02. analog try yourself_2016_ec

Electronics EngineeringAnalog Circuits

WORKBOOKWORKBOOKWORKBOOKWORKBOOKWORKBOOK

2016

Detailed Explanations ofTry Yourself Questions

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Diode Circuits and VoltageRegulator Circuits1

T1.T1.T1.T1.T1. Solution:Solution:Solution:Solution:Solution:

+ –C1

5 v D2

D1 C210 v–

+R

Negativeclamper

Negative Peakdetector

5 sin tω

Vc1 = 5v VC2 = –10 v

T2.T2.T2.T2.T2. (a)(a)(a)(a)(a)In this question we need to determine whichdiode is on and which diode is OFF, clearly diodeD3 is OFF because if it is on then current fromcurrent source will flow from n to p terminal ofthe diode D3 and this is not possible, hence D3

is OFF.Applying the same concept, we can say diodeD2 is also OFF.Diode D1 is on because it is forced by thebattery of 10 V.

T3.T3.T3.T3.T3. (c)(c)(c)(c)(c)Assume D1on, D2off, D3on

0.6 V 0.6 V

10 V

– 10 V

10 kΩ

10 kΩD2 off

10 kΩ

– 20 V

ID1 =

= 1.47 mA

ID2 = 0

ID3 =

= 0.94 mA

T4.T4.T4.T4.T4. (d)(d)(d)(d)(d)When E = 1.0 V D1 on D20ff

I1 =

Ω + Ω = 0.033 mA

3300 Ω

5600 Ω

0.7 V

1 VI1

When E = 1.4 V D1on, D2off

© Copyright www.madeeasypublications.org

3Workbook

3300 Ω

5600 Ω

0.7 V

1. 4 VI1

I1 =

=Ω + Ω

When E = 2 V, D1on, D2off

3300 Ω

5600 Ω

0.7 V

2 VI1

I1 =

=Ω + Ω

T5.T5.T5.T5.T5. Solution:Solution:Solution:Solution:Solution:Here there are two different hire constant involvedDuring T1 vo = v1e

–t/RC

At t = T1 = T

vo =

=

=

⎡ ⎤⎢ ⎥⎣ ⎦

=

α⎡ ⎤⎢ ⎥⎣ ⎦

During T2

Vo =

=

=

⎡ ⎤⎢ ⎥⎣ ⎦ = v2[1 – α]

+ = 20

α + = 20

α+ = 20 ...(i)

+ = 20

v2 – αv2 + v1 = 20v1 + v2 – αv2 = 20 ...(ii)

13.33 v

– 6.67 vFrom (i) and (ii)

α

= α

From equation with α neglected2v2 + v2 = 20

v2 =

= –6.67 v

v1 = +13.33

T6.T6.T6.T6.T6. (a)(a)(a)(a)(a)

– 10 V

10 V

5 mA

5 kΩ

+

–0.7 V

Vo1

+

–25 V

Vo1 = 0.7+ 25 V + 10 V = 35.7 V

5 mA

V02

5 kΩ

–10V

+10V

+

–+

–

25 V

20 V

Vo2 = –10 V – 25 V – 20 V = –55 V

T7.T7.T7.T7.T7. Solution:Solution:Solution:Solution:Solution:Io = 5 A μ

2 mΩ

IR

I

10 v

4 Electronics Engineering • Analog Circuits


I × 7 = Io × 11

I =

μ × = 7.85 μA

I = Io + IR

IR = I – Io = 7.85 μA – 5 μA= 2.85 μA

R =

I =

μ = 350 kΩ

T8.T8.T8.T8.T8. (a)(a)(a)(a)(a)Vin – (–0.7) = Vout

Vin + 0.7 = 1.1Vin = 1.1 – 0.7 = 0.4 v

T9.T9.T9.T9.T9. (b)(b)(b)(b)(b)Positive half cycle

D1on, D2off

vo =

× +

=

50 v

115 v2

17.9° 180°

Negative half cycleD1off, D2 off

V0 = 50v


Transistor Biasing and ThermalStabilization2

T1.T1.T1.T1.T1. Solution:Solution:Solution:Solution:Solution:

15 V

Q1

5 kΩ

3 kΩ

100 k

50 k

VTH =

×+

= 5 V

RTH = 100 k 50 k = 33.33 kΩ

5 kΩ

5 V

15 V

3 kΩ

33.33 k

IB1

+0.7

VTH – IB1RTH – VBE – IERE = 0

IB1 = ( )

+ + β

=

+ ×IC1 = βIB1 = 1.28 mAIE1 = (1 + β)IB1 = 1.29 mA

VC1 = 15 – IC1 RC1

= 15 – 1.28 mA × 5 = 8.6 V

2 kΩ5 kΩ

15 V15 V

2.7 kΩRE

+0.7–

Q2Q1

VC1 = 8.6 V

VE2 = VC1 + VEB2

= 8.6 + 0.7 = 9.3 V

IE2 =

=

= 2.85 mA

IC2 = β ×+ β

I

=

× = 2.82 mA

VC2 = IC2 × RC2

= 2.82 mA × 2.7 kΩ = 7.62 V



T2.T2.T2.T2.T2. Solution:Solution:Solution:Solution:Solution:VCC – IC1R2 – VCE1 = 06 – 1.5 mA × R2 – 3 = 0

R2 = 2 kΩ

IB1 =

=βI

= 0.01 mA

IB2 will be equal to IB1 as there is no change inR1

IC2 = β2 IB2

= 200 × 0.01 mA = 2 mAVCE2 = VCC – IC2R2

= 6 – 2 mA × 2kΩ = 2 VThe new operating point is

Q(2 V, 2 mA)

T3.T3.T3.T3.T3. Solution:Solution:Solution:Solution:Solution:Assume Q is in active region

50 kΩ

10V

2 kΩ

3 kΩ

5 V

IB active

IBactive =

+ ×

=

= 17 μA

ICactive = β IBactive

= 100 × 17 μA = 1.7 mAKVL in output loop

10 V

2 Ω

3 kΩ

VCEsat = 0.2 VVCsat = 2.5 V

VE = 2 k × 1.7 mAΩ

ICsat

VCsat = 3.6 VVCEsat = 0.2 V

VE = 2 kΩ × 1.7 mA = 3.4 V

ICsat =

Ω = 2.13 mA

ICsat > ICactive Active region

T4.T4.T4.T4.T4. Solution:Solution:Solution:Solution:Solution:Since I1 = 0.2 mA and I2 = 0.3 mA. So n1 = 2and n2 = 3, because we need to find minimumnumber of BJT required.


Small Signal BJT Amplifiers3

T1.T1.T1.T1.T1. (b)(b)(b)(b)(b)

VS 50 k 2 kΩ

VO

1 kΩ10 k

12 V

β = 100VA = ∞

(IE)Q =

Ω = 1.3 mA

DC Circuit

VTH 2 kΩ

1 kΩ

12 V

+

–VEC

RTH

VTH =

×+

= 10 v

RTh = 50 k ⎥⎥ 10 k = 8.33 kΩ

KVL in output loop12 – IE × 1 k – VEC – 2k × IE = 012 – 1.3 mA × 1k – VEC – 2 k × 1.3 mA = 0

VEC = 8.1 vGiven range 1 ≤ VEC ≤ 11 v

ΔVEC = 11 – 8.1 = 2.9 vThe voltage swing is

2ΔVEC = 2 × 2.9 = 5.8 v

T2.T2.T2.T2.T2. (d)(d)(d)(d)(d)

270 Ω

VTh

Vi RL

gm = 2 mS ; ro = 250 kΩ

rπ = βre =

β = = 50 kΩ

VTH RL

ro

270 Ω E C

Vπ

IB

IE

Vi

+ –g vm π

+

–rπ

+

–

B



Vπ =

π

π×

+i

=

i = – 0.994 vi

VTh + rogmvπ – vπ= 0VTh = –rogmvπ – vπ

= –(1 + gmro) (–0.994vi)= – (1 + 2mS × 250 kΩ) × 0.994vi

VTh = 497.9 vi

T3.T3.T3.T3.T3. (a)(a)(a)(a)(a)

Q2

Q1

Vo

Vi

B1

vi Vπ1 rπ1+

–g Vm1 1π

C1

E2

C2B2

Vπ2

+

–rπ2

Vo

g Vm2 2 π

Vi = Vπ1

Vo = – Vπ2

KCL at the output node

π

π+ = gm2 Vπ2

π+ = gm2Vo

π

⎡ ⎤+⎢ ⎥

⎣ ⎦= – gm1Vi

Av =

π

π=

+i

T4.T4.T4.T4.T4. (d)(d)(d)(d)(d)

Vi

220 k

12 V

3.9 k

C1C2

DC circuit

+ –VBE

220 k

IB

12 V

3.9 k (1 + )β IB

IB =( )

+ β +

= 0.0163 mAIE = (1 + β)IB = 1.97 mA

re =

II

=

= 13.15 Ω

BVi

βre

1.578 kΩ220 k1 – Av

Vo

3.9 kβIB

C

220 kAVAv – 1

= 220 k

Vo = – (220 k ⎥⎥ 3.9 k) βIB

Av =

=

= –291.41


9Workbook

Zi = i

iI =

β

= 0.752 k ⎥⎥ 1.578 k= 0.509 kΩ = 509.4 Ω

T5.T5.T5.T5.T5. (b)(b)(b)(b)(b)

hie =

Δ = = ΩΔI

hfe =

Δ = =Δ

II

T6.T6.T6.T6.T6. (?)(?)(?)(?)(?)Since collector current at operating point is givenin the question we do not need to solve the circuitat DC.The parameters of BJT are

gm =

= =

I

ro =

= ∞I

re =

α = Ω

The circuit for AC analysis,

iS

VS V0

4 k = Ω RLRS=100 kΩ

3 kΩ 25.6 kΩ

4 kΩ

10.4 kΩ

0.56 V117.71 Ω

+– V1

Assuming, base current is zero,V1 = –Vs

and Vo = –0.56 V1 × (4 kΩ || 4 kΩ)

= 112.30


Field Effect TransistorAmplifiers4

T1.T1.T1.T1.T1. (a)(a)(a)(a)(a)10 v

v1

v2

M1

M2

M3

Ad VD = VG ∴ we conclude that each MOSFETis in saturation.

ID = kn1 (VGS – VT)2

MOSFET M1

ID = kn1 (VGS1 – VT)2

VGS1 = 10 – 5 = 5 v

0.5 mA = 36μ ×

⎛ ⎞⎜ ⎟⎝ ⎠ × (5 – 1)2

⎛ ⎞⎜ ⎟⎝ ⎠ = 1.73

MOSFET M2


0.5 mA = 36μ ×

⎛ ⎞⎜ ⎟⎝ ⎠ 2 (3 – 1)2

⎛ ⎞⎜ ⎟⎝ ⎠ = 6.94

MOSFET M3


0.5 mA = 36μ ×

⎛ ⎞⎜ ⎟⎝ ⎠ 3 (2 – 1)2

⎛ ⎞⎜ ⎟⎝ ⎠ = 27.8

T2.T2.T2.T2.T2. (c)(c)(c)(c)(c)If VTH = 0.4 vPMOS in depletion mode

VS = 1.5 V VD = 0

1.5 V

VG = 0.5 V

0.5 V+–

+–

VSD > VSG + VTH → current saturation

VSD < VSG + VTH → Triode region

VSD = VS – VG = 1.5 – 0.5 = 1 v

VSD = VS – VD = 1.5 – 0 = 1.5 v1.5 > 1 V + 0.4 current saturation region

+–

+–

VS = 0.9 V VD = 0.9 V

0.9 V VG = 0 0.9 V

PMOS in depletion mode.VSD = VS – VD = 0.9 – 0.9 = 0


11Workbook

VSG = VS – VG = 0.9 – 0 = 0.90 < 0.9 + 0.4 triode region

T3.T3.T3.T3.T3. (b)(b)(b)(b)(b)3 v

vD1

v2

M1

M2

I1

Given VTH = 1 vSo MOSFET is an n channel enhancementMOSFET in both transistors

VD = VG

M1 & M2 are in current saturation

3 – VDS1 – VDS2 = 0

VDS1 + VDS2 = 3 V

VDS2 =

= 1.5 v (VDS1 = VDS2)

VGS2 = VDS2 = 1.5 v = V2

VGS1 = VDS1 = 1.5 v

I1 = ID1 = ID2

= ( )

μ −

= ( )

× × − μ = 7.5 μ A

T4.T4.T4.T4.T4. (b)(b)(b)(b)(b)It is common drain amplifier.

Av =

+=

Ω+ Ω

= 0.95

gm = 4.75 m

gm = 2 kn (VGS – VT)

= 2 kn

⎛ ⎞+ −⎜ ⎟

⎝ ⎠

I

gm = I

gm = 2

⎛ ⎞× μ ⎜ ⎟⎝ ⎠I

= 47

T5.T5.T5.T5.T5. (a)(a)(a)(a)(a)VDD

iD

v0

M1

M2

vGSD

rOL (load)

rOLg VmD gsvi

D

rOD

G

Vgs

SV0 = gm Vgs rOD ⎥⎥ rOL

Vi = VgsV0 = –gm vi (rOD⎥⎥ rOL)

Av =

i= – gmD (rOD⎥⎥ rOL)

gmD = i= × = 0.874 mA/v

Since λD = λL

rOD = rOL =

=λ ×i = 500 kΩ

Av = – 0.894 (500 k ⎥⎥ 500 k) = – 224

T6.T6.T6.T6.T6. (a)(a)(a)(a)(a)DC circuit

–5 v +5 v

100 kΩ 4 kΩ1 mA

ID

ID = IS = I mAn channel enhancement MOSFETAssume Q in saturation.

ID = kn (VGS – VT)2

I = 1 (VGS – 1)2

VGS = 2 V

gm = 2 kn (VGS – VT)= 2 mA/V



T7.T7.T7.T7.T7. (b)(b)(b)(b)(b)In the previous part we calculatedtransconductance of the device (gm = 2 mS).Now, drawing the circuit for AC analysis

V0

50 kΩ

VS

g Vm gs

4 kΩ 10 kΩ

500 Ω

Vgs+–i1

From the figure,V0 = – gm VGS (RD⎥⎥ RL)

gm Vgs = –i1

⇒

i= RD || RL

⇒ Vo = 0.283 sinωt V

T8.T8.T8.T8.T8. (b)(b)(b)(b)(b)

CMOSV

VV

0 = +

(PMOS)

i

T

VV

V

0 = +

(NMOS)

i

T

Vt

Amplifier

3

T9.T9.T9.T9.T9. (a)(a)(a)(a)(a)

v0

VGS2 VGS1

Iin

I mA

IG = 0

⎛ ⎞ =⎜ ⎟⎝ ⎠

3

⎛ ⎞ =⎜ ⎟⎝ ⎠

⎛ ⎞ =⎜ ⎟⎝ ⎠

1

2 I mA

Current Mirror

As

⎛ ⎞⎜ ⎟⎝ ⎠ = 2

Iin = 2 × I mA= 2 mA


BJTs & FETs at High Frequencies5

T1.T1.T1.T1.T1. (a)(a)(a)(a)(a)

To calculate the

we will draw the small

signal model of the two circuit (we will useT-model of the device).

re

RC

V0

CE

Amplifier-1

g Vm be

Re

Vbe

+

–

T-Model

re

RC

V0

Amplifier-2

g Vm be

Re

Vbe

+

–

T-Model

CE

Since β = ∞ and IE = 1 mA so re = 25 Ω.

So for =

where, Req is equivalent resistance acrosscapacitance CE.

Req = Re + re in Amplifier 1and Req = (Re ||re) in Amplifier 2

So,

=

+

=

||

So,

=

Ω Ω=

+ Ω + Ω

=

+

= 2.55 × 10–3

T2.T2.T2.T2.T2. (a)(a)(a)(a)(a)

To find fH and fL we draw the small signal model:The small signal paramters of the BJT are

gm =

= =I

re = 25.0 ΩFrom the circuit we can see that the capacitance(C = 10 µF) has high pass characteristic so it



decide fL and capacitance (C = 15 pF) has lowpass characteristic so it decide fH.To find out fL ⇒

fL =

×

Here, C is 10 µF and Req is the equivalentresistance seen by the capacitance.Lets draw the small signal model to find Req:

0.1 kΩ C = 10 Fμ

VS re = 25 kΩ

0.5 kΩ

5 kΩ

10 kΩ

V0

g Vm be

T-Model

Rib

40 kΩ5.7 kΩ

Rib = (re + 0.5 kΩ)[When emitter resistance is seen through

bas it get multiplied by (β + 1)]

So, Rib = 53.025 kΩand Req = 0.1 kΩ + (53.025 || 40 || 5.7) kΩNow at high frequency the capacitance(C = 15 pF) will play its role. The small signalmodel for high frequency will be

0.5 kΩ

5 kΩ

C = 15 pF

V0

g Vm be

40 kΩ5.7 kΩ

10 kΩ

re =40 kΩ

0.1 kΩ

So, fH =

π

=

−π × × × ×

= 3.183 MHzSo, fH – fL = 3.18 MHz


Multistage Amplifiers6

T1.T1.T1.T1.T1. (d)(d)(d)(d)(d)Given that ID = 0.4 mA and

= 0.8 mA/V2

= 0.8 mA/V2

So

= I = 1.13 mA/V

and =

=

The gain of the cascade connection will be ! − ×

gain = –1.13 mA/V × 2.5 kΩ = –2.85


Feedback Amplifiers7

T4.T4.T4.T4.T4. (b)(b)(b)(b)(b)

Given that β >>

So the voltage gain

≈

+ β β

From the circuit we see that feedback factor βis

+=

=

+

So, β =

Voltage gain = –5 (because theamplifier is an inverting amplifier)

T5.T5.T5.T5.T5. (b)(b)(b)(b)(b)The feedback element is Rf it samples voltageand mix current so shunt-shunt feedback.

T1T1T1T1T1 (a)(a)(a)(a)(a)The overall forward gain is 1000 and close loopgain is 100. Thus, β = 0.009.Now, when gain of each stage increase by 10%then overall forward gain will be 1331 and usingthe previous value of β the close loop will be102.55.⇒ Close loop Voltage gain increase by 2.55%.

T2T2T2T2T2 It is voltage shunt

IS

–

+

90 Ω

Rf

v0

vf

+

–IfRf

β =

=

I

T3T3T3T3T3 It is voltage series ckt can be called as seriesshunt feedback.


Oscillators8

T2.T2.T2.T2.T2. (a)(a)(a)(a)(a)

We need to find !""

x

Vx

C

C

R

R Vy

A

Va B

Nodal analysis at node A

# # #

−−+ +x = 0

Nodal analysis at node-B

#

−+ = 0

From abouve two equations

"

⎡ ⎤+ + −⎢ ⎥⎣ ⎦

= Vx

So, "

x=

+ +

T3.T3.T3.T3.T3. (a)(a)(a)(a)(a)

Now the value of

in the given circuit in same

as "

x

T1.T1.T1.T1.T1. (a)(a)(a)(a)(a)Consider the following circuit

A

2 R 2 L

RL

+

–

+

–

Vf Vo

First of all we will find the feedback factor

β =

$

$ $

⎛ ⎞ω= ⎜ ⎟ω + + ω⎝ ⎠

=

$

ω⎛ ⎞− −⎜ ⎟⎝ ⎠ω

For Barkhausen criteria we needAβ = 1

and imaginary part of Aβ should be equal to 0

So,

ω−ω

= 0

So, ω =

and f =

π



So, β =

+ +

=

$

$

ω− ω + ω

For Barkhausen criteria, we needAβ = 1 ∠0°

So, phase of β = 0So imaginary part of β = 0So 1 – ω2 C2 R2 = 0

ω =

T4.T4.T4.T4.T4. (a)(a)(a)(a)(a)The output can be ±12 V only,when output is 12 V then

R1

–

+

C

R2

P

10 kΩ

2 kΩ

10 kΩ

Vo

So, Vp = 6 Vwhen output is –12 V then

R1

–

+

C

R2

P

10 kΩ

2 kΩ

10 kΩ

Vo

So, Vp = –10 V


Power Amplifiers9

T1.T1.T1.T1.T1. (a)(a)(a)(a)(a)Solving the given question at DC, assuming both the devices are in forward active region.

Q2

Q1

+9 V

12 kΩ

43 kΩ

9.1 kΩ

–9 V

24 Ωk

100 Ωk

β = 80

IIE2

IC2

–9 V

IC1

IE1

loop 1

IB1

IB2

loop 2

So KVL in loop-1

Ω + + ΩI I = 9

# Ω + ΩI I = 8.3

⇒ I = 4.06 µA

So, I =

=I

Now in loop-2

− Ω + Ω + − −I I I = 0

So,

Ω −I =

Ω + ΩI I



⇒ I = 2.006 µA

⇒ I = 0.184 mA and

I = 0.186 mA

So, now we can find = 0.186 mA and

= 7.85 V

So, Q-points will be (325 µA, 7.16 V) and (184 µV, 7.85 V).

T2.T2.T2.T2.T2. (c)(c)(c)(c)(c)The DC power = VCC × ICQ = 13 V × 5 mA = 65 mW

Now, at DC inductor act as short circuitSo, VCEQ = 8 V

PAC =

$

× × =

So, maximum efficiency = % &'(

)'(*+( × = ×

=

,

× =


Operational Amplifiers10

T1.T1.T1.T1.T1. (c)(c)(c)(c)(c)

when diode is ON it is replaced by short circuitand the circuit can be drawn as shown below:

V0

–

+

0.5 kΩ

1 kΩ

VS

10 V

1 kΩ

Vo = –0.5 Vs – 5when diode is OFF then it is replaced by opencircuit and circuit can be drawn as shown below

v0

–

+

1 kΩ

VS

1 kΩ

vi=

Vo = –Vs

Thus, in 1 phase the slope of the transfercharacteristic should be –1 and in another phaseit should be –1/2. Hence, option (c) is correct.

T2.T2.T2.T2.T2. (b)(b)(b)(b)(b)output of op-amp 1

–R

V1

V2+

–

+Vin R

R

R

It is connected to schmitt trigger (invertingmode) → clockwise.But inverting amplifier + inverting schmitt trigger→ anticlockwise.

12

6–6

T3.T3.T3.T3.T3. (b)(b)(b)(b)(b)

Rif = i

+ β=

i

β Ab >> 1

v0

–

+

20 kΩ

Vf

If = 10 kΩ

vin

–+

2 kΩ



voltage shunt

v0

–

+

20 kΩ Vf

vin –

+

2 kΩ

10 kΩIf

10 kΩ

β =

= –1

β =I

=

−

β =

Rif =i

β =

×

=

× ×

Rif = 1 kΩ

T4.T4.T4.T4.T4. (b)(b)(b)(b)(b)Redrawing the circuit by replacing amplifier withits block diagram from the given propertiesRi = ∞ ; R0 = 0 ; voltage gain = AV

vin vin

AV

v0 vin= AV

iin

Rf

iin =

−

iin =

−=

[ ]

− −

Rin =i

=

−

T5.T5.T5.T5.T5. (b)(b)(b)(b)(b)

–

+

–

+

+–vin

10 kΩ

10 kΩ 10 kΩ

10 kΩ

Rf

RL

io

AB

C

i1i2

i3D

E

i3

From the circuit,VE = io RL

VE = VA (Virtual short concept)i1 = i2 = i3

If we apply KVL between node B and C,∴ VB = VC (Virtual short concept)

i1 = i2 = i3 =

Ω

VC – VD = i3 × 10 kΩ =

and VA – VB = i1 × 10 kΩ =

∴ VB = VC

⇒ VD – VE = –vin

∵ io =

−

T6.T6.T6.T6.T6. (c)(c)(c)(c)(c)If diode or BJT it is logarthmic amplifier.If MOSFET is kept in feedback then it is squareroot amplifier.

T7.T7.T7.T7.T7. (a)(a)(a)(a)(a)In the given circuit the op-amp diodecombination form a super diode and thiscomplicated question can be simplified byreplacing the op-amp diode combination by asingle ideal diode as shown below.


23Workbook

RL

C

vin–

+Vo

Now this circuit is a simple clamper, thus answeris option (a).

T8.T8.T8.T8.T8. (c)(c)(c)(c)(c)It is inverting op-amp and there is no feedbackinthe ckt op amp will saturate to ± Vsat.

T9.T9.T9.T9.T9. (d)(d)(d)(d)(d)

V0 (t) = ( )

− ∫

=

− × ×× μ

At t = 2 mS

=

− × ×× μ

=

−−× ×

× 5 × 2 ×1 0–3

= – 0.5

2 5 7 10 12 150

v to( )

t S(m )–0.5

–1.0

–1.5

T10.T10.T10.T10.T10. (a)(a)(a)(a)(a)From the figure we can see that when input ispositive then diode is off and op-amp works inopen-loop with output equal to +Vsat.When the input is negative then the diode turnon and it get replaced by a 0.7 V battery, sonow output is equal to Vin – 0.7.So option (a) is correct.


Function Generators, PLL &Wave Shaping Circuits11

T1.T1.T1.T1.T1. (c)(c)(c)(c)(c)From the Hysteresis voltage transfercharacteristic we can see that the Schimitttrigger has to be non-inverting.The saturation level is ±10 V and the upper andlower threshold are +3 V and –3 V respectively,so option (c) is correct according to allrequirements.

T2.T2.T2.T2.T2. (b)(b)(b)(b)(b)The saturation level of output are ±10 V.When output is –10 V then threshold voltage is–5 V so

–

+

ViVo

1

2 kΩ

R

Threshold voltage is

×

+

5 =

+

So, R1 = 2 kΩSimilarly when output is +10 V then threshold is+8 V so

–

+

ViVo

2 kΩR2

So,

×+

= s

20 = sR2 + 164 = sR2

R2 = 0.5 kΩ

T3.T3.T3.T3.T3. (c)(c)(c)(c)(c)First op-amp form a high pass filter with cut-offfrequency equal to 79.7 Hz.Second op-amp form a low pass filter with cut-offfrequency equal to 318.30 Hz. So the seriescombination of low pass filter and high passfilter form a band pass filter.


25Workbook

T4.T4.T4.T4.T4. (d)(d)(d)(d)(d)

I1

I2

Q

Q

I Q

Q

M1 M2

Vi OutputC

B

A

Vi

t (ns)

t (ns)

t (ns)

t (ns)

A

B

C

40 43

43

433

3

83

86

02. analog try yourself_2016_ec

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