Bipolar Junction Transistor (BJT) Lecture notes: Sec. 3 Sedra & - PowerPoint PPT Presentation

Bipolar Junction Transistor (BJT) Lecture notes: Sec. 3 Sedra & Smith (6 th Ed): Sec. 6.1-6.4* Sedra & Smith (5 th Ed): Sec. 5.1-5.4* * Includes details of BJT device operation which is not covered in this course F. Najmabadi, ECE65, Winter 2012

A BJT consists of three regions NPN transistor Simplified physical structure An implementation on an IC  Device construction is NOT symmetric  Device is constructed such that o “Thin” base region (between E & C) BJT does NOT act as two o Heavily doped emitter diodes back to back (when o Large area collector voltages are applied to all three terminals). F. Najmabadi, ECE65, Winter 2012

BJT iv characteristics includes four parameters NPN transistor  Six circuit variables: (3 i and 3 v )  Two can be written in terms of the other four: = + KCL : i i i E C B = − KVL : v v v BC BE CE  BJT iv characteristics is the relationship Circuit symbol and among ( i B , i C , v BE , and v CE ) Convention for current directions (Note: v CE = v C – v E )  It is typically derived as = ( ) i f v B BE = i g ( i , v ) C B CE F. Najmabadi, ECE65, Winter 2012

BJT operation in the “active” mode As Emitter is heavily doped, a large number of BE junction is forward biased electrons diffuse into the base (only a small ( v BE = V D 0 ) fraction combine with holes) / v BE V The number of these electrons scales as e T  If the base is “thin” these electrons get near the depletion region of BC junction and are swept into the collector if v CB ≥ 0 ( v BC ≤ 0 : BC junction is reverse biased !) = / v BE V i I e T C S  In this picture, i c is independent of v BC (and v CE ) as long as Active mode: = − = − ≤ v v v V v 0 i I BC BE CE D 0 CE = = v / V C S i e BE T ≥ β β B v V CE D 0 = v / V i I e BE T C S  Base current is also proportional to ≥ v V CE D 0 and therefore, i C : i B = i C / β v BE V / e T F. Najmabadi, ECE65, Winter 2012

BJT operation in saturation mode BE junction is forward biased Similar to the active mode, a large number of ( v BE = V D 0 ) electrons diffuse into the base.  For v BC ≥ 0 BC junction is forward biased and a diffusion current will set up, reducing i C . 1. Soft saturation: v CE ≥ 0.3 V (Si)* v BC ≤ 0.4 V (Si), diffusion current is small and i C is very close to its active-mode level. 2. Deep saturation region: 0.1 < v CE < 0.3 V (Si) or v CE ≈ 0.2 V = V sat (Si), i C is smaller than its active-mode level ( i C < β i B ). o Called saturation as i C is set by outside circuit & does not respond to changes in i B . “Deep” Saturation mode: I = v / V S i e 3. Near cut-off: v CE ≤ 0.1 V (Si) BE T β B Both i C & i B are close to zero. < β i i C B ≈ v V CE sat * Sedra & Smith includes this in the active region, i.e., BJT is in active mode as long as v CE ≥ 0.3 V. F. Najmabadi, ECE65, Winter 2012

BJT iv characteristics includes four parameters NPN transistor Simplified physical structure  BJT iv characteristics is the relationship Circuit symbol and among ( i B , i C , v BE , and v CE ) Convention for current directions (Note: v CE = v C – v E )  It is typically derived as = ( ) i f v B BE = i g ( i , v ) C B CE F. Najmabadi, ECE65, Winter 2012

BJT iv characteristics: i B = f ( v BE ) & i C = g ( i B , v CE ) Saturation: BE is forward biased, BC is forward biased ≤ ≤ ≈ β Active: 1. Soft saturation: 0 . 3 v 0 . 7 V , i i CE C B BE is forward biased ≤ ≤ < β 2. Deep saturation: 0 . 1 0 . 3 V , v i i BC is reverse biased CE C B ≤ ≈ = β 3. Near cut-off: v 0 . 1 V , i 0 i i CE C C B i B Cut-off : BE is reverse biased = = i 0 , i 0 B C F. Najmabadi, ECE65, Winter 2012

Early Effect modifies iv characteristics in the active mode  i C is NOT constant in the active region.  Early Effect: Lines of i C vs v CE for different i B (or v BE ) coincide at v CE = − V A  +  v   = v / V CE i I e 1 BE T   C S   V A F. Najmabadi, ECE65, Winter 2012

NPN BJT iv equations “Linear” model = = Cut-off : = = i 0 , i 0 i 0 , i 0 B C B C BE is reverse biased < v V BE D 0 i I = = v / V C S i e BE T = ≥ Active: β β v V , i 0 B BE D 0 B BE is forward biased = β ≥ i i , v V  +  v C B CE D 0 BC is reverse biased =   v / V CE i I e 1 BE T   C S   V A I = ≥ = v / V v V , i 0 (Deep) Saturation: S i e BE T BE D 0 B β B BE is forward biased = < β v V , i i ≈ < β CE sat C B BC is reverse biased , v V i i CE sat C B = = For Si, V 0 . 7 V , V 0 . 2 V D 0 sat F. Najmabadi, ECE65, Winter 2012

PNP transistor is the analog to NPN BJT PNP transistor “Linear” model = = Cut-off : i 0 , i 0 B C EB is reverse biased < v V EB D 0 Active: = ≥ v V , i 0 EB D 0 B EB is forward biased = β ≥ i i , v V CB is reverse biased C B EC D 0 = ≥ (Deep) Saturation: v V , i 0 EB D 0 B EB is forward biased = < β v V , i i EC sat C B CB is reverse biased Compared to a NPN: 1) Current directions are reversed 2) Voltage subscripts “switched” F. Najmabadi, ECE65, Winter 2012

Notations Voltage sources are DC voltages: identified by node Use “Double subscript” of BJT voltage! terminal: V CC , V BB , V EE . Resistors: Use “subscript” of BJT terminal: R C , R B , R E . F. Najmabadi, ECE65, Winter 2012

Transistor operates like a “valve:” i C & v CE are controlled by i B Controlled part: i C & v CE are set by transistor state (& outside circuit) Controller part: Circuit connected to BE sets i B  Cut-off ( i B = 0 ): Valve Closed i C = 0  Active ( i B > 0 ): Valve partially open i C = β i B  Saturation ( i B > 0 ): Valve open i C < β i B i C limited by circuit connected to CE terminals, increasing i B does not increase i C F. Najmabadi, ECE65, Winter 2012

Recipe for solving BJT circuits (State of BJT is unknown before solving the circuit) 1. Write down BE-KVL and CE-KVL: 2. Assume BJT is OFF, Use BE-KVL to check: BJT OFF: Set i C = 0, use CE-KVL to find v CE (Done!) a. BJT ON: Compute i B b. Assume BJT in active. Set i C = β i B . Use CE-KVL to find v CE . 3. If v CE ≥ V D 0 , Assumption Correct, otherwise in saturation: BJT in Saturation. Set v CE = V sat . Use CE-KVL to find i C . 4. (Double-check i C < β i B ) NOTE: o For circuits with R E , both BE-KVL & CE-KVL have to be solved simultaneously. F. Najmabadi, ECE65, Winter 2012

Example 1: Compute transistor parameters (Si BJT with β = 100 ). = × + 3 BE - KVL : 4 40 10 i v B BE = + 3 CE - KVL : 12 10 i v C CE = < = Assume Cut - off : i 0 and v V 0 . 7 V B BE D 0 = × × + → = 3 BE - KVL : 4 40 10 0 v v 4 V BE BE = > = → v 4 V V 0 . 7 V Assumption incorrect BE D 0 = = ≥ BE ON : v V 0 . 7 V and i 0 BE D 0 B = × × + → = µ > 3 BE - KVL : 4 40 10 i 0 . 7 i 8 . 25 A 0 B B = β ≥ = Assume Active : and 0 . 7 V i i v V C B CE D 0 = β = × × − = 6 i i 100 8 . 25 10 8 . 25 mA C B = × × − + → = 3 3 CE - KVL : 12 10 8 . 25 10 v v 3 . 75 V CE CE = > = → 3 . 75 V 0 . 7 V Assumption correct v V CE D 0 F. Najmabadi, ECE65, Winter 2012

BJT Transfer Function (1) = + BE - KVL : v R i v i B B BE = + CE - KVL : V R i v CC C C CE = < Cut - off : i 0 and v V B BE D 0 = × + → = BE - KVL : v R 0 v v v i B BE BE i = i 0 C = × + → = CE - KVL : V R 0 v v V CC C CE CE CC < → For v V BJT in Cutoff i D 0 = = = i 0 , i 0 , v V B C CE CC = ≥ BE ON : v V and i 0 BE D 0 B − v V = × + → = i D 0 BE - KVL : v R i V i i B B D 0 B R B ≥ → ≥ i 0 v V B i D 0 F. Najmabadi, ECE65, Winter 2012

BJT Transfer Function (2) − v V = = i D 0 BE ON : and v V i BE D 0 B R B = + CE - KVL : V R i v CC C C CE = β ≥ Active : i i and v V c B CE D 0 − v V = β × 0 i D i C R B = + → = CE - KVL : V R i v v V - R i CC C C CE CE CC C C − V V ≥ → ≤ + CC D 0 v V v V β CE D 0 i D 0 R / R C B − V V ≤ ≤ + → CC D 0 For BJT in active V v V β D 0 i D 0 / R R C B F. Najmabadi, ECE65, Winter 2012

BJT Transfer Function (3) − v V = = i D 0 BE ON : v V and i BE D 0 B R B = + CE - KVL : V R i v CC C C CE = < β Saturaatio n : v V and i i CE sat c B V - V = + → = CC sat CE - KVL : V R i V i CC C C sat C R C − V V < β → > = + CC sat i i v V V β c B i IH D 0 R / R C B − V V + < → 0 CC D For V v BJT in saturation β D 0 i R / R C B F. Najmabadi, ECE65, Winter 2012