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1 1 EE105 – Fall 2014 Microelectronic Devices and Circuits Prof. Ming C. Wu 511 Sutardja Dai Hall (SDH) 2 Ideal vs Non-ideal Op Amps Ideal Op Amp A = R id = R o = 0 Non-ideal Op Amp A < R id < R o > 0 Other non-ideal characteristics: Finite common-mode rejection ratio (CMRR) Output voltage is not just proportional to the difference input voltage but also their absolute values Common-mode input resistance dc error sources (offset voltage, input bias current, offset current) Output voltage, slew rate, maximum current limits

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### Transcript of Lecture04-Non-ideal Op Amps (Feedback circuit).pptx

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EE105 – Fall 2014 Microelectronic Devices and Circuits

Prof. Ming C. Wu

[email protected]

511 Sutardja Dai Hall (SDH)

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Ideal vs Non-ideal Op Amps

Ideal Op Amp A =∞ Rid =∞ Ro = 0

Non-ideal Op Amp A <∞ Rid <∞ Ro > 0

Other non-ideal characteristics: •  Finite common-mode rejection ratio

(CMRR) •  Output voltage is not just proportional to

the difference input voltage but also their absolute values

•  Common-mode input resistance •  dc error sources (offset voltage, input

bias current, offset current) •  Output voltage, slew rate, maximum

current limits

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Classic Feedback Systems

•  A = transfer function of open-loop amplifier or open-loop gain.

•  β = transfer function of feedback network.

•  T = loop gain = Aβ –  For negative feedback: T(s) > 0 –  For positive feedback: T(s) < 0

vd = vi − v f

vo = Avd

v f = βvo

Av =vo

vi

=A

1+ Aβ=1β

Aβ1+ Aβ

#

\$ %

&

' ( = Av

Ideal T1+ T

#

\$ %

&

' (

4

Gain Error and Fractional Gain Error

•  Gain Error: –  GE = (ideal gain) - (actual gain)

•  For non-inverting amplifier,

•  Fractional or percentage error:

GE = 1β− 1β

T1+T

\$

%

& &

'

(

) ) = 1β

11+T

\$

%

& &

'

(

) )

FGE =GE

AvIdeal =

11+ T

#

\$ %

&

' (

=1

1+ T≅1T

for T >>1

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Gain Error Example: Noninverting Amplifier

•  Problem: Find actual gain and gain error for an amplifier

•  Given data: Ideal closed-loop gain of 200 (46 dB), open-loop gain of op amp is 10,000 (80 dB).

•  Approach: Amplifier is designed to give ideal gain and deviations from ideal case are determined.

•  Note: R1 and R2 are not normally designed to compensate for finite open-loop gain of amplifier.

•  Analysis:

Av = AvIdeal T

1+ T

"

# \$

%

& ' with T = Aβ =104 1

200"

# \$

%

& ' = 50

AvIdeal =

= 200 Av = 200 501+ 50"

# \$

%

& ' =196 FGE =

200 −196200

= 0.02

Note : FGE ≅1T

=150

= 0.02

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Gain of Noninverting Amplifier

For i− = 0, v1 =R1

R1 + R2vo = βvo

β =R1

R1 + R2

vo = Avid = A vi − v1( ) = A vi −βvo( )

Av =vovi=

A1+ Aβ

=1β

Aβ1+ Aβ"

#\$

%

&'=

T1+T"

#\$

%

&'

T = Aβ is called the loop gainFor T >>1

Av → AvIdeal =

1β=1+ R2

R1

vid = vi − v1 = vi −βvo

vid = vi −T

1+Tvi =

vi1+T

Although no longer zero, vid is smallfor large loop gain T.Feedback factor

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Output Resistance (Both Noninverting and Inverting Amplifiers)

The output terminal is driven by test source vx and current ix is calculated to determine output resistance (all independent sources are turned off). Note that the equivalent circuit is same for both inverting and non-inverting amplifiers.

Rout =vx

ix

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Output Resistance (cont’d)

Analysis : ix = io + i2

io =vx − Avid

Ro

and i2 =vx

R2 + R1

Also, vid = −v1 and v1 =R1

R2 + R1vx = βvx

1

Rout

=ix

vx

=1+ Aβ

Ro

+1

R2 + R1

Rout =Ro

1+ TR2 + R1( )

Thus, shunt feedback reduces Rout .

Typically, Ro 1+ T( ) << R2 + R1( ) and Rout ≅Ro

1+ T. For infinite T, Rout = 0.

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Nonideal Op Amps Open-loop Gain: Design Example

•  Problem: Design non-inverting amplifier and find the required open-loop gain

•  Given Data: Av = 35 dB, Rout ≤ 0.2 Ω, Ro = 250 Ω

•  Analysis:

AvIdeal =10

35dB

20dB = 56.2 | β =1

AvIdeal =

156.2

Rout =Ro

1+ Aβ≤ 0.2 Ω

∴ A ≥1β

Ro

Rout

−1(

) *

+

, - | A ≥ 56.2 250

0.2−1

(

) *

+

, - = 7.02x104 or 96.9 dB

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Input Resistance of Noninverting Amplifier

Test voltage source vx is applied to input and current ix is calculated:

ix =vx − v1

Rid

| v1 = i1R1 ≅ i2R1

Assuming i− << i2 gives i1 ≅ i2

v1 =R1

R1 + R2

vo = βvo

v1 = β Avid( ) = Aβ vx − v1( )

v1 =Aβ

1+ Aβvx =

T1+T

vx

∴ ix =vx − v1

Rid=vx −

T1+T

vx

Rid=

vxRid 1+T( )

Rin = Rid 1+T( )

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Gain of Inverting Amplifier

Let i1 be the current going from the input source through R1 and i2 be the current going through R2 into the amplifier output.

i1 =vi − −vid( )

R1

and i2 =−vid( )− voR2

i2 = i1 and vo = Avid

After some algebra,

Av =vovi= −

R2

R1

"

#\$

%

&'Aβ

1+ Aβ= Av

Ideal T1+T"

#\$

%

&'

with β = R1

R1 + R2

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Input Resistance of Inverting Amplifier

R1 removed Rin =

vxix=ixR1 + v−

ix= R1 +

v−ix

i 1= i− + i2 =v1

Rid+v1 − voR2

=v1

Rid+v1 + Av1

R2

∴ G1 =i1v1

=1Rid

+1+TR2

Rin = R1 + Rid || R2

1+ A!

"#

\$

%& ≅ R1 +

R2

1+ AFor large A, Rin ≅ R1

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Nonideal Amplifier Summary Inverting and Noninverting Amplifier

β =R1

R1 + R2 and T = Aβ

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Feedback Amplifier Categories Voltage Amplifiers - Series-Shunt Feedback

•  A voltage amplifier should have a high input resistance to measure the desired voltage and a low output resistance to drive the external load. To achieve the desired behavior, the input ports of the amplifier and feedback network are connected in series, and the output ports are connected in parallel (shunt).

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Feedback Amplifier Categories Transresistance Amplifiers - Shunt-Shunt Feedback

•  A transresistance amplifier converts an input current to an output voltage. Thus it should have a low input resistance to sink the desired current and a low output resistance to drive the external load. To achieve the desired behavior, the input ports of the amplifier and feedback network are connected in parallel (shunt), and the output ports are connected in parallel (shunt).

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Feedback Amplifier Categories Current Amplifiers - Shunt-Series Feedback

•  A current amplifier should provide a low resistance current sink at the input and a high resistance current source at its output. The input ports of the amplifier and feedback network are connected in parallel, and the output ports are connected in series.

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Feedback Amplifier Categories Transconductance Amplifiers - Series-Series Feedback

•  The transconductance amplifier converts an input voltage to an output current. This amplifier should have a high input resistance and a high output resistance. The input ports and the output ports of the amplifier and feedback networks are connected in series.

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Feedback Amplifier Category Summary