c8 – electronic amplifiers. amplifiers withopamp.. electronic...electronic amplifiers modeling the...

ELECTRONIC DEVICESAssist. prof. Laura-Nicoleta IVANCIU, Ph.D.

C8 – Electronic amplifiers. Amplifiers with OpAmp.

2Laura-Nicoleta IVANCIU, Electronic devices

Contents

Electronic amplifiers Types of supply Power transfer and power balance Types of amplifiers VTC Modeling the voltage amplifier Amplifier performances Frequency response

Amplifiers with OpAmp Non-inverting amplifier Inverting amplifier



Electronic amplifiers



Amplifier = active three-port network that delivers an output signal xo(t) (voltage or current) with the same shape as the input signal xi(t) and can provide more power, on an adequate load.

)()( tAxtx io =

Linear circuit: x0 proportional with xi

A0 non-inverting

A – amplification, gain



Types of supply


Single source supply Two source supply (symmetric differential)

The supply is provided by dc voltage and/or current sources. Voltage sources are most commonly used.



Power transfer and power balance


The average power of the output signal is greater than the average power of the input signal:

Pout > PinThe excess of the output power is taken from the supply sources.

Psupply + Pin = Pout + Pdissip

Psupply ≈ Pout + Pdissip

η =Pout/Psupply

!A step up transformer is not an amplifier!

Efficiency:



Types of amplifiers


Based on the types of input/output signals (voltage/current):

voltage amplifier Av = vo/vI - dimensionless

current amplifier Ai = io/iI - dimensionless

transconductance amplifier Ai/v = io/vI – [S], [mS]

transresistance amplifier Av/i = vo/iI – [Ω], [kΩ]



VTC


( )OHOLOv

OH

v

OLI

VVvA

VA

Vv

;

;;

∈

∈

amplification (active) region:

VOL = -VPS ; VOH = +VPS

rail-to-rail OpAmp: ( )PSPSO VVv +−∈ ;

general-purpose OpAamp V)2...V1;V2...V1( −++−∈ PSPSO VVv

ideal amplifier:

voltage-to-voltage, non-inverting amplifier, symmetric differential supply



VTC


Signal transfer

Note: if the input signal is low enough for the amplifier to work in a linear region around the OP: small signal approximation



Modeling the voltage amplifier


two-port networks: only the behavior of the input and output ports is explicitly taken into account, and the input-output signal transfer

valid regardless of the internal complexity of the amplifiers valid in the bandpass frequency domain

Linear controlled sources

active two-port network only one finite, non-zero parameter: forward transfer parameter (gain) the output signal is controlled by the input signal pseudo-sources

VCVS (voltage controlled voltage source)

vO = avvi





av – open circuit gainRi – draws current from viRo – deteriorates vo in the presence of loadi

ov v

va =

s

ov v

vA =

voL

L

is

iv aRR

RRR

RA++

=

Ideal amplifier?

Amplifier model

Amplifier model connected in a circuit





voL

L

is

iv aRR

RRR

RA++

=

s

ov v

vA =

Av is closer to the open circuit gain av when the voltage losses at the input (across Rs) and at the output (across Ro) are reduced.

Ri >> Rs – the source voltage is transferred to the input

Ro



Amplifier performances


Gain (forward transfer factor) Analyze the circuit, by using circuit theorems and equation (Kirchhoff, Ohm, Millman) Express the output signal as a function of the input signal, then compute the gain

Input resistance

i

ii i

vR =



Amplifier performances


Output resistance Set the input signal source to zero Connect a test source at the output

test

testo i

vR =

sc,o

open,oo i

vR =

open circuit short-circuit

#1

#2



Frequency response


analyze the equivalent model of the amplifier including capacitive components too, by means of their complex impedances: 1/jωC

complex transfer function for the gain:

input and output impedances are now complex

)()(

)(ωω

ωjvjv

jAi

o=

)( ωjZi ( )ωjZo

Band-pass amplifier

Why does the gain decrease at low/high frequencies?



Frequency response


internal capacitances (parasitic) of active devices => Av @ f

coupling/decoupling capacitors (tens of µF) => Av @ f (zero dc gain)

capacitive coupling at input

capacitive coupling of two amplifier stages

Direct coupling amplifier (LPF)

I. Utilization as comparator, in switching modevO ∈ {VOL ; VOH}

vD > 0, vO→ +∞, vO limited by the positive supply vO = VOH ≈ +VPSvD< 0, vO→ -∞, vO limited by the negative supply vO = VOL ≈ -VPS

II. Utilization as amplifiervO ∈ (VOL ; VOH)

It is mandatory that vD = 0, so then vO = a·vD = ∞·0 - indeterminationvD is kept at 0 by means of external components (R) arranged in a negative feedback (NF) configuration


Amplifiers with OpAmp



vO = avD = ∞·vD

C6 + C7





NI

NF automatically keeps vD at zero

↓↑↑↑= − DODD vvvvv ,,,0

vO = avD

NI II Amplifier

vI ground non-inverting

ground vI inverting

vI1 vI2 differential


Non-inverting amplifier

C8 – Electronic amplifiers. Amplifiers with OpAmp. Amplifiers with OpAmp

OvRRRv

21

1

+=−

021

1 =+

−=−= −+ OID vRRRvvvv

OI vRRRv

21

1

+= 1

21RR

vvA

I

Ov +==




Alternative method for computing Av

i

i

gain is set only by the ratio of two resistors (external components) the gain value: precise and stable the gain is independent of the OpAmp, it is not influenced by the technological

spread of the OpAmp’s parameters

0=Dv

Ivv =+ Ivv =

−

1

21RR

vv

AI

Ov +==The same current goes through R1 and R2

Direct consequences of NF for an OpAmp with very high intrinsic gain (a → ∞ for ideal op-amp):




Input and output resistances

0=∞

== open

sc

open O

O

Oo

vi

vRvI sees an open circuit, so Ri = ∞

Computed on the equivalent model




Example

vO(t) for triangular vI(t), 3 V amplitude, zero dc component




Adjustable gain

1

21R

PRAmaxv

++=

PRRA

minv ++=

1

21

1

21max R

RAv +=

0min

=vA




Voltage follower

total (full) NF no voltage gain (Av = 1) infinite current gain (Ai = ∞)

Voltage followers are used as a buffer stages between a source (or the output of a circuit) with high RO (can only supply low current) to a low RL (needs high current) – impedance matching.

IO vv = the output voltage follows the input voltage


Inverting amplifier


0=+v

OI vRRRv

RRRv

21

1

21

2

++

+=−

0021

1

21

2 =+

−+

−=−= −+ OID vRRRv

RRRvvv

1

2

RR

vvA

I

Ov −==


Inverting amplifier


Alternative method for computing Av

21 ii =1

10

Rvi I −=

21 Rv

Rv OI −=

22

0R

vi O

−=

1

2

RR

vv

AI

Ov −==

0=+v0=−v

virtual ground

−+ = vv=>

0=DvNF =>


Inverting amplifier


Input and output resistances

Noninverting amplifier: Ri→∞

Inverting amplifier: Ri→ finite (units, tens of kΩ)

1RRi =

0=oR

The input source “sees” only R1 (the inverting input is virtual ground)


Inverting amplifier


ExampleRi, Ro, AvvI range for active regionequivalent model

k101 == RRi0=oR

1010

100

1

2 −=−=−=RRAv

vI range: )V2.1;V2.1( +−

R1 = 10 K, R2 = 100K,

supply ±12 V


Inverting amplifier


Voltage follower

IO vv −=

1−==I

Ov v

vA

RRi =


Design example


Design an inverting amplifier with Ri > 8 kΩ and |Av| adjustable in the range [10, 18].

101

2min

==RRAv

181

2max

=+

=R

PRAv

12 10RR = 12 18RPR =+

What next?


Design example



12 10RR = 12 18RPR =+

kΩ81 ≥= RRi kΩ101 =RChoose

kΩ10010102 =⋅=RkΩ801001018 =−⋅=P

P = 100 kΩ will be used. Keeping R2 = 100 kΩ will result in:

kΩ1,1118

100100182

1 =+

=+

=PRR R2 = 100 kΩ P = 100kΩ

Verification: 1.9min

=vA 18max =vA Acceptable?

Solution 1


Design example



12 10RR =

Select

Verification: 18max

=vA

What if we began w/ P = 10kΩ?

Solution 2Ω= k100P

12 10RR =

12 18k100 RR =+Ω= k5.121RΩ= k1252R

10min

=vA

Ω>Ω== k8k5,12RRi

12 18RPR =+


High gain and high input resistance


OPTIONAL

1RRi =

10201.01

10101

1010−=

⋅⋅

++

−=vA

same current through R1 and R21 same voltage across R3 and R21

+

+−==

31

2221

1

2221

RRRR

RRR

vvA

I

Ov

For

k1.0k,10k10k,1

322

211

====

RRRR

k1=iR


SummaryThe little black bug (OpAmp) is also able to make: Amplifiers with OpAmp

Non-inverting amplifiers Inverting amplifiers

Next week: Summing and differential amplifiers with OpAmp.

To do: Homework 6


Slide Number 1Slide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18 Alternative method for computing AvSlide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24 Alternative method for computing AvSlide Number 26Slide Number 27Slide Number 28Slide Number 32

c8 – electronic amplifiers. amplifiers withopamp.. electronic...electronic amplifiers modeling the...

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