Bipolar Transistors

T

he transistor is a crystal with three doped regions. The emitter is heavily doped; its job is to emit or inject electrons into the base. The base is lightly doped and very thin; it passes most of the emitter-injected electrons on to the collector. The doping level of the collector lies between that of the emitter and the base. The collector is the largest of the three regions because it must dissipate more heat than the emitter or base. With an npn transistor, free electrons are the majority carriers in the emitter and collector.

npn

(arrow: not pointing in)

Note: For pnp discussion simply reverse all polarities of npn discussion.

pnp

Additional information:

· The doping material in n-type semiconductors is called donor atoms since they “donate” extra electrons to the material. The main charge carriers in n-type semiconductor are the extra electrons and they are the majority carriers.

· The doping material in p-type semiconductors is called acceptor atoms since they “accept” extra electrons from the material. The main charge carriers in p-type semiconductor are the holes and they are the majority carriers.

Regions and Transistor currents

0) Breakdown Region: region where Vce > Vz.

® the transistor will likely be destroyed or degraded

1) Cut-off Region: when I_B = 0

® region for which the emitter and collector junctions are both reverse-biased

2) Saturation Region: 0 < Vce < Vz (typically 1 V or less)

® both emitter and collector junctions are forward-biased

3) Active Region: represents normal transistor operation

® emitter diode is forward-biased; the collector diode is reverse-biased

® I_E = Ic + I_B = bI_B + I_B = (b+1)I_B

I_E @ Ic = bI_B

® a small current flowing into the base controls a much larger current

flowing into the collector

Note: V_cutin = 0.7 V or 0.6 V for Si; V_cutin = 0.3 V or 0.2 V for Ge

Steps in Determining Transistor State (Operating Region)

1) If 0 ³ I_B then CUT-OFF.

2) If I_B > 0 then either SATURATION or ACTIVE. Assume ACTIVE state and solve transistor parameters like I_B, Ic, or Vce.

3) If there’s an unlikely (weird) transistor parameter value then SATURATION otherwise ACTIVE.

Example: Find the operating regions:

where, V_g = 0.7 V; I_CE = 1mA; b = 100; and V_SAT = 0.2 V

1) For the cut-off region, assume cut-off region state.

I_B = (V_S – V_BE)/ R_B = (V_S- 0.7)/5k £ 0

Thus, V_S £ 0.7 V. This means that V_S has to be less than 0.7 V to be in the cut-off region.

b) For the saturated region, assume saturated region state.

I_B = (V_S – V_BE)/ R_B = (V_S- 0.7)/5k £ 0

I_c = (V_CC – V_SAT)/ R_C = (12- 0.2)/1000 = 11.8 mA

Since I_C < bI_B,

11.8 mA < 100 (V_S- 0.7)/5k £ 0

Thus, V_S > 1.29 V. This means that V_S has to be greater than 1.29 V to be in the saturation region.

c) The active regions lies between the cut-off and the saturation regions; thus, 0.7 V < V_S £ 1.29 V.

Applications

• The most important example of the transistor is that of being an active component; that is, a device that can amplify input, producing an output signal with more power in it than the input signal. The additional power comes from an external power source (power supply to be exact).
• Transistor switch.
• Essential ingredient of almost every electronic circuit.
• IC’s (Integrated Circuits) are constructed from discrete transistors.

Circuit Examples:

Emitter-Follower

V_out = V_in – V_BE = I_ER_E

Z_in = (b + 1)Z_L = (b + 1)R_E

Z_out = Z_source/(b + 1)

DV_in à DV_Base ; small DI_B

à DV_out ; large DI_E

Single-Supply Emitter

Assume V_CC to be 15 V.

Quiescent point:

• I_C and V_CE values when there is no input
• Assumptions: V_BE = 0.7 V; I_C ≡ I_E
• R_B = (R₁R₂) / (R₁+R₂) = 69.6 k
• V_BB = V_CC [R₂ / (R₁+R₂)] = 8 V
• V_E = V_BB – V_BE = 8 – 0.7 V = 7.3 V
• V_CE = V_C – V_E = 15 – 7.3 V = 7.7 V
• I_C ≈ I_E = V_E / R_E = 7.3 / 7.5 k ≈ 1 mA
• Thus, Q pt: (7.7 V, 1 mA)

Common-Emitter Amplifier

Assume V_CC to be 12 V.

Quiescent point:

• I_C and V_CE values when there is no input
• Assumptions: V_BE = 0.7 V; I_C ≡ I_E
• V_B = V_CC [R₂ / (R₁+R₂)] = 2.4 V
• V_E = V_B – V_BE = 1.7 V
• I_E = V_E / R_E = 1.7 mA
• I_C ≈ I_E ≈ 1.7 mA
• V_C = V_CC – I_CR_C = 5.2 V
• V_CE = V_C – V_E = 3.5 V
• Thus, Q pt: (3.5 V, 1.7 mA)

Gain:

• ∆V_in à ∆V_B à ∆V_E
• ∆V_in = ∆V_B = ∆V_E
• I_{E =} ∆V_E / R_E ≈ ∆I_C
• ∆V_C = -∆I_CR_C = -∆V_ER_C / R_E = - ∆V_in (R_C/R_E)
• ∆V_out = -∆V_in (R_C/R_E)
• gain = ∆V_out / ∆V_in = -R_C / R_E ≈ -4
• thus, the output is –4 times the input

Web page created by: Beverley Joy Ang, Gail Chua, Katrina Mona Santiago, Mailyn Terrado,

and Kurt Lester Yu.

Lecture Notes by: Mr. Carlos Oppus

February 21, 2003.