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10V. 2 k. 2 k. B = 100. I E0. I E0. 4.7 k. 4.7 k. 1.5 k. 2I E0. -10 V. Class 3. # 2. Given circuit:. 12V. 100 k. R C. B = 100. Class 3. # 3. Given circuit:. Class 3. The significance of the resistance R E Thumb-rule for the required value of the emitter resistance:
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10V 2 k 2 k B = 100 IE0 IE0 4.7 k 4.7 k 1.5 k 2IE0 -10 V Class 3 • #2. Given circuit:
12V 100 k RC B = 100 Class 3 • #3. Given circuit:
Class 3 • The significance of the resistance RE • Thumb-rule for the required value of the emitter resistance: • REIE~ 1 V. • Stabilizing the operation point by means of a parallel voltage feedback :
VS = 10 V RC = 10 k R2 = 70 k IC0 I1 IB0 R1= 20 k RE = 1.4 k Class 3 • #4. Given circuit:
Class 3 • From equation 1: • From equations 2,3 and 4: • Solution: IC0 = 0.41 mA ; I1 = 0.055 mA
Iout = IC Iin 2IB IC IB IB Class 3 • The current mirror(theory) • Basic connection
V+ Iin Iout T3 T1 T2 R Class 3 • Improvement #1