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SPRING 2003 EE40 MIDTERM 2
Problem 1: 15 Points Possible (1.5 Points for each question)
- A gallium atom would most likely be found in
a) n-type material
b) p-type material
- An arsenic atom which has 5 valence electrons shares 4 of the electrons to form bonds with neighbors, and retains (has) 1 leftover valence electron. Therefore, this arsenic atom is
a) negatively charged
b) positively charged
c) neutrally charged (no charge)
- When a diode is first formed (when p-type and n-type material are brought together) with no external voltage applied, electrons cross the p-n junction to fill holes on the other side. This is called
a) reverse breakdown
b) forward bias
c) drift
d) diffusion
- This movement of electrons creates many
a) negative ions in the n-type material
b) positive ions in the n-type material
c) holes in the n-type material
d) free electrons in the n-type material
- This movement of electrons across the p-n junction generally continues until
a) the potential drop across the p-n junction becomes too large for the electrons to cross
b) all the holes on the other side are filled
- The magnitude of the potential drop across the p-n junction is largest under
a) forward bias
b) reverse bias
c) open-circuit condition (diode left alone)
d) short-circuit condition (diode ends connected)
- The (average) strength of the chemical bonds keeping electrons in place is strongest in the
a) p-type material
b) n-type material
c) metal contacts
- When the diode is forward biased, electrons move
a) from n-type to p-type because the electrons' chemical attraction to holes overcomes the electric force keeping them from crossing
b) from p-type to n-type because the electrons' chemical attraction to holes overcomes the electric force keeping them from crossing
c) from n-type to p-type because the electromotive force created by the potential rise across the junction is stronger than the chemical bonds
d) from p-type to n-type because the electromotive force created by the potential rise across the junction is stronger than the chemical bonds
Problem 2: 15 Points Possible
For the circuit below, make a sketch of VO(t) using the ideal diode model, using the large-signal diode model, and using the small-signal diode model (5 Points possible for each).
Assume VF = 0.7 V and RD = 20 ohms
Be sure to include enough detail in your sketch (peaks, waveform shapes, etc) to let the grader know that your answer is completely correct!
Problem 3: 10 Points Possible
Consider the operational amplifier circuit below. Determine the upper bound on the input amplitude B that will keep the amplifier from hitting a rail.
Problem 5: 15 Points Possible + 2 Bonus Points Possible
A B C F
a) Use the sum-of-products method to write a Boolean function for F in terms of A, B, and C. ( Points Possible)
b) Draw the circuit for your function from part a) using only inverters and (multiple-input) NAND gates. (5 Points Possible)
c) Simplify your function from part a) to use fewer gates and/or fewer inputs to each gate. Use any method you like. Any correct simplification will result in full credit. 2 Bonus Points will be awarded for the simplification that uses the fewest total number of logic gates. (5 Points + 2 Bonus Points Possible)
Problem 6: 15 Points Possible
The inputs A, B, and C simultaneously transition to logic 1 at t=0 after being at logic 0 for a long time. Create a timing diagram showing the logic transitions for output F in the circuit below.
Assume that each gate has a propagation delay tp.
Alternate Easier Problem 7 for 5 Points Possible (10 Points will be lost)
Design a 1-bit analog to digital converter.
Given Vin < 2 V, design a circuit that will compare Vin to 1 V.
Vout should be logic 1 (4 V) if Vin > 1 V, and Vout should be logic 0 (0 V) if Vin < 1 V.
Don't be concerned about the operation of the circuit for Vin very close to 1 V.
