E field lines are terminated in the e log n e nd na n

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (W B ) JC # e - leaving depregion = - # e entering depregion J n (W B ) and I C o = (- I E ) Jason Woo = MT J C J n (W B ) J En = J n (W B ) J En (- J E ) 38 EE121B Spring 2010 Including the reversed junction currents, we can write I E = I EBo (exp(qVBE / kT ) - 1) + oR IC IC = - ICBo (exp(qVBC / kT ) - 1) + oF I E where ICB0 and IEB0 are the saturation currents of the collectorbase and the emitter-to base junctions when the other terminals (emitter and collector respectively) are open. In the common emitter case under normal mode operation, when IB=0 IC = I E ICBO = 1 - OF Jason Woo 39 EE121B Spring 2010 Let ICEO ICBO /(1 - OF ) i.e., if IB = 0 I E = IC = ICEO Jason Woo 40 EE121B Spring 2010 BVCB0 BVCB0 is the (collector) breakdown voltage when the emitter is open and the base is grounded. It is therefore just the junction breakdown of the base-collector junction. That is, when M= J n (W B + X BC ) J n (W B ) Experimentally, it was found that: 1 M= 1 - (V / BV )m i.e., for the case of BVCB0 with 3 M 6 1 M (VCB ) = 1 - (VCB / BVCBO )m Jason Woo 41 EE121B Spring 2010 BVCE0 BVCE0 is the breakdown voltage of the collector when the base is open and the emitter is grounded. In this case, there are interaction between the 2 junctions and BVCE0 happens when:...
View Full Document

This note was uploaded on 06/01/2010 for the course EE EE161 taught by Professor Wang during the Spring '09 term at UCLA.

Ask a homework question - tutors are online