c° Copyright 2008. W. Marshall Leach, Jr., Professor, Georgia Institute of Tec hnology, School of Electricaland Computer Engineering.The BJT Bias EquationBa sic Bia s Eq u at ion(a) Look out of the 3 terminals of the BJT and make Thévenin equivalent circuits as shown in Fig. 1.Figure 1: Basic bias circuit.(b) Write a loop equation for the base-emitter loop.VBB− VEE= IBRBB+ VBE+ IEREE(c) Use the relation IC= βIB= αIEto express IBand IEas functions of the current desired. Let ICbe the current.VBB− VEE=ICβRBB+ VBE+ICαREE(d) Solve for IC.IC=VBB− VEE− VBERBBβ+REEαAn “educated guess” of the value of VBEmust be made to evaluate this. Typical values are 0.7V for ICtransistors, 0.6−0.65 V for low-power discrete transistors, and 0.5−0.6Vfor higher-power discrete transistors.(e) Chec k for the active mode. For the active mode, VCB> 0.VCB= VC− VB=(VCC− ICRCC) − (VBB− IBRBB)or=(VCC− ICRCC) − (VEE+ IEREE+ VBE)Example 1VBB=V+R2+ V−R1R1+ R2RBB= R1kR2VEE= V−REE= REVCC= V+RCC= RC1Figure 2: Circuit for Example 1.Figure 3: Circuit for Example 2.2Example 2VBB= V+R2RC+ R1+ R2− ICRCRC+ R1+ R2R2RBB=(R1+ RC) kR2VCC= V+R1+ R2RC+ R1+ R2− IBR2RC+ R1+ R2RCRCC= RCk (R1+ R2)VEE=0 REE= REThe base-emitter loop equation for ICisV+R2RC+ R1+ R2− ICRCRC+ R1+ R2R2=ICβRBB+ VBE+ICαREThiscanbesolvedforICand it can be determined if the BJT is in the active mode.Example 3Figure 4: Circuit for Example 3.VBB=V+R2+ V−R1R1+ R2RBB= R1kR2VEE= V+RERC+ R3+ RE− ICRCRC+ R3+ REREREE= REk (RC+ R3)VCC= V+R3+ RERC+ R3+ RE+ IERERC+ R3+ RERCRCC= RCk (R3+ RE)The base-emitter loop equation for ICisV+R2+ V−R1R1+ R2−µV+RERC+ R3+ RE− ICRCRC+ R3+ RERE¶=ICβRBB+ VBE+ICαREEThiscanbesolvedforICand it can be determined if the BJT is in the active mode.3Figure 5: Circuit for Example 4.Example 4VBB1= V+R2R1+ R2RBB1= R1kR2VEE1=0 REE1= RE1VCC1= V+− IB2RC1RCC1= RC1The base-emitter loop equation for IC1isV+R2R1+ R2=IC1β1RBB1+ VBE1+IC1α1REThiscanbesolvedforIC1.For Q2VBB2= V+− IC1RC1RBB2= RC1VEE2=0 REE2= RE2VCC2= V+RCC2= RC2The base-emitter loop equation for IC2isV+− IC1RC1=IC2β2RC1+ VBE2+IC2α2RE2This can be solve for IC2.Giv en IC1and IC2, it can be determined if the two BJTs are in the active mode.Example 5VBB1= V+R2R1+ R2RBB1= R1kR2VEE1= −IB2RE1= −IC2β2RE1REE1= RE1VBB2= IE1RE1=IC1α1RE1RBB2= RE1VEE2=0 REE2= RE2VCC2= V+RCC2= RC2Let the currents to be solved for be IC1and IC2. The two base-emitter loop equations areV+R2R1+ R2−µ−IC2β2RE1¶=IC1β1R1kR2+ VBE1+IC1α1RE1IC1α1RE1=IC2β2RE1+ VBE2+IC2α2RE24Figure 6: Circuit for Example 5.These can be rewritten as follows:IC1µR1kR2β1+RE1α1¶− IC2µRE1β2¶= V+R2R1+ R2− VBE1−IC1µRE1α1¶+ IC2µRE1β2+RE2α2¶= −VBE2The above two equations require simultaneous solution. The determinant solutions areIC1=1∆·µV+R2R1+ R2− VBE1¶µRE1β2+RE2α2¶− VBE2RE1β2¸IC2=1∆·−µR1kR2β1+RE1α1¶VBE2+RE1α1µV+R2R1+ R2− VBE1¶¸where ∆ is the determinant given by∆ =µR1kR2β1+RE1α1¶µRE1β2+RE2α2¶−µRE1α1¶µRE1β2¶In the event that IE1À IB2,theIC2/β2term in the first equation can be neglected so that the first equationbecomesV+R2R1+ R2=IC1β1R1kR2+ VBE1+IC1α1RE1In this case, the approximate solutions areIC1'V+R2R1+ R2− VBE1R1kR2β1+RE1α1IC2=IC1µRE1α1¶−
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