Chapter 4 problems: Difference between revisions
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| .2 || 12 || 4 | | .2 || 12 || 4 | ||
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*<math>A_v \cong (19-12)/.4=12.5</math> | |||
===4.34=== | ===4.34=== | ||
===4.45=== | ===4.45=== | ||
===4.64=== | ===4.64=== |
Latest revision as of 12:36, 5 March 2010
4.20
- We want to establish a DC operating point for the BJT. Writing a loop equation gives us -VBB-Vin(t)+iBRB+vBE=0. To form a load line, we will let vin(t)=0 & iB=0. Thus vBE=VBB+vin(t) and iB(t)=(VBB+vin(t))/RB
- for the minimum, Q-point and maximum values.
V V uA -.2 .45 2 0 .58 5 .2 .62 10
- Writing a loop equation for the output: .
V V mA -.2 19 .8 0 16 2 .2 12 4