{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Electrical Engineering 105 - Fall 1993 - Howe - Midterm 2

# Electrical Engineering 105 - Fall 1993 - Howe - Midterm 2

This preview shows pages 1–3. Sign up to view the full content.

EECS 105, Midterm #2, Fall 1993 EECS 105, Fall 1993 Midterm #2 Professor R. T. Howe Ground Rules: Closed book and notes; one formula sheet (both sides) Do all work on exam pages Answers accurate to within 10% will receive full credit Default bipolar transistoe parameters: npn : (beta) n = 100, V An = 50 V, I Sn = 10 -16 A. pnp : (beta) p = 50, V Ap = 25 V, I Sp = 10 -16 A. Default MOS transistor parameters: NMOS : (mu) n C ox = 50 (mu) AV -2 , (lambda) n = 0.02V -1 , V Tn = 1 V. PMOS : (mu) p C ox = 25 (mu) AV -2 , (lambda) p = 0.02V -1 , V Tp = -1 V. Problem #1. Matched Complementary Bipolar Transistor Design [12 points] The cross sections, minority carrier concentrations, and circuit schematics are shown for matched npn and pnp vertical BJTs, operated in the forward-active region. file:///C|/Documents%20and%20Settings/Jason%20Raft...20-%20Fall%201993%20-%20Howe%20-%20Midterm%202.htm (1 of 5)1/27/2007 4:43:34 PM

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

View Full Document
EECS 105, Midterm #2, Fall 1993 Given : all doping levels are matched and the emitter areas are identical N dE (npn) = N aE (pnp) N aB (npn) = N dB (pnp) N dC (npn) = N aC (pnp) A E (npn) = A E (pnp) Given
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

Electrical Engineering 105 - Fall 1993 - Howe - Midterm 2

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online