4/6/2011 Example Amplifier Ditortion 1/9 Example: Amplifier Ditortion Recall thi circuit from a previou handout: 15.0 R C =5 K v ( t) = v ( t) o R B =5 K β = 100 _ vi( t ) 58. R E =5 K CUS We found that the mall-ignal voltage gain i: A vo = vo ( t) 66.7 v ( t) = i
4/6/2011 Example Amplifier Ditortion 2/9 Say the input voltage to thi amplifier i: v ( t) = coωt i Q: What i the larget value that can take without producing a ditorted output? A: Well, we know that the mall-ignal output i: v ( t) = A v ( t) o vo i = 66.7 coωt BUT, thi i not the output voltage! The total output voltage i the um of the mall-ignal output voltage and the DC output voltage! Note for thi example, the DC output voltage i the DC collector voltage, and we recall we determined in an earlier handout that it value i: = = 10 Thu, the total output voltage i : C v( t) = vo( t) = 10.0 66.7 coωt
4/6/2011 Example Amplifier Ditortion 3/9 It i very important that you realize there i a limit on both how high and how low the total output voltage v ( t ) can go! That right! If the total output voltage v ( t ) trie to exceed thee limit even for a moment the BJT will leave the active mode. And leaving the active mode reult in ignal ditortion!
4/6/2011 Example Amplifier Ditortion 4/9 Let break the problem down into two eparate problem: 1) If total output voltage v ( t ) become too mall, the BJT will enter aturation. 2) If total output voltage v ( t ) become too large, the BJT will enter cutoff. We ll firt conider problem 1. For the BJT to remain in active mode, vce ( t ) mut remain greater than 0.7 for all time t (or equivalently vcb( t ) > 0.0). From an earlier handout, we know that E = 5.05. The large capacitor on the emitter keep thi voltage contant with repect to time. Therefore, the voltage vce ( t ) will remain greater than 0.7 only if the collector voltage vc ( t ) remain greater than 5.05 0.7 = 5.75. Note 5.75 i the bae voltage B. f coure, the collector voltage i alo the output voltage ( v( t) = vc( t) ), o that we can conclude that the output voltage mut remain larger than B =5.75 to remain in active mode:
4/6/2011 Example Amplifier Ditortion 5/9 5.75 < v ( t) = 10 66.7 coωt In other word, the lower limit on the total output voltage i: L = 5.75 Note that we can olve thi equation to determine the maximum value of mall-ignal input magnitude : 5.75 < 10 66.7 coωt 66.7 coωt < 4.25 coωt < 0.064 Since coωt can be a large a 1.0, we find that the magnitude of the input voltage can be no larger than 64 m, i.e., < 0.064 If the input magnitude exceed thi value, the BJT will (momentarily) leave the active region and enter the aturation mode! Now let conider problem 2 For the BJT to remain in active mode, the collector current mut be greater than zero (i.e., i C > 0). therwie, the BJT will enter cutoff mode. Applying hm Law to the collector reitor, we find the collector current i:
4/6/2011 Example Amplifier Ditortion 6/9 i C v 15 v = = R 5 CC C it i evident that collector current i poitive only if v < 15. In other word, the upper limit on the total output voltage i: Since: L = 15.0 v ( t) = 10 66.7 coωt we can conclude that in order for the BJT to remain in active mode: 10 66.7 coωt > 15.0 Therefore, we find: 5.0 coωt > = 0.0075 66.7 Since coωt 1, the above equation mean that the input ignal magnitude can be no larger than: < 75 m If the input magnitude exceed 75 m, the BJT will (momentarily) leave the active region and enter the cutoff region!
4/6/2011 Example Amplifier Ditortion 7/9 In ummary: 1) If > 64 m, the BJT will at time enter aturation, and ditortion will occur! 2) If > 75 m, the BJT will at time enter cutoff, and even more ditortion will occur! To demontrate thi, let conider three example: 1. < 64 m The output ignal in thi cae remain between CC =15.0 and B =5.75 for all time t. Therefore, the output ignal i not ditorted. L = = CC 15 v ( t ) = 10 L = = B 5.75 t 2. 64 m < < 75 m
4/6/2011 Example Amplifier Ditortion 8/9 The output ignal in thi cae remain le than CC =15.0 for all time t. However, the mall-ignal output i now large enough o that the total output voltage at time trie to drop below B = 5.75 (i.e., CE drop below 0.7 ). For thee time, the BJT will enter aturation, and the output ignal will be ditorted. L = = CC 15 v ( t ) = 10 L = = B 5.75 3. > 75 m In thi cae, the mall-ignal input ignal i ufficiently large o that the total output will attempt to exceed both limit (i.e., CC = 15.0 and B = 5.75 ). Therefore, there are period of time when the BJT will be in cutoff, and period when the BJT will be in aturation. t
4/6/2011 Example Amplifier Ditortion 9/9 L = = CC 15 v ( t ) = 10 L = = B 5.75 t For a given amplifier voltage gain, you mut determine the larget poible input vi ( t ) that will produce a ditortion-free output ignal. To do thi, you mut determine the limit of the total output voltage. There will be two limit one for aturation (L - ) and one for cutoff (L ).