EE105 Fall 2011Lecture 6, Slide 1Prof. Salahuddin, UC Berkeley Lecture 6 OUTLINE PN Junction Diodes – Reverse Breakdown – Large and Small signal models Reading: Chapter , Bipolar Junction Transistors: Chapter 4

EE105 Fall 2011Lecture 6, Slide 2Prof. Salahuddin, UC Berkeley Depletion Width at Equilibrium V(x)V(x) x a-b V0V0 0 --(1) --(2) --(3) V(a)=0 V(-b)=V 0; Built in potential

EE105 Fall 2011Lecture 6, Slide 3Prof. Salahuddin, UC Berkeley Reverse Breakdown Mechanisms a)Zener breakdown occurs when the electric field is sufficiently high to pull an electron out of a covalent bond (to generate an electron-hole pair). b)Avalanche breakdown occurs when electrons and holes gain sufficient kinetic energy (due to acceleration by the E-field) in-between scattering events to cause electron- hole pair generation upon colliding with the lattice.

EE105 Fall 2011Lecture 6, Slide 4Prof. Salahuddin, UC Berkeley Reverse Breakdown As the reverse bias voltage increases, the electric field in the depletion region increases. Eventually, it can become large enough to cause the junction to break down so that a large reverse current flows: breakdown voltage

EE105 Fall 2011Lecture 6, Slide 5Prof. Salahuddin, UC Berkeley Parallel PN Junctions Since the current flowing across a PN junction is proportional to its cross-sectional area, two identical PN junctions connected in parallel act effectively as a single PN junction with twice the cross-sectional area, hence twice the current.

EE105 Fall 2011Lecture 6, Slide 6Prof. Salahuddin, UC Berkeley Small and Large Signal Models- Basic Concept

EE105 Fall 2011Lecture 6, Slide 7Prof. Salahuddin, UC Berkeley Example: Diode DC Bias Calculations V x =1 V

EE105 Fall 2011Lecture 6, Slide 8Prof. Salahuddin, UC Berkeley Constant-Voltage Diode Model for Large-Signal Analysis If V D < V D,on : The diode operates as an open circuit. If V D  V D,on : The diode operates as a constant voltage source with value V D,on.

EE105 Fall 2011Lecture 6, Slide 9Prof. Salahuddin, UC Berkeley Example: Diode DC Bias Calculations V x =1 V Say, V D =V 0 ND=10 17 NA=10 16

EE105 Fall 2011Lecture 6, Slide 10Prof. Salahuddin, UC Berkeley Large signal model: key points This example shows the simplicity provided by a constant-voltage model over an exponential model. Using an exponential model, iteration is needed to solve for current. Using a constant-voltage model, only linear equations need to be solved.

EE105 Fall 2011Lecture 6, Slide 11Prof. Salahuddin, UC Berkeley Small-Signal Analysis Small-signal analysis is performed at a DC bias point by perturbing the voltage by a small amount and observing the resulting linear current perturbation. – If two points on the I-V curve are very close, the curve in- between these points is well approximated by a straight line:

EE105 Fall 2011Lecture 6, Slide 12Prof. Salahuddin, UC Berkeley Diode Model for Small-Signal Analysis Since there is a linear relationship between the small-signal current and small-signal voltage of a diode, the diode can be viewed as a linear resistor when only small changes in voltage are of interest. Small-Signal Resistance (or Dynamic Resistance)

EE105 Fall 2011Lecture 6, Slide 13Prof. Salahuddin, UC Berkeley Diode Model Summary

EE105 Fall 2011Lecture 6, Slide 14Prof. Salahuddin, UC Berkeley Bi-Polar Junction Transistors

EE105 Fall 2011Lecture 6, Slide 15Prof. Salahuddin, UC Berkeley Reverse-Biased PN Junction as a Current Source PN junction diode current is ~independent of the reverse-bias voltage. It depends only on the rate at which minority carriers are introduced into the depletion region. Key Insight: We can increase the reverse current by injecting minority carriers near to the depletion region.

EE105 Fall 2011Lecture 6, Slide 16Prof. Salahuddin, UC Berkeley BJT Structure and Circuit Symbol A bipolar junction transistor consists of 2 PN junctions that form a sandwich of three doped semiconductor regions. The outer two regions are doped the same type; the middle region is doped the opposite type.

EE105 Fall 2011Lecture 6, Slide 17Prof. Salahuddin, UC Berkeley NPN BJT Operation (Qualitative) In the forward active mode of operation: The collector junction is reverse biased. The emitter junction is forward biased. current gain:

EE105 Fall 2011Lecture 6, Slide 18Prof. Salahuddin, UC Berkeley Base Current The base current consists of two components: 1)Injection of holes into the emitter, and 2)Recombination of holes with electrons injected from the emitter.

EE105 Fall 2011Lecture 6, Slide 19Prof. Salahuddin, UC Berkeley BJT Design Important features of a well-designed BJT (large  ): – Injected minority carriers do not recombine in the quasi-neutral base region.  – Emitter current is comprised almost entirely of carriers injected into the base (rather than carriers injected into the emitter). 

EE105 Fall 2011Lecture 6, Slide 20Prof. Salahuddin, UC Berkeley Carrier Transport in the Base Region The length of the quasi-neutral base region (W B = x 2 -x 1 ) is much smaller than the minority-carrier diffusion length, so that very few of the carriers injected (from the emitter) into the base recombine before they reach the collector-junction depletion region.  Minority-carrier diffusion current is ~constant in the quasi-neutral base The minority-carrier concentration at the edges of the collector- junction depletion region are ~0.

EE105 Fall 2011Lecture 6, Slide 21Prof. Salahuddin, UC Berkeley Minority-Carrier Diffusion in the Quasi-Neutral (P-type) Base Region “Long Base” L B ≡ diffusion length of minority carriers in the quasi-neutral base “Short Base” W B ≡ width of quasi-neutral base

EE105 Fall 2011Lecture 6, Slide 22Prof. Salahuddin, UC Berkeley Collector Current The equation above shows that the BJT is indeed a voltage-dependent current source; thus it can be used as an amplifier.

EE105 Fall 2011Lecture 6, Slide 23Prof. Salahuddin, UC Berkeley Common-Emitter Current Gain,  Assuming that no minority-carrier recombination occurs within the quasi-neutral base region: – The collector current is equal to the current due to minority-carrier injection from the emitter into the base: – The base current is equal to the current due to minority-carrier injection from the base into the (short) emitter: The current gain  can thus be expressed as a function of the BJT physical parameters:

EE105 Fall 2011Lecture 6, Slide 24Prof. Salahuddin, UC Berkeley Emitter Current Applying Kirchhoff’s Current Law to the BJT, we can easily find the emitter current.

EE105 Fall 2011Lecture 6, Slide 25Prof. Salahuddin, UC Berkeley Summary of BJT Currents