ece231s – introductory electronics midtermexams.skule.ca/exams/ece231h1_20151... · midterm...

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ECE231S – Introductory Electronics

Midterm

Lecturers – A. Liscidini, O. Trescases, B. Wang

Date – Thursday, Feb. 26, 2015

Duration: 110 minutes

1. Answers should be written in pen. Answers written in pencil may be considered ineligible for

remarking.

2. The marks for each question are indicated within brackets [ ]. Use the back side of sheets for

your rough work. Place your final answers in the boxes where given.

3. Show your work: answers without justification will not receive full marks!

4. Aids: calculator only. (Only non-programmable calculators are allowed).

Last Name: ________________________

First Name: ________________________

Student #: (write on the back of this cover page)

Lecture Section:

��LEC101 ��LEC102 ��LEC103 (Liscidini) (Trescases) (Wang)

Question 1 /12

Question 2 /5

Question 3 /8

Question 4 /6

Question 5 /4

Question 6 /10

Total /45

SOLUTIONS

Question 1 [12 marks]:

This question has four unrelated parts.

Part I. [3 marks] For each statement, circle TRUE or FALSE:

a) Mobile carriers in a semiconductor will drift according to a concentration gradient:

b) Leakage current in a PN junction diode increases at high temperature due to increased thermal generation of electron-hole pairs: c) The depletion region gets wider as the forward voltage is increased in a PN junction.

d) An n-type semiconductor can be converted to p-type by doping it with a sufficient concentration of boron atoms.

e) A p-type semiconductor has some free electrons in the conduction band.

t) Some diodes are intentionally designed to operate in the breakdown region.

Part II. The following circuit was used in Lab # 1.

R12 10k!)

.........

a) TRUE (:ALS:J

FALSE

c) TRUE e FALSE

FALSE

FALSE

a) [1 mark] Based on your observations during the lab, which of the passive components is physically the largest on the PCB (circle your answer)?

R" SW, 0 c, c, c,

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Note: very important to pay attention in the labs

Question 1 (cont.)

Part IV. [2 marks] How does the 'virtual short' concept apply in this circuit (clearly explain your reasoning)?

..rla., e 1\)+

,,('r

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h e k.. Vs) a "'-.ol tiDe,; VLCf Q

Note: Important to notice that the op amp terminals are reversed

Question 2 [5 marks]

The circuit shown below is used as a basic over-temperature protection alarm, where the diode, D, is used as a temperature sensor.

VDD = 3.3 V, R2 = 10 kQ, 1/ = 300 fAA.

[5 marks] Design the value of R/ so that the alarm is activated (ie: V a/arm = 3.3 V) when the diode temperature exceeds the trip-point of 75°C. An excerpt from the diode datasheet is shown below. Assume both op amps to be ideal. Clearly show your steps and reasoning.

[3 marks] Bonus: Based on your design, estimate the maximum percentage error in the temperature trip-point if the op amps have an offset voltage of up to +/- 10 mY.

1000m .... ...... ..

1u

10n

1n

10u l OOp ... ...1._ ... _ ... ...1._-'-_ ... ....1_ ... o 0.2 o 10 20 30 40 50 60 70 80

Forward voltage (V ) Reverse voltage (V)

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Note: Important to get the correct voltage at exactly 300 uA

Question 2 (cont.)

A \S

-::;.. C \ +- ) VD

y :. \fCD ,(«teL

---\ - \<:::

.I o ,b V

... -

+ h) (0:0 \I OD

(2.\

O,l./;Lb4 \C

Answer for Question 2:

Bonus:

Error = <j • 5" %

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( ! + i2.,) (VOJ, f Vu ) = f

V +

t YK Df aM..(' '2.

\I b l \J f -+ \J os J . I, - V (J S (

VJ) :=. o.S9:."12..V

Yt) - o'b :;: -(4,:L5M V

9.r% erro-r

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Question 4 [6 marks]:

Consider the circuit shown below. VB = 5V (DC source), R = 1kΩ.

a) [2 marks] For each diode, circle the operating region and determine the DC current and voltage. You may assume the Constant-Voltage-Drop model for the diodes, where VD = 0.7 V in the on-state.

D1: ON / OFF VD1 = ID1 =

D2: ON / OFF VD2 = ID2 =

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Question 4 (cont.)

b) [4 marks] Draw the small-signal circuit and evaluate the small-signal gain, Av = vout/vin.

Av = [V/V]

0.5

0.33

0.33

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Question 5 [4 marks]: An op amp with a gain-bandwidth product of 10 MHz and a slew-rate of 10 V/μs is used in a non-inverting configuration with a closed-loop gain of 6 dB. a) [2 marks] For a sinusoidal input with an amplitude of 100 mV, find the frequency that results an

output amplitude of 141 mV.

b) [2 marks] Under the conditions of part a), determine if the output is slew-rate limited.

Bonus: [3 marks] If the input is a step from 0 to Vstep, determine the maximum value of Vstep such that the op amp does not experience slew-rate limiting.

Answers for Question 5

Part a) Frequency = kHz

Part b) Slew-rate limiting? YES / NO

Bonus: Vstep = V

Question 6 [10 fSl 1- S

Consider the circuit shown below, where both op amps are ideal.

R = 1 kQ.

c

R

The measured bode plot for G(s) = vo(s) IvJ(s) is shown below.

IGI

o 0.1 100/

a) [3 marks] Calculate the value of c. 3ookHe--'h"j

-.. i+

Page 12 of 14

nF

Was marked according to the circuit as it is drawn (ie: not an inverting integrator)

Question 6 (cont.)

b) [4 marks] Accurately sketch the output voltage, vo(t) on the provided axis for the provided input waveform. What is the value of the output voltage after 2

5V

o

-5 V

v

c) [3 marks] Show how the circuit can be modified such that the closed-loop dc gain of G(s) is limited to 60 dB (indicate the component values), and sketch the new magnitude response of G(s) on the same bode plot.

- ArJ..

Page 13 of 14

-6v R P J) C G ,,) I +

R - 10<.:)0

95% of the class did not attempt this, and we decided to leave it out

d) [1 mark] What is the main motivation for limiting the DC gain in the integrator circuit?

-rW Q.lAy . y/6Y\ - t;c-(O ;:Pc.. eft }l.. (Aflt} S wW ch LSo notre> c h u...Q ·

Page 14 of 14

Note that this questions applies even if the circuit is not an ideal integrator - it would still saturate for any small offset voltage in the opamp

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