Lecture 12 Bipolar Amplifiers (1)
Operating Point Analysis and Design
Simple Biasing Assumed VBE and β → IB→IC →VC Comments: IB is sensitive to VBE. IC is very sensitive to β.
Iteration Solution (Kick off iteration with VBE=0.7) IC=βIB
Example 1 Iteration VBE IB (uA) IC (mA) 1 0.7 11.3 1.8 2 0.6834 11.317 Assumed: RB=1 Mohms RC=2 Kohm IS=6.734e-15 Beta=155
ADS Simulation Parameter ADS Iteration VBE 674 mV 683.4 IB 11.3 uA IC 1.76 mA 1.8 mA IB=11.3 uA IC=1.76 mA Beta=155
Execute Iterations in Matlab
β Variation Parameter Β=100 Β=155 VBE 672.1 mV 683.4 mV IB 11.328 uA IC 1.1 mA 1.8 mA gm 43.6 mS 69.2 mS Variation of gm implies that the gain of the amplifier will vary significantly as a function of beta.
Resistive Divider Biasing
Back of the Envelope Calcuation Assume the base current is negligible, i.e. negligible compared to the current in R1. IC does not depend on Beta!
Negligible Base Current VBE=R3/(R1+R3)VCC VBE=5.75/(5.75+17)2.5V=0.631 V I(R1)/IB=110uA/2.10=52.38
Sensitivity to Component Variation Nom. 1% 5% R2 (KOhm) 5.75 5.8075 6.037 VBE (mV) 623 626 639 IB (uA) 2.10 2.35 3.59 IC (uA) 229 262 424 Gm (mS) 8.8 10.1 16.3 1% error in R2 leads to 14 % error in IC. 5% error in R2 leads to 85 % in IC.
Biasing with Emitter Degeneration If R2↑→VP↑→IE↑→VRE↑→small ∆VBE →Small ∆IC An error in VX due to inaccuracies in R1, R2 and VCC is absorbed by RE, leading to a smaller change in VBE.
Design Rules VRE should be > 100 mV.
ADS Simulation IB=12.4 uA IC=1.91 mA Beta=154
Sensitivity to Component Variation Nom. 1% 5% R3 (KOhm) 6.8 6.868 7.140 VBE (mV) 0.68 IB (uA) 12.4 12.5 13.1 IC (mA) 1.91 1.93 2.02 Gm (mS) 73.5 74.2 77.7 1% error in R2 leads to 1.05 % error in IC. 5% error in R2 leads to 5.7 % error in IC.
Misc.
What if base current is not ignored!
Iterative Method Initial assumption: VBE=0.7 V