Lecture 27 OUTLINE The BJT (cont’d) Small-signal model Cutoff frequency Transient (switching) response Reading: Pierret 12; Hu 8.8-8.9.

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Lecture 27 OUTLINE The BJT (cont’d) Small-signal model Cutoff frequency Transient (switching) response Reading: Pierret 12; Hu

Small-Signal Model Transconductance: Common-emitter configuration, forward-active mode: “hybrid pi” BJT small signal model: EE130/230M Spring 2013 Lecture 27, Slide 2

Small-Signal Model (cont.) where Q F is the magnitude of minority-carrier charge stored in the base and emitter regions forward transit time EE130/230M Spring 2013 Lecture 27, Slide 3

Example A BJT is biased at I C = 1 mA and V CE = 3V.  dc = 90,  F = 5ps, T = 300K. Find (a) g m, (b) r , (c) C . Solution: (a) (b) r  =  dc / g m = 90/0.039 = 2.3 k  (c) EE130/230M Spring 2013 Lecture 27, Slide 4

Cutoff Frequency, f T The cutoff frequency is defined to be the frequency (f =  /2  ) at which the short-circuit a.c. current gain equals 1: EE130/230M Spring 2013 Lecture 27, Slide 5

f T is commonly used as a metric for the speed of a BJT. Si/SiGe HBT by IBM For the full BJT equivalent circuit: To maximize f T : increase I C minimize C J,BE, C J,BC minimize r e, r c minimize  F EE130/230M Spring 2013 Lecture 27, Slide 6

Base Widening at High I C : Kirk Effect At very high current densities (>0.5mA/  m 2 ), the density of mobile charge passing through the collector depletion region exceeds the ionized dopant charge density: EE130/230M Spring 2013 Lecture 27, Slide 7 increasing I C For a NPN BJT:  The base width (W) is effectively increased (referred to as “base push out”)   F increases and hence f T decreases. This effect can be avoided by increasing N C  increased C J,BC, decreased V CE0

Summary: BJT Small Signal Model Hybrid pi model for the common-emitter configuration, forward-active mode: EE130/230M Spring 2013 Lecture 27, Slide 8

BJT Switching - Qualitative EE130/230M Spring 2013 Lecture 27, Slide 9

Turn-on Transient Response The general solution is: Initial condition: Q B (0)=0 since transistor is in cutoff where I BB =V S /R S EE130/230M Spring 2013 Lecture 27, Slide 10

Turn-off Transient Response The general solution is: Initial condition: Q B (0)=I BB  B EE130/230M Spring 2013 Lecture 27, Slide 11

Reducing  B for Faster Turn-Off The speed at which a BJT is turned off is dependent on the amount of excess minority-carrier charge stored in the base, Q B, and also the recombination lifetime,  B. – By reducing  B, the carrier removal rate is increased Example: Add recombination centers (Au atoms) in the base EE130/230M Spring 2013 Lecture 27, Slide 12

Schottky-Clamped BJT When the BJT enters the saturation mode, the Schottky diode begins to conduct and “clamps” the C-B junction voltage at a relatively low positive value.  reduced stored charge in quasi-neutral base EE130/230M Spring 2013 Lecture 27, Slide 13