《半导体物理与器件》第四版答案第十章(共17页).doc

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1、精选优质文档-倾情为你奉上Chapter 10专心-专注-专业10.1 (a) p-type; inversion (b) p-type; depletion (c) p-type; accumulation (d) n-type; inversion_10.2(a) (i) V cm or m (ii) V cm or m(b) V so cm (i) V cm or m (ii) V cm or m_10.3(a) 1st approximation: Let V Then cm 2nd approximation: V Then cm(b) V V_10.4 p-type silicon

2、 (a) Aluminum gate We have V Then or V (b) polysilicon gate or V (c) polysilicon gate or V_10.5 V V_10.6(a) cm(b) Not possible - is always positive.(c) cm_10.7 From Problem 10.5, V (a) F/cm V(b) F/cm V_10.8(a) V V(b) F/cm (i) V (ii) V(c) V F/cm (i) V (ii) V_10.9 where V Then or V Now or We have or F

3、/cm So now C/cm or cm_ 10.10 V cm C/cm F/cm (a) n poly gate on p-type: V V(b) p poly gate on p-type: V V(c) Al gate on p-type: V V_10.11 V cm C/cm F/cm (a) n poly gate on n-type: V V(b) p poly gate on n-type: V V(c) Al gate on n-type: V V_10.12 V The surface potential is V We have V Now We obtain or

4、 cm Then or C/cm We also find or F/cm Then or V_10.13 F/cm C/cm By trial and error, let cm. Now V cm C/cm V Then Then VV_10.14 F/cm C/cm By trial and error, let cm Now V cm C/cm V Then Then V, which is within the specified value._10.15 We have F/cm C/cm By trial and error, let cm Now V cm C/cm V The

5、n V Then V V which meets the specification._10.16(a) V F/cm Now V(b) V cm C/cm Now or V_10.17 (a) We have n-type material under the gate, so where V Then or cmm (b) For an polysilicon gate, or V Now or C/cm We have C/cm We now find or V_10.18 (b) where V and V Then or V (c) For We find or cmm Now or

6、 C/cm Then or VV_10.19 Plot_10.20 Plot_10.21 Plot_10.22 Plot_10.23(a) For Hz (low freq), F/cm F/cm Now V cm Then F/cm (inv)F/cm(b) MHz (high freq), F/cm (unchanged) F/cm (unchanged) F/cm (unchanged) (inv)F/cm(c) V Now C/cm V_10.24(a) Hz (low freq), F/cm F/cm Now V cm Then F/cm (inv)F/cm(b) MHz (high

7、 freq), F/cm (unchanged) F/cm (unchanged) F/cm (unchanged) (inv)F/cm(c) V Now C/cm Then V_10.25 The amount of fixed oxide charge at x is C/cm By lever action, the effect of this oxide charge on the flatband voltage is If we add the effect at each point, we must integrate so that _10.26 (a) We have T

8、hen or or V (b) We have Now or or V (c) We find or Now which becomes Then or V_10.27 Sketch_10.28 Sketch_10.29 (b) or V (c) Apply V, V For V, n-side: at , then so for In the oxide, , so constant. From the boundary conditions, in the oxide In the p-region, at , then At , So that Since , then The pote

9、ntial is For zero bias, we can write where are the voltage drops across the n-region, the oxide, and the p-region, respectively. For the oxide: For the n-region: Arbitrarily, set at , then so that At , which is the voltage drop across the n-region. Because of symmetry, . Then for zero bias, we have

10、which can be written as or Solving for , we obtain If we apply a voltage , then replace by , so We find which yields cm Now or V We also find or V_10.30 (a) n-type (b) We have F/cm Also or cmnm (c) or which yields C/cmcm (d) which yields F/cm or pF_10.31 (a) Point 1: Inversion 2: Threshold 3: Deplet

11、ion 4: Flat-band 5: Accumulation_10.32 We have Now let , so For a p-type substrate, is a negative value, so we can write Using the definition of threshold voltage , we have At saturation which then makes equal to zero at the drain terminal._10.33(a) mA(b) mA(c) Same as (b), mA(d) mA_10.34(a) mA(b) m

12、A(c) mA(d) Same as (c), mA_10.35(a) (b) mA(c) mA_10.36(a) Assume biased in saturation region V Note: VV So the transistor is biased in the saturation region.(b) mA(c) or mA_10.37 F/cm A/V=1.111 mA/V(a) , V, V, mA V, V, mA V, V, mA V, V, mA(c) for V V, mA V, mA V, mA V, mA_10.38 F/cm A/V=0.961 mA/V(a

13、) , V, V mA V, V mA V, V mA V, V mA(c) for V V mA V mA V mA V mA_10.39(a) From Problem 10.37,mA/V For V, , V mA V, V mA V, V mA_10.40 Sketch_10.41 Sketch_10.42 We have so that Since , the transistor is always biased in the saturation region. Then where, from Problem 10.37, mA/Vand V Then (mA) 0 1 2

14、3 4 5 0 0.336 2.67 7.22 14.0 23.0_10.43 From Problem 10.38, mA/V For V, For V, For V, mA/V_10.44(a) mA/V(b) mA(c) mA_10.45 We find that V Now where or F/cm We are given . From the graph, for V, we have , then or or which yields cm/V-s_10.46 (a) or V (b) so which yields A/V (c) V so or A (d) or A_10.

15、47(a) F/cm (i) A/V or A/V (ii) (b) (i) A/V or A/V (ii) _10.48 From Problem 10.37, mA/V(a) so mA/V(b) so mA/V_10.49 From Problem 10.38, mA/V(a) or mA/V(b) or mA/V_10.50(a) Now F/cm Then V(b) V (i) cm C/cm V A/V or mA/V For , V For V(c) (i) For , V(ii) V, V V (iii) V, V V (iv) V, V V_10.51 V or V Now V_