put your final answers on this sheet and attach any...

5
EE 230 Name__________________________________________ Exam 2 – Nov. 5, 2015 Put your final answers on this sheet and attach any additional sheets behind. You must include your work to get full credit. Problem 1 A comparator transfer characteristic is shown at right. Sketch a circuit that would produce this characteristic. Be sure to specify the details of your circuit: V REF , R a , R b , and the output saturation levels. Note: There are many possible values of R a and R b , but the ratio R a /R b is specific. R a = ____________________ R b = ___________________ V REF = ____________________ V L+ = _________________________ V L– = __________________________ –8 V +8 V +6 V +2 V v i v o

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EE 230 Name__________________________________________Exam 2 – Nov. 5, 2015

Put your final answers on this sheet and attach any additional sheets behind. You must include your work to get full credit.

Problem 1

A comparator transfer characteristic is shown at right. Sketch a circuit that would produce this characteristic. Be sure to specify the details of your circuit: VREF, Ra, Rb, and the output saturation levels. Note: There are many possible values of Ra and Rb, but the ratio Ra/Rb is specific.

Ra = ____________________ Rb = ___________________ VREF = ____________________VL+ = _________________________ VL– = __________________________

–8 V

+8 V

+6 V+2 V

vi

vo

EE 230 Name__________________________________________Exam 2 – Nov. 5, 2015

Put your final answers on this sheet and attach any additional sheets behind. You must include your work to get full credit.

Problem 2

For the oscillator circuit shown at right, determine the Ra, Rb, and R and the high and low output limits for the op amp that will give the output and capacitor waveforms shown below.

Note: There are many possible values of Ra and Rb, but the ratio Ra/Rb is specific.

Ra = ____________________ Rb = _____________________ R = ____________________VL+ = _________________________ VL– = __________________________

10 V

vo

THTH = TL = 0.5 ms

TL0 V

7.5 V

vc

2.5 Vtt

+–+

–VREF

= 5 V

Ra

Rb

RC 0.47 µF

vo

vc

EE 230 Name__________________________________________Exam 2 – Nov. 5, 2015

Problem 3

Two identical non-inverting op-amp circuits are cascaded, as shown below. The op-amps each have gain-bandwidth of 1 MHz. What will be the corner frequency of the cascaded amp? (Recall: Corner frequency is the point at which the magnitude of the transfer function of the cascaded system is down by the square-root of 2 from its low-frequency value.)

f1 = _________________________________________

+–

4 k!1 k!

+–

4 k!1 k!

vivo

EE 230 Name__________________________________________Exam 2 – Nov. 5, 2015

Problem 4

An op-amp will be used in an inverting configuration to amplify a sinusoid, vi = 5 mV sin(ωt) The op-amp has an offset voltage of VOS = 7.5 mV and output voltage limits of VL+ = +10 V and VL– = –10 V. What is the largest value of (the magnitude of the) gain that the op amp can have without having the output hit the limits?

R2 / R1 (max) = ___________________________________

+–

+–

vi

R1

R2

vo

R3 = R1�R2

EE 230 Name__________________________________________Exam 2 – Nov. 5, 2015

Possibly useful information

Op amp non-ideal effects:

Bias current and offset voltage add DC voltage to the output.

Gain-bandwidth limitation: closed-loop Go·fcl = GBW of op amp

slew rate: for sinusoidal output:

Comparators:

Non-inverting:

Inverting:

Non-linear oscillators:

Non-inverting:

Inverting:

First-order filters:

low-pass:

high-pass:

vo =

�R2 � R3 � R2R3

R1

�Ibias +

�1+

R2R1

�Vos

SR =dvodt

��max ωVA < SR

VTL = VREF

�1+

RaRb

��

�RaRb

�VL+ ; VTH = VREF

�1+

RaRb

��

�RaRb

�VL�

VTL =VREF + Ra

Rb VL�

1+ RaRb

; VTH =VREF + Ra

Rb VL+

1+ RaRb

TL = RC�VTH � VTL��VL�

��

�TH = RC

�VTH � VTL

VL+

TL = RC ln

�VL� � VTHVL� � VTL

�TH = RC ln

�VL+ � VTLVL+ � VTH

Tlp (s) = Goωo

s + ωo=

Go1+ s

ωo

; Tlp (jω) =Go

1+ jωωo

;��Tlp

�� =Go�

1+�

ωωo

�2

Thp (s) = Gos

s + ωo=

Go1+ ωo

s; Thp (jω) =

Go1+ ωo

jω;��Thp

�� =Go�

1+�ωoω

�2