop across collector emitter will be zero. Therefore Vcc=IcRc Q 7) Explain the design of RC coupled amplifier?A 7)Design a two stage RC coupled BJT amplifier to meet the following specifications. Av 5000. Sico10 FL =20Hz Vo =2V and Vcc=15V. Calculate Av,Ri and Ro of the designed circuit. What would be the gain of the circuit if both the capacitors are removed. Given data: Av FL =20Hz Sico FL =20Hz and Vo=2V Step 1: Circuit Diagram of amplifier circuit As the input resistance or impedance is large in given data Ri 1 MHz we need to use FET amplifier rather than BJT as first stage. BJT cannot provide higher gain so that we use BJT for second stage hence it is CS-CE configuration.
Vcc=6V Step 2: Voltage gain calculations FET has low voltage gain Av1=10------- First stage Av2 = 150 -----Second Stage AVT = Av1 x Av2 = 10 x 150 = 1500.We first design second stage Using hybrid h-parameter Step3: Calculation of collector resistor Rc and selection of transistor. Since transistor is not specified we take BCS 47A Av2 = - hfe (Rc||RL) / hie RL =∞Av2 = -hfe(Rc)/hie Rc = Av2.hie /hfe(min) From the data sheet hie =2.7 khfe = 125 Rc2 = 150 x 2.7k/125 =3.24 kΩHence the approximate value of Rc = 3.6kΩ Step3: Calculation of Q point Vcc is given as 6V Take VCEQ as 50% Vcc to have Q point at the centre and accommodate full swing of output voltage. Take VRE as 20% Vcc VceQ = ½ Vcc = 1.2 x 6 = 3V Vre = 20% Vcc = 1.2 V Apply KVL to output loop Vcc – VRC – VCEQ -VRE =0 VRC = Vcc – VceQ – Vre= 6-3-1.2 Vrc = 1.8V We know that VRC = Ic Rc Ic = VRC/Rc = 4.6 /3.6 kΩIcQ = 0.5 mA Q point(3V,0.5mA) Step4: Calculation of RE(emitter resistance )Vre = IE . RE IE = (1+β) IB IB = Ic/βIB = 0.5mA/180 IB = 2.7 μA IE = (1+β) IB IE = 0.48mA RE = VRE/IE = 1.2 / 0.48mA = 2.5kΩSelect RE = 2.4 kΩStep 5: Calculation of R1 and R2 for second stage Given Sico10 VB = VBE + VRE = 0.7 + 1.2 = 1.9 V But VB = Vcc.R2/ (R1+R2)VB/Vcc = R2/R1+R2Using Thevenin’s theorem Rth = R1R2/ (R1+R2)Rth/R1 = R2/(R1+R2) For voltage divider bias stability factor is given by S / 1+β (Re/Rth + Re) Take β=hfe(max)S =8 hfe(max) for BC 547-A hfe max =260. Substituting in the above formula we get 8 = 1+260/ 1+260 (Rth / Rth +RE)8/ 261 = 1/1+260(RE /Rth + RE) 261/8 = 1/1+260 (RE/Rth + RE) 32.62 = 1+260(RE/Rth+Re) 31.62 = 260(RE/Rth+RE) 0.122 = RE/Rth+RE0.122 Rth + 0.122 RE = RE Solving for Rth we get Rth = 17.27 kΩBy using eq(1) and eq(2) we get VB/Vcc = Rth/R1 R1 = Rth x Vcc/VB = 17.27 x 10 3 x 6/ 1.9 R1=54.53 kΩSelect the higher standard value of R1 = 56kΩ.Rth = R1R2/R1+R2 17.27 x 10 3 = 56kΩ R2 / 56kΩ+ R2 R2= 24.97 kΩSolving for R2 we get R2=24.97kΩTherefore R2 =22kΩ Q8) Write a short note on design of direct coupled amplifier?A 8)The design of a single stage RC coupled amplifier is shown below.Design of Re and Ce:Let voltage across Re: VRe = 10%Vcc ………….(1) Voltage across Rc: VRc = 40% Vcc. ……………..(2) The remaining 50% will drop across the collector-emitter .From (1) and (2) Rc =0.4 (Vcc/Ic) and Re = 01(Vcc/Ic).Design of R1 and R2:Base current Ib = Ic/hfe.
Let Ic ≈ Ie .
Let current through R1; IR1 = 10Ib.
Also voltage across R2: VR2 must be equal to Vbe + VRe. From this VR2 can be found.
Therefore VR1 = Vcc-VR2. Since VR1 ,VR2 and IR1 are found we can find R1 and R2 using the following equations.
R1 = VR1/IR1 and R2 = VR2/IR1. Finding Ce: Impedance of emitter by-pass capacitor should be one by tenth of Re.XCe = 1/10 (Re)XCe = 1/2πFCe Finding Cin: Impedance of the input capacitor(Cin) should be one by tenth of the transistors input impedance (Rin).i.e. XCin = 1/10 (Rin)Rin = R1 parallel R2 parallel (1 + (hfe re))re = 25mV/Ie.Xcin = 1/2πFCin.From this Cin can be found. Finding Cout: Impedance of the output capacitor (Cout) must be one by tenth of the circuit’s output resistance (Rout).XCout = 1/10 (Rout).Rout = Rc.XCout = 1/ 2∏FCout.From this Cout can be found. Setting the gain: Introducing a suitable load resistor RL across the transistor’s collector and ground will set the gain.Expression for the voltage gain (Av) of a common emitter transistor amplifier is as follows.Av = -(rc/re)re = 25mV/Ieand rc = Rc parallel RLFrom this RL can be found Q 9) Explain the application of multisatges amplifiers?A 9)The multistage amplifier applications can be found in various industries in various scenarios and those are:Employed in the conditions when perfect impedance matching is required Used in the applications when correct frequency response is necessary These amplifiers are also used for DC isolation purposes Applications those need enhanced gain, and good flexibility Enhanced bandwidth Multistage amplifiers designed with MOSFET devices are also employed in many applications Audio transformers Microphones Multistage amplifier cascading is used for high-voltage and high-speed applications Q 10) Explain the advantages of multistage amplifier?A 10)The advantages of the multistage amplifier are flexibility within input & output impedance and higher gain. The multistage amplifier applications are, it can be used to increase extremely weak signals to utilizable levels. The distortion can be reduced by changing the signal within stages.
Let Ic ≈ Ie .
Let current through R1; IR1 = 10Ib.
Also voltage across R2: VR2 must be equal to Vbe + VRe. From this VR2 can be found.
Therefore VR1 = Vcc-VR2. Since VR1 ,VR2 and IR1 are found we can find R1 and R2 using the following equations.
R1 = VR1/IR1 and R2 = VR2/IR1. Finding Ce: Impedance of emitter by-pass capacitor should be one by tenth of Re.XCe = 1/10 (Re)XCe = 1/2πFCe Finding Cin: Impedance of the input capacitor(Cin) should be one by tenth of the transistors input impedance (Rin).i.e. XCin = 1/10 (Rin)Rin = R1 parallel R2 parallel (1 + (hfe re))re = 25mV/Ie.Xcin = 1/2πFCin.From this Cin can be found. Finding Cout: Impedance of the output capacitor (Cout) must be one by tenth of the circuit’s output resistance (Rout).XCout = 1/10 (Rout).Rout = Rc.XCout = 1/ 2∏FCout.From this Cout can be found. Setting the gain: Introducing a suitable load resistor RL across the transistor’s collector and ground will set the gain.Expression for the voltage gain (Av) of a common emitter transistor amplifier is as follows.Av = -(rc/re)re = 25mV/Ieand rc = Rc parallel RLFrom this RL can be found Q 9) Explain the application of multisatges amplifiers?A 9)The multistage amplifier applications can be found in various industries in various scenarios and those are:
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