{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw11_su10_sol - ECE 442 Summer 2010 HW 11 Solutions 6.5 iDn...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
i Dn = i Dp 1 2 μ n C ox ( W L ) n V 2 ovn = 1 2 μ p C ox ( W L ) p V 2 ovp g mn = g mp g m = 2 I D V ov V ovn = V ovp ( W/L ) p ( W/L ) n = μ n μ p = 460 160 = 2 . 875 I 0 = I REF = 50 μA, L = 0 . 5 μm, W = 5 μm, V t = 0 . 5 V, k 0 n = 250 μA/V 2 I 0 = I D = 1 2 k 0 n W L V 2 ov 50 = 1 2 × 250 × 5 0 . 5 ( V GS - 0 . 5) 2 V GS = 0 . 7 V or 0 . 3 V V GS = 0 . 3 V < V t V GS = 0 . 7 V I D = I RE = V DD - V GS R 1 . 8 - 0 . 7 R = 0 . 05 R = 22 k Ω V GS V 0 V DS 2 V DS = V GS - V t = 0 . 7 - 0 . 5 = 0 . 2 V V Dmin = 0 . 2 V. r o = V A I D = V 0 A × L I D = 20 × 0 . 5 0 . 05 = 200 k Ω Δ I 0 = Δ V o r o = 1 200 k = 5 μA Δ I 0 = 5 μA
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
I REF = I S e V BE /V T V BE = V T ln I REF I S I REF = 10 μA V BE = 0 . 025 ln 10 × 10 - 6 10 - 15 = 0 . 576 V I REF = 10 mA V BE = 0 . 025 ln 10 × 10 - 3 10 - 15 = 0 . 748 V 10 μA I REF 10 mA 0 . 576 V V BE 0 . 748 V β I B I 0 = I REF : 10 μA I 0 10 mA I 0 = I REF 1 1+2 I REF = 10 μA I 0 = 10 μ 1 + 2 100 = 9 . 8 μA I REF = 0 . 1 mA I 0 = 0 . 1 m 1 + 2 100 = 0 . 098 mA I REF = 1 mA I 0 = 1 m 1 + 2 100 = 0 . 98 mA I REF = 10 mA I 0 = 10 m 1 + 2 70 = 9 . 72 mA β → ∞ ⇒ I 0 = I REF = 2 mA r o 2 = V A 2 I 0 = 90 2 m = 45 k Ω r o 2 = Δ V 0 Δ I 0 10 - 1 Δ I 0 = 45 k Δ I 0 = 0 . 2 mA Δ I 0 I 0 = 0 . 2 2 = 10% change
Background image of page 2
I
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}