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SEMICONDUCTOR ELECTRONICS. situation when two hydrogen atoms are brought together. interaction between the electrostatic fields of the atoms split each.

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Presentation on theme: "SEMICONDUCTOR ELECTRONICS. situation when two hydrogen atoms are brought together. interaction between the electrostatic fields of the atoms split each."— Presentation transcript:

1 SEMICONDUCTOR ELECTRONICS

2 situation when two hydrogen atoms are brought together. interaction between the electrostatic fields of the atoms split each energy level into two, gives one level slightly higher than before and another one slightly lower Band theory of solids

3 1 ATOM2 ATOMSMANY ATOMS BAND GAP

4 Insulators, Semiconductors and Metals VB CB EGEG EGEG VB

5 The Bond Theory 1. Ionic bonds: electrostatic attraction between the ions. Atoms donate/accept electrons to become +vely or -vely charged ions. 2. Covalent Bonds: Atoms joined by sharing valence electrons. 3. Metallic bonds: exhibited by electrons with single valence electrons, e.g. Cu, Na, Ag, Au. substances find minimum energy configuration when they pool their valence electrons. The electrons are no longer tied to specific sites (atoms) but are unlocalised and are free to travel throughout the metal forming an electron cloud.

6 Intrinsic and Extrinsic semiconductors intrinsic Extrinsic (with P) Extrinsic with Al (Gp 4 elements)

7 III DIODE CIRCUITS AND CHARACTERISTICS

8 P-n junction: 1. depletion layer established on both sides of the junction. 2.barrier (junction) potential is developed across the junction. 3.Formation of junction and diffusion capacitances. DIODE CIRCUITS AND CHARACTERISTICS

9 P-n junction (a) depletion region + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - pn + + + - - - - - + + Depletion layer with fixed ions Free (mobile charges)

10 (b) Junction or barrier potential 0 np + + + ------ ++++++ - - - Depletion layer VBVB

11 e ( I e )O ( I h ) + + + + + - - - - - VAVA 0 I V B - V A q(V B – V A ) qV A ECEC EFEF EVEV Forward biased p-n junction.

12 Forward I-V characteristics. Ge Si I V VγVγ

13 Reverse Biased p-n Junction. + + + ---- ++++ - - - V A + V B q(V B + V A ) qV A O e

14 Reverse I-V characteristics Forward voltage Reverse voltage I (μA)

15 Diode Law For ideal diode: i D = 0, v D  0, v D = 0, i D > 0, With eqn

16 Diode Law 1. 2V A = 0, I = 0 3 4 1 2 3

17 Rectifier circuits i RL v vi

18 BIPOLAR JUNCTION TRANSISTORS

19 Bipolar j. transistor NPN NPN PNP PNP C B E B CE

20 Biasing PN junctions. B-E junction forward biased by V EE -ve terminal of V EE connected to n-side flow of i d across due to flow of majority carriers (electrons) from the N-type emitter become minority carriers in the base VE E B C E Forward biased B-E (input) junction

21 C-B reverse biased by V CC +ve of V CC connected to N-type collector depletion region at the junction widens current flowing from B-E due to minority electrons crossing the junction from p-type base. constitute flow of reverse current in the junction. VCC B E C Reverse biased C-B (output) junction

22 VEE VC C ICIC IB Simultaneous biasing B is ground (0V) E is -ve wrt B & C is +ve wrt B e- flow constitutes the dominant type (in npn) I E = I C + I B IE e npn

23 I C sum of the injected & thermally generated minority carriers. if VEE is left open, & C- B has normal reverse bias Then I CBO will flow Hence total I is I C = I C(Inj) + I CBO or (I C = I CBO -  I E ) Where I C(Inj) is I C due to carriers injected into the base. IE IB VEEVCC IC pnp

24 portion of I C that survives after passing through the B to become I C

25 6/21/2016 CB connection IE IB IC VCB = Output voltage B EC VBE = Input Voltage - - + + IE IB IC VBC = Output voltage B EC VEB = Input Voltage + + - - NPNPNP SCI 2010/2011simiyuj@uonbi.ac.ke

26 CB Input Characteristics F-B diode (input is across the forward- biased B-E junction) greater the value of V CB, the more readily minority carriers in the base are swept through the B-E junction. IE (mA) VCB = 0 VCB = 10V VCB = 25V VBE (V) Increasing output bias (V BE = Φ(VCB, IE))

27 CB Output characteristics Active: I E = 0, I C = I CO I C rises with V CB & I C is slightly less than I E since I C = -  I E and  ≈ 1 Saturation V CB is +ve and the junction is forward biased I.e C-B & E-B are forward biased satura tion Active region Cut off IE ICO (I C = Φ(V CB, I E ))


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