121B_1_bjt

E je is not too small therefore to maximize we need to

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Unformatted text preview: sh-out effect. 0.8 0.7 0.6 5A 4A 3A 2A IB=1A 1 2 3 4 5 6 IC (mA) 0.5 0.4 0.3 0.2 0.1 0 0 VCE Jason Woo 28 EE121B Spring 2010 Output conductance (Early effect) At fixed VBE, IC increaes with VCB since WB = WBM - W DBE - W DCB and WDCB as VCB WB I C Now, ignoring the recombination in the base AqDB ni2 qVBE IC exp N BWB kT Jason Woo 29 EE121B Spring 2010 In this case I C VCE But VBE I C - I C W B = = VCB W B VCB W B = W BM - W DBEp - W DCBp W B VCB W DCBp =- VCB Now, recall from pn junctions: Vbi + VCB = N ABW DCBp Jason Woo q 2 2 N ABWDCBp + N DCW DCBn 2 = N DCWDCBn 30 ( ) EE121B Spring 2010 Solving for WDBC, we have: W DBCp = - 2 (Vbi + VCB ) N qN AB 1 + AB N DC That is: W B VCB W DCBp =- VCB =- 2qN * (Vbi + VCB ) -1/ 2 where Jason Woo N N = N AB 1 + AB N DC * EE121B Spring 2010 31 Recall C DBC =W = DBC = 2 (Vbi + VCB ) N N q AB DC N AB + N DC = 2 (Vbi + VCB ) 2 N AB q N 2 (Vbi + VCB ) N 2 qN AB (Notes: Capacitance/A) with W DBC = W DBCn + W DBCp W B VCB W DCBp =- VCB C DBC =- qN AB 32 EE121B Spring 2010 Jason Woo or I C VCE VBE - I C W B = W B VCB IC C DBC IC C DBC = = W B qN AB qWB N AB Define Early voltage, VA , via IC I C = VCE VA we have VA = Jason Woo qWB N AB C DBC 33 EE121B Spring 2010 IC IB6 IB5 IB4 IB3 IB2 IB1 -VA ~0.7V VCE As the equation implies, VA is independent of IC Jason Woo 34 EE121B Spring 2010 In advanced bipolar transistors, the base is doped reasonably high and the base does not reaches high level injection. However, the collector is only moderately dop...
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This note was uploaded on 06/01/2010 for the course EE EE161 taught by Professor Wang during the Spring '09 term at UCLA.

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