Bipolar Junction Transistor (BJT)

1 Bipolar Junction Transistor (BJT). Lecture notes: Sec. 3. Sedra & Smith (6th Ed): Sec. *. Sedra & Smith (5th Ed): Sec. *. * 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).

2 Two can be written in terms of the other four: KCL : iE = iC + iB. KVL : vBC = vBE vCE. Circuit symbol and BJT iv characteristics is the relationship Convention for current directions among (iB , iC , vBE , and vCE ). (Note: vCE = vC vE) It is typically derived as iB = f (vBE ). iC = g (iB , vCE ). 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 (vBE = VD0) fraction combine with holes). The number of these electrons scales as e vBE / VT. If the base is thin these electrons get near the depletion region of BC Junction and are swept into the collector if vCB 0.

3 (vBC 0 : BC Junction is reverse biased!). iC = I S e vBE / VT. In this picture, ic is independent of vBC. (and vCE ) as long as Active mode: iC IS vBC = vBE vCE = VD 0 vCE 0. iB = = e v BE / VT. vCE VD 0. iC = I S e vBE / VT. Base current is also proportional to vCE VD 0. e vBE / VT and therefore, iC : iB = iC/ . F. Najmabadi, ECE65, Winter 2012. BJT operation in saturation mode BE Junction is forward biased Similar to the active mode, a large number of (vBE = VD0) electrons diffuse into the base. For vBC 0 BC Junction is forward biased and a diffusion current will set up, reducing iC . 1. Soft saturation: vCE V (Si)*. vBC V (Si), diffusion current is small and iC is very close to its active-mode level.

4 2. Deep saturation region: < vCE < V (Si). or vCE V = Vsat (Si), iC is smaller than its active-mode level (iC < iB). o Called saturation as iC is set by outside Deep Saturation mode: circuit & does not respond to changes in iB. IS. iB = e vBE / VT 3. Near cut-off: vCE V (Si).. Both iC & iB are close to zero. iC < iB. vCE Vsat * Sedra & Smith includes this in the active region, , F. Najmabadi, ECE65, Winter 2012 BJT is in active mode as long as vCE V. BJT iv characteristics includes four parameters NPN Transistor Simplified physical structure Circuit symbol and BJT iv characteristics is the relationship Convention for current directions among (iB , iC , vBE , and vCE ). (Note: vCE = vC vE) It is typically derived as iB = f (vBE ).

5 IC = g (iB , vCE ). F. Najmabadi, ECE65, Winter 2012. BJT iv characteristics: iB = f(vBE) & iC = g(iB , vCE). Saturation: BE is forward biased, BC is forward biased 1. Soft saturation: vCE V, iC iB Active: BE is forward biased 2. Deep saturation: vCE V, iC < iB BC is reverse biased 3. Near cut-off: vCE V, iC 0 iC = iB. iB. Cut-off : BE is reverse biased iB = 0, iC = 0. F. Najmabadi, ECE65, Winter 2012. Early Effect modifies iv characteristics in the active mode iC is NOT constant in the active region. Early Effect: Lines of iC vs vCE for different iB (or vBE ) coincide at vCE = VA. v . iC = I S e vBE / VT 1 + CE . VA . F. Najmabadi, ECE65, Winter 2012. NPN BJT iv equations Linear model Cut-off : iB = 0, iC = 0 iB = 0, iC = 0.

6 BE is reverse biased vBE < VD 0. iC IS. Active: iB = = e vBE / VT vBE = VD 0 , iB 0.. BE is forward biased vCE iC = iB , vCE VD 0. BC is reverse biased iC = I S e v BE / VT. 1 + . VA . IS vBE = VD 0 , iB 0. (Deep) Saturation: iB = e vBE / VT. BE is forward biased . vCE = Vsat , iC < iB. BC is reverse biased vCE Vsat , iC < iB. For Si, VD 0 = V, Vsat = V. F. Najmabadi, ECE65, Winter 2012. PNP Transistor is the analog to NPN BJT. PNP Transistor Linear model Cut-off : iB = 0, iC = 0. EB is reverse biased vEB < VD 0. Active: vEB = VD 0 , iB 0. EB is forward biased CB is reverse biased iC = iB , vEC VD 0. (Deep) Saturation: vEB = VD 0 , iB 0. EB is forward biased vEC = Vsat , iC < iB.

7 CB is reverse biased Compared to a NPN: 1) Current directions are reversed 2) Voltage subscripts switched . F. Najmabadi, ECE65, Winter 2012. Notations DC voltages: Voltage sources are Use Double subscript of BJT identified by node terminal: VCC , VBB , VEE . voltage! Resistors: Use subscript of BJT. terminal: RC , RB , RE . F. Najmabadi, ECE65, Winter 2012. Transistor operates like a valve: . iC & vCE are controlled by iB. Controlled part: iC & vCE are set by Transistor state (&. Controller part: outside circuit). Circuit connected to BE sets iB. Cut-off (iB = 0): Valve Closed iC = 0. Active (iB > 0): Valve partially open iC = iB. Saturation (iB > 0): Valve open iC < iB. iC limited by circuit connected to CE terminals, increasing iB.

8 Does not increase iC. 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: a. BJT OFF: Set iC = 0, use CE-KVL to find vCE (Done!). b. BJT ON: Compute iB. 3. Assume BJT in active. Set iC = iB . Use CE-KVL to find vCE . If vCE VD0 , Assumption Correct, otherwise in saturation: 4. BJT in Saturation. Set vCE = Vsat . Use CE-KVL to find iC . (Double-check iC < iB ). NOTE: o For circuits with RE , both BE-KVL & CE-KVL have to be solved simultaneously. F. Najmabadi, ECE65, Winter 2012. Example 1: Compute Transistor parameters (Si BJT with = 100).

9 BE - KVL : 4 = 40 103 iB + vBE. CE - KVL : 12 = 103 iC + vCE. Assume Cut - off : iB = 0 and vBE < VD 0 = V. BE - KVL : 4 = 40 103 0 + vBE vBE = 4 V. vBE = 4 V > VD 0 = V Assumption incorrect BE ON : vBE = VD 0 = V and iB 0. BE - KVL : 4 = 40 103 iB + iB = A > 0. Assume Active : iC = iB and vCE VD 0 = V. iC = iB = 100 10 6 = mA. CE - KVL : 12 = 103 10 3 + vCE vCE = V. vCE = V > VD 0 = V Assumption correct F. Najmabadi, ECE65, Winter 2012. BJT Transfer Function (1). BE - KVL : vi = RB iB + vBE. CE - KVL : VCC = RC iC + vCE. Cut - off : iB = 0 and vBE < VD 0. BE - KVL : vi = RB 0 + vBE vBE = vi iC = 0. CE - KVL : VCC = RC 0 + vCE vCE = VCC. For vi < VD 0 BJT in Cutoff iB = 0, iC = 0, vCE = VCC.

10 BE ON : vBE = VD 0 and iB 0. vi VD 0. BE - KVL : vi = RB iB + VD 0 iB =. RB. iB 0 vi VD 0. F. Najmabadi, ECE65, Winter 2012. BJT Transfer Function (2). vi VD 0. BE ON : vBE = VD 0 and iB =. RB. CE - KVL : VCC = RC iC + vCE. Active : ic = iB and vCE VD 0. vi VD 0. iC = . RB. CE - KVL : VCC = RC iC + vCE vCE = VCC - RC iC. VCC VD 0. vCE VD 0 vi VD 0 +. RC / RB. VCC VD 0. For VD 0 vi VD 0 + BJT in active RC / RB. F. Najmabadi, ECE65, Winter 2012. BJT Transfer Function (3). vi VD 0. BE ON : vBE = VD 0 and iB =. RB. CE - KVL : VCC = RC iC + vCE. Saturaation : vCE = Vsat and ic < iB. VCC - Vsat CE - KVL : VCC = RC iC + Vsat iC =. RC. VCC Vsat ic < iB vi > VIH = VD 0 +. RC / RB. VCC VD 0.