"COMMON EMITTER AMPLIFIER CIRCUIT ANALYSISAn amplifier is a circuit which is used to increase the magnitude of a signal. The input signal toan amplifier will be a current or voltage and the output will be an amplified version of the inputsignal.1) AMPLIFICATION-The impure signal from the sourceis separated intoalternating current anddirect currentbythe coupling capacitor.Directcurrent is sent to a variable resistor also known as transistor. Thetransistor varies the amount of current flowing through the circuit based on the input signal fromthe source. A signal from input source causes the transistor to reduce its resistance and allowingcurrent to flow. The amount of current allowed to flow is based on the size of the input signalfrom the source. A large signal causes more current to flow and results in more amplificationthan the smaller signal. The frequency of the input signal also determines how quickly thetransistor operates and the higher the frequency the higher the gain. Hence the transistor controlslevel and frequency of the electric current sent to the speaker like a valve and this is howamplification is achieved.- For a Common Emitter amplifier (CE), there’s a positive gain which is greater than unity henceused for audio amplification.- It’s basically made up of the following elements:a) Input coupling capacitor (C ) which blocks any DC component from reaching the base1 terminal.b) R1 and R2 which are biasing resistors.They provide the transistor base with thenecessary bias voltage to drive it into the active region.1 | P a g e c) Output coupling capacitor (C ) which prevents any DC voltage from clamping the2 output of the amplifier at the collector terminal.d) Collector Resistor (R ) and Emitter Resistor (RE) which are designed to maintain theC operating the point of the amplifier at the center.e) By-pass capacitor (C3) which acts as a short to AC signal hence prevent distortion in theoutput signal.f) Load Resistor(Rl)which is connected in the collector circuit so that the change in thecollector current causes a change in the voltage developed across it hence the voltagegain .i.e the larger the value of the resistor, the larger the amplification.g) AC voltage source (V ) which provides the input signal.1 h) Collector voltage supply (VCC) which biases the circuit.i) NPN transistor.1) the working of the amplifiera)Positive half cycleThe input and output of the common emitter amplifier are coupled to other circuits through C1and a power amplifier through C2.When the amplifier is quiescent (receiving no signal), thevoltage at a point A is fixed by the potential divider network, R1 and R2. A current flows from Ainto the base of E. A larger collector current passes through R3,the transistor and R4.Values ofR3,R4 and the collector current are such that the voltage at C2 is about half of the supply ,V/2.A positive-going signal voltage, V in, applied to C1, causes the voltage at A V to rise as it isBE already positive to the ground. This causes an increase in forward bias at the emitter basejunction which consequently increases the base current I , which leads to an increase in theB collector current. The pd across R3 increases, leading to a fall in the output voltage this is calledinverting amplifier..The collector current I becomesC I ßI C= BThis leads to a higher voltage drop at the load resistance (R ).As a result V decreases sinceL CE V =V - I R CE CC C LThis drop causes a negative output half cycle.b)Negative half cycleIf a negative half cycle is applied between in the base- emitter circuit;2 | P a g e V decreases because it’s positive with respect to the ground. There will be a decrease inBE forward bias at the emitter base junction causing a decrease in the base current I . As a result theB collector current will decrease causing a decrease in voltage drop at the load resistor (R ) henceL V increases. A positive output signal is collected in collector-emitter circuit.CEFig.b Amplifier circuit showing amplification of signal. PROCEDURE FOR DESIGNING THE CIRCUITa. Deciding on the DC supply voltage VCCThis should be less than the maximum V voltage for the transistor you intend to use and willcc also depend on the available supply; this may be a bench power supply or a battery. We used a10V supply. b. Calculating a value for the collector resistor R c The transistor quiescent collector voltage needs to be about half of V so that the output signalCC can swing by equal amounts above and below this value without driving the transistor intosaturation (0V and maximum collector current) or cut off (zero current and V equal to thec supply voltage). R will therefore be half of V (0.5 ×10=5V) divided by I . c cc c 5 ? ?? = = 0.5? ?0.01? Thus in our design, we used a value of 0.5k3 | P a g e c. Calculating the value of RETo provide efficient bias stabilization, the emitter voltage VE should be about 10% to 15% ofVCC. Assuming that I is the same as I (It is only different by the small amount of the baseE Ccurrent), ??????? ??????? , =RE ??????? ??? ? ??? ,?? 15 V =* 10 = 1.5?E 100 1.5 R== 0.15? E 0.01 d. Estimating the value for base current I B This can be found by dividing the collector current IC by the transistor’s current gain hfeobtained from the data sheet.?? 0.01 I = = B h?? 200 =50µA e. Calculating VB The base voltage should be about 0.7V (700mV) higher than VE to ensure that the input signal isbiased on the linear part of the transistor input characteristic. VE=1.5+0.7=2.2Vf.Calculating the DC bias network current. To ensure adequate bias stability, the current flowing through R and R should be about 10 to 201 2 times greater than the base current I so the current flowing through R and R will be simply B 1 2 I x 20. B Thus I = 20 × 10µA DC bias= 1mA.g. Calculating the resistance for R1??? - R =1? ?? ???? 10-2.2 R = = 7.8?1 0.001 4 | P a g e?? ??W e chose an available value of 7.8K.h. Calculating the resistance for R2 R =2 ??? ???? 2.2 R ==2.2?2 0.001 i. Choosing a value for C1 and C2 The main consideration in choosing these capacitor values is to ensure that their capacitivereactance is low enough, compared with the input impedance of the amplifier, or any loadconnected to the output, to allow signals at all the required frequencies to pass.the general designrule is “Make the reactance atleast 10 times smaller than the total resistance I series with thecapacitor ,at the lowest frequency of the signal.”Use the small signal equivalent circuit and superposition to estimate the input resistanceof the transistor.r’ = 25mV/I e B25 = = 12.5?/ -3 2 × 10 ? ?? ' ? = = ? + (? + 1)? ˜ (? + 1)??? ? ? ? ? ? =101x1k=101k11.68×101 r // R = = 10.47in b 11.68+101 Voltage v will be given by; c 1 ? = ? ; where ic is the current through the capacitor and vc is the voltage drop? ? ??? across the capacitor.? 1? =? (? +10.47+ ) ? ? ??? ? ? ? ˜? 1 ? + ? ??? 5 | P a g e?? ??Choose capacitance for negligible voltage drop at vc for f=fmin? ? ? ˜ =v /10 for f=fmins ? ??? ? ? 10 10 ? = = = 7.6??3 2? ? ? 2? × 20 × 10.47 × 10 ??? ? C1 or C2 >7.6??We chose 10?? for the two coupling capacitors.In our design we chose the following valuesC 6.74µF1 =C = 1.52 µF2j. Choosing a value for C3 C3 mustremove as much of the AC from across RE as possible, and so must have a lowreactance at all audio frequencies. As the lowest frequency is going to be around 20Hz, C3 musthave a reactance (XC) that is small compared to the value of RE at all frequencies above 20Hz.A 0.47µF capacitor was more convenient,as seen1 ? = ? =? ? 2? ? ? ??? Where f =20Hz and ? = 150? (calculated above)min ? 1 ? = = 31.83??2? × 150 × 20 C > 31.83?? we chose 40?? for the bypass capacitor.SIMULATED CIRCUIT6 | P a g eFig c.Simulated circuit2. Procedure for dc analysisFor drawing DC equivalent circuit the following procedure is adapted:1. Short all AC sources 2. Open circuit all capacitors7 | P a g e DC EQUIVALENT CIRCUITV2 30V 2 2 R1 Rc 248K 10.5K 1 1 Q1 2N3904 2 2 R2 RE 52K 4.5K 1 1 Fig.d DC Equivalent Circuit.CALCULATIONSRin = 248//52= 42.99 K? 52 2 V = × ? = × 30 = ? . ??2 ? +? 300 1 2 V = 5.2 – 0.7 = 4.5VE3. Procedure for ac analysisFor drawing AC equivalent circuit the following procedure is adapted:1. Short all the DC sources i.e. treat them as ground2. Short all coupling and bypass capacitors8 | P a g e??AC EQUIVALENT CIRCUIT2 2 2 2 2 V3 1mVac R1 R2 Bre Rc Rl 0Vdc 248K 52K 5000k 10.5k 250k 1 1 1 1 1 Fig.e AC Equivalent Circuit.CALCULATIONS25 25 ?? = = = 25?? 1 ?10.5×250 3 ? = ? //? = × 10 = 10.077K ? ? ? 10.5+250 ß=200 (This was derived from the specifications of the transistor used)ßRe’=25×200=5000?? =R //R //ßre1 2 ?? 1 1 1 1 = + +? 52 248 5 ?? ? =4.479K?? 3 1 1×10 -7 ? = = =2× 10 A=0.? µ?? ? 5000 ?? (???? ) -7 ? = ßI = 2 × 10 × 200=40µA? ? Current Gain? =ß = 200 (This was derived from the specifications of the transistor used)? Voltage Gain3 ? 10.077×10 ? = = = 403.06?? 25 Power Gain9 | P a g e?? ??? = ? × ?? ? ? =200 × 403.06 = 80612? = 10??? ?? 10 ? =10???80612 = 49.06410 4.Is the AC resistance same as the DC collector resistance?They are not equal since the AC collector resistance is 3.333K? while that of the DC is 10K?5.Draw curves at different points of the circuit and explain themThe Base is the driven element. The input signal is injected into the base-emitter circuit while theoutput signal is taken out from the collector-emitter circuit.The emitter-base junction is forwardbiased while the collector-emitter junction is reversed biased.The following diagrams show different waveforms at different points on the amplifierAnalogue practical transistor currents, waveforms and gains10 | P a g e "