Single Stage H Biased Amplifier Biology Essay

Single Stage H Biased Amplifier Biology EssayThe objective of this laboratory exercise was to frame, simulate, build and test a single order H- aslant amplifier to chassis specifications and requirements. The design specified a emf gain of 50 be noticeed, a lower cut off frequency of 100Hz and ut intimately regular swing. An NPN 2N3904 electronic transistor was to be use, with a supply potential of 15volts. A 100k load opposition was also required in the design. The esteems for the components snarled in the design were first metric, using logical deductions, and then simulated using the software Multisim. The H- biased amplifier was then built on a solder less(prenominal) bread board. The various jimmys were measured in the laboratory using the appropriate equipment. Analysis of the calculated, simulated and measured orders was done. The dissolvers were discussed and used to determine whether the specifications were met.This laboratory exercise was an introduction to the design of electronic devices. Valuable k directlyledge and practical skills were gained in performing this exercise. This knowledge would prove useful in future designs.table of contentsTable of physiquesList of TablesList of Symbols/ AbbreviationsA Amperes AC Alternative CouplingAv potential Gain BJT Bipolar Junction TransistorCi input Capacitor Co Output CapacitorCE Emitter Capacitor (By pass) dB DecibelsDC Direct Coupling f Frequencyhfe Current Gain rush Emitter Input ImpedanceHz Hertz I CurrentIB Base Current IC accumulator register CurrentIE Emitter Current I1 Current with R1I2 Current through R2 K.V.L Kirchhoffs emf Lawk- Kilo m milliR fortress R1 Resistor 1R2 Resistor 2 RE Emitter ResistorRe Unbypassed Emitter Resistor RE* Recalculated Emitter ResistorRC Collector Resistor RL Load ResistorRTH Thevenins Equivalent shelter V- potential differenceVB Voltage crosswise Base VBE Base Emitter VoltageVCC Supply Voltage VCE Voltage acros s Collector and EmitterVRC Voltage across Collector Resistor VRE Voltage across Emitter ResistorVTH Thevenins Equivalent Voltage VR1- Voltage across R1VR2 Voltage across R2 XCE Reactance of CEXCi Reactance of Ci XCo Reactance of CoZi Input Impedance Zo Output Impedance OhmsIntroductionThe following gives a draft description of the Bipolar Junction Transistor (BJT).A Bipolar Junction Transistor is an active semiconductor device formed by joining 2 P-N junctions whose function is amplification of an electric current. (Seale 2003). The transistor can also be used for the purposes of switching. However, in this exercise, the focus is centered on the application of amplification.A bipolar junction transistor consists of three regions of drugged semiconductors. P- type and N- type semiconductor materials are alternatively joined together to form the transistor. This therefore results in 2 PN junctions. The transistor consists of three regions, namely, the emitter, the low an d the collector. The diagram to a lower place illustrates the basic structure of a transistor, showing the PN junctions and the emitter, base and collector.BJT layersFigure Transistor fount http//encyclobeamia.solarbotics.net/articles/bip_junct_trans.html (accessed November 17th 2010 at 544pm.)From the diagram it is easily seen that one P-N junction is between the emitter and the base and the other P-N junction is between the collector and the base. Since the emitter and collector are usually doped somewhat differently, they are rarely electrically interchangeable. (Seale 2003). The base also forms the mechanical base for the structure. (Seale 2003). The base region is made as thin as realistic (about 10-6m) to get a reasonable good levels of current gain.(Seale 2003). Furthermore, it is made thin for easier passage of electrons through the base region to the collector.Bipolar transistors are either NPN or PNP, based on the arrangement of their N-type and P-type materials. An NPN transistor is formed by sandwiching a very thin region of P-type between two regions of N-type materials.Figure 2 below shows an NPN transistor, while manakin 3 shows the symbol used to represent it.NPN layersFigure NPN Transistor acknowledgment http//encyclobeamia.solarbotics.net/articles/bip_junct_trans.html (accessed November 17th 2010 at 544pm.)Figure NPN Transistor SymbolImageSource http//encyclobeamia.solarbotics.net/articles/bip_junct_trans.html (accessed November 17th 2010 at 544pm.)Similarly, a PNP transistor is formed by sandwiching a very thin region of N-type between two regions of P-type materials.Figure 4 below shows the PNP transistor, while insure 5 shows the symbol used to represent it.Figure PNP TransistorPNP layersSource http//encyclobeamia.solarbotics.net/articles/bip_junct_trans.html (accessed November 17th 2010 at 544pm.)Figure PNP Transistor SymbolImageSource http//encyclobeamia.solarbotics.net/articles/bip_junct_trans.html (accessed November 17th 2010 at 544pm.)(It should be noted that the circle around the transistor usually not seen when the transistor is drawn in circuit diagrams.) It is complete that the distinguishing characteristic of the two transistor symbols is the direction of the arrow.A transistor in a circuit will be in one of three conditions bang off (no collector current). In this region it can be used as switch.In the active region (some collector current, more than a hardly a(prenominal) tenths of a volt above the emitter. In this region, it can be used for amplifier applications.In saturation (collector a few tenths of a volt above emitter), outsized current useful for switch on applications.(Nave)The figure below illustrates these regions of transaction.http//hyperphysics.phy-astr.gsu.edu/hbase/solids/imgsol/tran6.gifFigure Regions of Operation of the TransistorSource http//hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.htmlc4 , accessed November 17th 2010 at 540 pm.A NPN transistor was used to in this laboratory exercise to act as an amplifier. The transistor must therefore be in the active region.Background possiblenessThe following describes the reasons for the choice of biasing arrangement and type of configuration. Possible applications of the design project are also included.For transistors to produce amplification, an operating point must first be established. Since a transistor can act as an amplifier in the active region, it is here that the operating point is established. This operating point is fixed, and so it is referred to as a quiescent point (Q point). This is known as biasing. This operating point is the point where the applied signal would be amplified. therefore it is required that this point be constant. Biasing is the operation of setting an operating point within an operating range and the purpose of bias design is to set the collector current of a transistor to a specific value and keep deviations due to temperature and beta variations to a specified mini mum.(Kuhn 2008)Biasing can take two forms, fixed biasing and H biasing. The fixed bias design works with a transistor with an appropriate current gain, beta. Temperature changes results in a change of beta. This results in a variation of Ic and consequently, the Q point changes. Hence, the fixed biased arrangement is thermally unstable. In the H biased design, the Q points are independent of beta and therefore the Q points are more stable. Two resistors form a potential divider and fix the base potential. Since the base potential is kept nearly constant, if Ic changes, an increase potential drop would develop across RE and VBE would decrease. This results in Ic dropping to its original value. This biased arrangement is therefore thermally stable. Due to this thermal stability, the H biased arrangement was chosen in gustatory perception to the fixed biased arrangement.Now, though the H biasing arrangement was chosen, the type of configuration was still undecided. Three types of conf igurations exist. These include the common emitter, common base and common collector. fit to Kuhn 2008, in general, the best amplifier to use is the one whose input resistance is comparable to the source resistance and whose turnout resistance is comparable to the load resistance.Source Resistance/ Load resistanceGood Choice of Amplifier to considerGreater than 10 earthy Collector2 -10Common Collector or Common Emitter0.5 -2Common Emitter0.1 -0.5Common Emitter or Common BaseLess than 0.1Common BaseTable Showing Choice of Amplifier based on Source/Load ResistanceSource http//www.kennethkuhn.com/students/ee351/text/bjt_general_design.pd, accessed 17th November 2010 at 517 pm.It can be shown that the common-emitter amplifier is capable of acrunving the highest possible place gain. He concludes by saying that, overall, the common-emitter amplifier is the most flexible in terms of input and output resistance while also achieving reasonable power gain.Therefore, the common emitter conf iguration was chosen. Figure 7 shows this configuration.Figure 7 Common Emitter ConfigurationSource electronic Workbench Software- MultisimFigure 7 Common Emitter ConfigurationQ12N3904VccR1R2ReRE*RcRLCoCiCEThis design project would serve as an introduction to the practical application of electronics. With the experience, knowledge and understanding gained in designing and the building the circuit, more advanced and complex circuits could perhaps be built. These circuits would then be used in everyday applications, e,g a microphone, speaker or perhaps a new innovation. This simple circuit in itself could be used to amplify signals with the appropriate adjustments.Design ObjectivesThe following shows the specifications required for the design of the amplifier.The specifications involved in the design of the single stage H-biased amplifier were as followsVoltage gain of 50Lower cut off frequency less than 100HzMaximum interchangeable swingSupply electromotive force of 15voltsIt was required that a micro signal NPN 2N3904 transistor be used in the design. A fixed load resistance of 100k was also required.MethodThe following shows the apparatus and materials used in the design project, the operating conditions, how the various resistor and condenser values were obtained and the basic laboratory procedure.Apparatus and MaterialsThe following shows a list of the apparatus and materials used in overall design and building of the amplifier.2N3904 NPN TransistorSolder less Bread boardResistors (56, 680, 3.6k, 62k,10k,100k)Capacitors (4.7uF, 47uF ,330uF)Connecting wiresPliersDual trace OscilloscopePower Supply carry GeneratorMulti-meterLRC meterAnalysis of Operating ConditionsThe following gives an analysis of the operating conditions for the circuit.The design of the amplifier consisted of two types of operating conditions. These included the DC conditions and the AC conditions. The DC and AC conditions were not mixed. Therefore each type of operating condition wa s considered individually.DC AnalysisThe following describes the operating conditions for the DC analysis.In DC circuits, condensers act as an open circuit. In an open circuit, no current flows. Therefore the condensers were omitted when considering the operating conditions for DC.For the transistor to function as an amplifier, the various quiescent currents and voltages associated with it need to be known. Also, the resistor values need to be known. Therefore, some calculations must first be made.The figure below shows the circuit required for DC analysis.Figure term of enlistment essential for DC AnalysisFigure Circuit Required for DC AnalysisVccR2RERcQ12N3904R1Figure 8 Circuit Required for DC AnalysisSource Electronic Workbench Software MultisimNote that this circuit is same to figure 7. However, the capacitors were omitted as they are part of the AC analysis.The supply voltage, , was given as 15 volts. ascertain IcThe following describes how the value of Ic , was chosen. The transistor datasheet ( set in the appendix) was used to choose an appropriate value for. Firstly, the represent of Typical Pulsed Current Gain vs Collector Current, , was used. Since the quiescent points are independent of beta, , (), then the exact value of becomes irrelevant. Hence, beta can be chosen as any appropriate value. Since the operating temperature for the transistor was room temperature, therefore, the graph at 25C was selected. From the graph it was seen that beta was relatively constant at an of 2mA. Due to this constant beta, a selection of 2mA for was made. It was found that beta was approximately 230. It should be noted that a value of 1mA could also have been chosen, based on the same criteria upon which was chosen. To add some justification to the value chosen, the graph of Base- Emitter ON voltage vs Collector Current was also considered. It was known that an approximate value for was 0.68V. Using again the graph at room temperature, (25C), the equal val ue for a voltage of 0.68 was 2mA. Since multiple graphs indicated that a value of 2mA would be appropriate, therefore, was chosen to be 2mA. counting ofThe following shows the steps involved in find the value of.A requirement was maximum biradial swing. That is, must be able to increase and decrease by the same value. From the circuit,( figure 8), is in series with the emitter of transistor, and therefore in series with the voltage. Since would cause a voltage drop, , across itself, this would therefore reduce the voltage . This voltage drop across, that is, would therefore limit the symmetrical swing. So, to obtain maximum symmetrical swing, it was desired that be measly, when compared to the voltage drop across. With a small , then would be large, and hence produce maximum symmetrical swing. Since 10% can be considered negligible in comparison to 90%, it was therefore assumed that was one tenth the supply voltage.That is,= 1/10..1Since was given as 15volts, then,= 1/10 (15)= 1.5 voltsOhms law was applied,Then, = 2In the circuit, the current which flows through , was the same current . (It is seen later that .20)Therefore, = 3The convention was rearranged to cause the subject,=.Since = 2mA,Therefore,= 1.5V2mASo, = 750Calculation forThe following shows the steps involved in determining the value of.The design for maximum symmetrical swing was considered here.Applying Kirchhoffs Voltage Law (K.V.L) to the circuit= ..4Equation 4 was rearranged giving the equation=So, = -..5 phone, for maximum symmetrical swing, that must increase and decrease by the same value. This implies that would therefore be equal to.6Therefore,= -..7So, 2=-8Hence, =-) 2 .9Where =15V and =1.5VTherefore = (15-1.5) 2=13.5 2= 6.75VOhms law was applied, = 10The formulation was rearranged to construct the subject giving,= ,Since = 2mATherefore, = (6.75V) 2mA= 3375Calculation for values of andThe following shows the steps involved in determining the value of and .The circuit in figure 8 was first off redrawn as shown in figure 9. This was done so that the Thevenins model of the circuit could be easily obtained. Figure 10 shows the Thevenins model for the circuit.Q12N3904VccR1R2RERcVccFigure Redrawn Circuit from Figure 8Source Electronic Workbench Software- MultisimQ12N3904VccRTHReRcVTHFigure 10 Thevenins Model of the CircuitSource Electronic Workbench SoftwareFrom figure 9, Thevenins Theorem was applied for the resistance.This gave, = // 11= ) .12Below shows how an expression for was found.Since was in parallel with, therefore, the voltage across is the same as the voltage across.Voltage Divider Rule was applied to obtain= * .13K.V.L was applied to the circuit in figure 10, this resulted in,= ++.14But = +.15So, by substitution of equation 4 into equation 3 gives,= +++16Also, = 17Substituting 17 in 16 gives,= ++ (+= ++ (1+= + 1+)+ ..18Rearranging 18 to make the subject of the formula gives,+1+)+ =+1+) =-Therefore, =- +1+)19Since hfe is high, to ensure that I B does not cause variations in VB, IB was chosen as one tenth .Also, Recall 15= +In transistor designs, the base current, , is made much smaller than the collector current. (recall from introduction).Hence, .20Since is therefore significantly smaller than, then, it can be said that is negligible when compared to. Therefore, it was assumed that was one tenth of. So =1/10 21Since = 2mAHence,=1/10 (2mA)So,= 0.2mASince = I2, then,I2 = 0.2mAWhen the transistor is on, also known as being forward biased, there is a voltage drop of 0.7 volts across the base-emitter.i.e. = 0.7 volts.K.V.L was applied to the circuit in figure,V2=+22Ohms law was used,I2R2= +.23The formula was rearranged to make R2 the subject,R2 =+) I2 24Since = 1.5V, = 0.7V and I2 = 0.2mAThen, I2R2= (1.5V +0.7V)Therefore, R2 = (1.5V +0.7V) 0.2mAR2 = 2.2V 0.2 mAHence, R2 = 11kThe Voltage divider rule was used to find the value of R1So, VB =(R2 * ) (R1 +R2)25The formula was rearranged to make R1 the subject.(R1 +R2) VB = (R 2 * )(R1 +R2) = (R2 * ) VBTherefore, R1 = (R2 * ) VB R2 ..26So, R1= (11k *15V) 2.2V 11kR1= (165000) 2.2 11kR1= 75000-11000Therefore, R1 =64000R1 = 64kThe following table shows the calculated and standard resistor values. The standard resistor values represent the resistors that are available. It is unlikely that stores have a 3.375k resistor, hence a value close to it (3.6k) would be chosen.Resistor calculate Value (k) touchstone Value(k)RE0.750.75RC3.3753.6R16462R21110Table Showing metric and Standard Resistor valuesNow that R1 and R2 have been calculated, VTH , RTH and IB , can now be calculated.Recall 12= ) 12Therefore, RTH = (64k*11k) (64k+11k)RTH = 9386.667Recall 13= * 13Therefore, = 11k (11k +64k)* 15= 2.2VTherefore, IB can now be found. Recall 19=- +1+19So, IB = 2.2 -0.7 (9386.667 + 1+230(750)IB = 0.0082mAAC AnalysisThe following describes the operating conditions for the AC analysis.The AC analysis was done using h-parameters.The figure below shows the circuit u sed for AC analysis.Q12N3904VccR1R2ReRE*RcRLCoCiCEFigure Circuit used for AC AnalysisSource Electronic Workbench Software- MultisimThe akin of this circuit is shown below.Figure 12 Equivalent CircuitSource Electronic Workbench Software-MultisimFigure Equivalent CircuithfeIBR1hieR2ReRcRLK.V.L was applied and the result was obtained. (Re and RE* were taken as a single resistor, RE)So, Vin= hie IB.27And Vo= hfeIB(Rc//RL)28Since voltage gain is given by, Av = .29Therefore, Av = -hfeIB(Rc//RL) hie IB.30Hence, Av =- hfe(Rc//RL) hie..31The value for Av was therefore found.In the transistor data sheet, using the graph of Input Impedance, it was seen that at a collector current, of 2mA, that the identical hie value was 2k. The graph of Typical Pulsed Current Gain vs , hfe was 230.Av = 230( 3375//100000) 2000Av = -230 ( 3265) 2000Av = -(750950) 2000Therefore, Av = 375.475However, the design specified a voltage gain of 50. Since 375.475 is greater than 50, the resistor RE was spli t into two smaller resistors, Re and RE*. The resistor RE* was bypassed with a capacitor. (Page 28 shows theoretically how the introduction of the bypass capacitor reduces the gain of the amplifier.)Calculation of Re and RE*The following shows the method in which Re and RE* were determinedK.V.L was applied to the circuit in figure 12.At the input, Vin = IBhie + (IB +hfeIB) Re32Vin = IB (hie + (1 +hfe) Re )33At the output, Vo = -IBhfe (Rc //RL)..34Therefore, voltage gain, Av = Vo Vin 29Av = -IBhfe(Rc //RL) IB(hie + (1 +hfe) Re )3535 Simplified was to,Av = hfe(Rc //RL) (hie + (1 +hfe) Re )..36The formula,36 was rearranged to make Re the subject.So, (hie + (1 +hfe) Re ) = hfe(Rc //RL) Av(1 +hfe) Re = hfe(Rc //RL) Av hieTherefore , Re = hfe(Rc //RL) Av hie (1+ hfe )37The appropriate values were substituted into the equation and the value of Re was determined.Since the desired gain is 50, hence, Av is substituted as 50.Re = 230(3375 //100) -50 2000 (1+ 230)Re = 2 30(3265) -50 2000 ( 231)Re = 15019 2000 ( 231)Re = 13019 ( 231)Re = 56.359Therefore, Re = 56Since RE = Re + RE*..38Then , 750 = 56 + RE*Therefore, RE* = 694The following table shows the calculated and standard resistor values for Re and RE*.The standard values represent the values of resistors that were available.ResistorCalculated value (k)Standard value (k)Re0.0563590.056RE*0.6940.680Table Showing Calculated standard resistor valueCalculation for Capacitor values, Ci ,Co and CEThe following shows the steps involved in the calculations of Ci , CO and CE.Ci and CO act as decoupling capacitors in the circuit. The power supply decoupling capacitors filter undesired electronic signals which have been coupled onto the power supply voltage. (Kelly and Emad 1998). Essentially, what this means is that the decoupling capacitors block the DC voltage while permitting the AC to flow. This prevents the DC signal from affecting the bias. Also, it blocks the DC from reaching the AC input source.The main(prenominal) function of the bypass capacitor was to reduce the gain of the amplifier to the desired value. Without the bypass capacitor, the gain of the amplifier was given by,Av =- hfe(Rc//RL) hie..31With the introduction of the bypass capacitor, the gain was now denoted by,Av = hfe(Rc //RL) (hie + (1 +hfe) Re )36From these two equations, since the numerator is the same, and the denominator is larger in 36, hence, the value of Av would be smaller. Therefore, the gain would be reduced with the introduction of the bypass capacitor.Also, the bypass capacitor is to filter out noise at high frequencies. Basically, as the frequency of a signal increases and the pulse width decreases, the impedance of the capacitor decreases and the bypass capacitor acts as a short circuit to these high frequency charges. The dissipation of high frequency charge is related to the value of the capacitor and the subsequent series resistance associated with it. (Wenzel et al.1997.)Firstly , when a particular capacitor was being analyzed, (to be calculated), the other two capacitors were considered as short circuits. This was done so that the effect of the individual capacitor on the circuit could be analyzed. Capacitors act as short circuits, allowing current to flow through them and bypassing the component.The reactance of a capacitor, Xc, is given by,Xc =1 (2fC).39where f is the frequency.There would be a voltage drop across the capacitor. With respect to the capacitor at the input, this voltage drop across the capacitor would therefore limit the voltage across the resistors R1 and R2 .Hence , reduce the voltage being sent to the input of the amplifier to be amplified. It is desired that the maximum voltage possible, be sent to the input to be amplified. Since this voltage drop across the capacitor limits the voltage being sent to the input, it is therefore desired that the voltage drop across the capacitor be negligible in comparison to the input impedance of the circuit. Hence, the reactance of the capacitor was assumed to be one tenth of the input impedance, Zi.So, for capacitor Ci, , Reactance Xci = 1/10 Zi40In figure 12 , the input impedance Zi is given by,Zi = R1 //R2//hie..41In the transistor data sheet, using the graph of Input Impedance, it was seen that at a collector current, of 2mA, that the corresponding hie value was 2k.Therefore, Zi = 64k//11k//2kZi =9.387k//2kSo, Zi = 1.649kNow, Xci = 1/10 ZiHence, Xci = 1/10(1.649k)So, Xci = 164.9Recall 39, Xc =1 (2fC)Therefore, Xci =1 (2fCi)The formula was rearranged to make Ci the subject.2fCi =1 XciTherefore, Ci = 1 (Xci *2f).41Since f represents the lower cut off frequency, which was given as 100Hz. Hence the value for Ci was found.C i= 1 (Xci *2f)Ci = 1 (164.9 *2100)Ci = 1 (103609.726)Ci = 0.00000965FSo, Ci = 9.65uFA similar method was used to find CO.The reactance of the capacitor was considered negligible in comparison the output impedance. This was done so that there would be a minima l voltage drop across the capacitor and the maximum output voltage dropped across the load resistor, RL.So, XCO = 1/10 ZO, (similar to 40 )Where ZO= (RC//RL)Then, ZO= (3.375k//100k)So ZO= 3.265kSince XCO = 1/10 ZOThen, XCO = 1/10(3.265kXCO =326.5Now, XCO =1(2fCO)The formula was rearranged to make CO the subject. The result was as follows.CO = 1 (XCO *2f)Therefore CO =1(326.5 *2*100)CO = 4.87uFSimilarly, for CE.The reactance of the capacitor was considered negligible in comparison the output impedance. This was done so that there would be a minimal voltage drop across the capacitor and the maximum output voltage dropped across the resistor, Re.XCE = 1/10 ReXCE = 1/10 (56)XCE = 5.6Since , XCE = 1 (2fCE)The formula was rearranged to make CE the subject. The result was as follows.CE= 1 (XCE*2f)CE = 1 (5.6*2*100)CE = 284uFThe following table shows the calculated, standard and the chosen capacitor values for Ci ,Co and Ce .CapacitorCalculated Value (uF)Standard Capacitor Value(uF)Capacito r value chosen (uF)Ci9.6510 ,4747Co4.874.74.7Ce284330330Table Showing Calculated and Standard Capacitor valuesThe standard capacitor values represent the capacitor values that were available in stores. It is generally rare to find a capacitor value of exactly 4.87uF, hence a value of 4.7uF was chosen.The value of Ci was chosen to be 47uF instead of 10uf to ensure that the frequency was not exceeded. From the equation, C i= 1 (Xci *2f), it was noticed that if that capacitor value was increased, the frequency would decrea