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Recall Lecture 13 Biasing of BJT Three types of biasing Fixed Bias Biasing Circuit Biasing using Collector to Base Feedback Resistor Voltage Divider Biasing Circuit Applications of BJT As digital logic gates NOT NOR CHAPTER 5 BASIC BJT AMPLIFIERS (AC ANALYSIS) The Bipolar Linear Amplifier Bipolar transistors have been traditionally used in linear amplifier circuits because of their relatively high gain. To use the circuit as an amplifier, the transistor needs to be biased with a dc voltage at a quiescent point (Q-point) such that the transistor is biased in the forward-active region. If a time-varying signal is superimposed on the dc input voltage, the output voltage will change along the transfer curve producing a timevarying output voltage. If the time-varying output voltage is directly proportional to and larger than the time-varying input voltage, then the circuit is a linear amplifier. The linear amplifier applies superposition principle Response – sum of responses of the circuit for each input signals alone So, for linear amplifier, DC analysis is performed with AC source turns off or set to zero AC analysis is performed with DC source set to zero EXAMPLE iC , iB and iE, vCE and vBE Sum of both ac and dc components Graphical Analysis and ac Equivalent Circuit From the concept of small signal, all the time-varying signals are superimposed on dc values. Then: and PERFORMING DC and AC analysis DC ANALYSIS Turn off AC SUPPLY = short circuit AC ANALYSIS Turn off DC SUPPLY = short circuit DO YOU STILL REMEMBER? Let’s assume that Model 2 is used VDQ = V IDQ DC equivalent rd id AC equivalent AC equivalent circuit – Small-Signal Hybrid-π Equivalent OR ib THE SMALL SIGNAL PARAMETERS The resistance rπ is called diffusion resistance or B-E input resistance. It is connected between Base and Emitter terminals The term gm is called a transconductance ro = VA / ICQ rO = small signal transistor output resistance VA is normally equals to , hence, if that is the case, rO = open circuit Hence from the equation of the AC parameters, we HAVE to perform DC analysis first in order to calculate them. EXAMPLE The transistor parameter are = 125 and VA=200V. A value of gm = 200 mA/V is desired. Determine the collector current, ICQ and then find r and ro ANSWERS: ICQ = 5.2 mA, r= 0.625 k and ro = 38.5 k CALCULATION OF GAIN Voltage Gain, AV = vo / vs Current Gain, Ai = iout / is Small-Signal Voltage Gain: Av = Vo / Vs ib Common-Emitter Amplifier Remember that for Common Emitter Amplifier, the output is measured at the collector terminal. the gain is a negative value Three types of common emitter Emitter grounded With RE With bypass capacitor CE STEPS OUTPUT SIDE 1. Get the equivalent resistance at the output side, ROUT 2. Get the vo equation where vo = - ib ROUT INPUT SIDE 3. Calculate Rib using KVL where Rib = vb / ib 4. Calculate Ri 5. Get vb in terms of vs – eg: using voltage divider. 6. Go back to vo equation and replace where necessary Emitter Grounded VCC = 12 V 93.7 k 0.5 k 6.3 k Voltage Divider biasing: Change to Thevenin Equivalent RTH = 5.9 k VTH = 0.756 V RC = 6 k β = 100 VBE = 0.7V VA = 100 V Perform DC analysis to obtain the value of IC BE loop: 5.9IB + 0.7 – 0.756 = 0 IB = 0.00949 IC = βIB = 0.949 mA Calculate the small-signal parameters r = 2.74 k , ro = 105.37 k and gm = 36.5 mA/V Emitter Grounded VCC = 12 V 93.7 k 0.5 k 6.3 k RC = 6 k β = 100 VBE = 0.7V VA = 100 V vb Follow the steps 1. Rout = ro || RC = 5.677 k 2. Equation of vo : vo = - ( ro || RC ) ib= - 567.7 ib 3. Calculate Rib using KVL: ib r - vb = 0 hence Rib = r 4. Calculate Ri RTH||r = 1.87 k 5. vb in terms of vs use voltage divider: vb = [ Ri / ( Ri + Rs )] * vs = 0.789 vs vb so: vb = 0.789 vs replace in equation from step 1 6. Go back to equation of vo vo = - 567.7 ib but ib = vb / Rib (from step 3) vo = - 567.7 [0.789 vs] / 2.74 vo = -163.5 vs AV = vo / vs = - 163.5 bring VS over TYPE 2: Emitter terminal connected with RE – normally ro = in this type β = 120 VBE = 0.7V VA = VCC = 5 V 250 k RC = 5.6 k 0.5 k 75 k RE = 0.6 k 7.46 k 0.5 k RC = 6 k vb 57.7 k RE = 0.6 k vb 1. Rout = RC = 6 k 2. Equation of vo : vo = - RC ib= - 720 ib 3. Calculate Rib using KVL: ib r + ie RE - vb = 0 but ie = (1+ ) ib = 121 ib so: ib [ 121(0.6) + 7.46 ] = vb Rib = 80.06 k 4. Calculate Ri RTH||Rib = 33.53 k 5. vb in terms of vs use voltage divider: vb = [ Ri / ( Ri + Rs )] * vs = 0.9853 vs vb so: vb = 0.9853 vs 6. Go back to equation of vo vo = - 720 ib = - 720 [ vb / Rib ] vo = - 720 [ 0.9853 vs / 80.06 ] vo = - 8.86vs AV = vo / vs = - 8.86 bring VS over TYPE 3: With Emitter Bypass Capacitor, CE Circuit with Emitter Bypass Capacitor ● There may be times when the emitter resistor must be large for the purpose of DC design, but degrades the small-signal gain too severely. ● An emitter bypass capacitor can be used to effectively create a short circuit path during ac analysis hence avoiding the effect RE vb CE becomes a short circuit path – bypass RE; hence similar to Type 1 β = 125 VBE = 0.7V VA = 200 V IC = 0.84 mA VCC = 5 V 20 k RC = 2.3 k 0 k 20 k RE = 5k Bypass capacitor β = 125 VBE = 0.7V VA = 200 V IC = 0.84 mA 3.87 k 0 k vb 10 k 238 k RC = 2.3 k Short-circuited (bypass) by the capacitor CE vb Follow the steps 1. Rout = ro || RC = 2.278 k 2. Equation of vo : vo = - ( ro || RC ) ib= - 284.75 ib 3. Calculate Rib using KVL: ib r - vb = 0 hence Rib = r = 3.87 k 4. Calculate Ri RTH||r = 2.79 k 5. vb in terms of vs use voltage divider: vb = [ Ri / ( Ri + Rs )] * vs = vs since RS = 0 k vb so: v b = vs 6. Go back to equation of vo vo = - 284.75 ib = - 284.75 [ vb / Rib ] vo = - 284.75 (vs) / 3.87 vo = -73.58 vs AV = vo / vs = - 73.58 bring VS over