THE NEGATIVE-FEEDBACK LOOP - College of Engineering ... · I V ICVS 0 0 i to v v o/i i r m...

EE 323 -Feedback and stability of 29

FEEDBACK AND STABILITY

THE NEGATIVE-FEEDBACK LOOP

• xS = input signal • xOUT = AxIN • xF = “feedback” signal • xIN = xS – xF • linear feedback: xF = βxOUT • β=feedback factor ( constant)

xOUT = AxIN = A(xS – xF) xOUT = A(xS – βxOUT)

xIN XOUT A Σ A

β

xS xIN xOUT

xF

+

_ Feedback network

Output Signal source

Open loop Closed loop


•••• xOUT depends upon itself – a property intrinsic to the nature of a feedback path

xOUT(1+Aββββ) =AxS or �A1A

xx

AS

OUTfb ++++

========

•••• Afb = closed-loop gain (gain with feedback) •••• Aββββ>>>>>>>>1

�

1�A

AAfb ====≈≈≈≈

•••• Closed-loop gain, Afb • is independent of A in the limit Aββββ>>>>>>>>1, • depends only on the feedback factor ββββ.

•••• This feature is important. It allows Afb to be precisely set regardless of the exact value of A.


• Feedback network is generally made from passive (and easy-to-control) circuit elements and factors that affect A (component variations, temperature, and circuit non-linearity) become much less important to the closed-loop circuit.

•••• Worth the price of reducing gain from AOL to ACL.

Example: Non-inverting op-amp configuration


GENERAL REQUIRMENTS OF FEEDBACK CIRCUITS

� Signals at summing node must be the same type (i.e., all voltages or currents).

� The output, xOUT, needs not be of the same signal type as its input. � The amplification factor, A, can have dimension units:

� Av=Volt/Volt or Ai=Ampere/Ampere � Ar=Volt/Ampere or Ag=Ampere/Volt.

� Feedback function, ββββ, must have units reciprocal to those of A, (i.e., product Aββββ is dimensionless -- ensures xF is the same signal type as xS and xIN).

� In general, feedback network is made from passive components only and ββββ never exceeds unity.

� In the feedback loop, xF is subtracted from xS, making the feedback negative.


� If xF is added to xS at the summation node, the feedback becomes positive (oscillator, active filters.)

� Negative feedback benefits (desirable in amp. design): � Reducing amplifier non-linearity, � Improving input and output impedance, � Extending amplifier bandwidth, � Stabilizing gain, and reducing amplifier sensitivity to transistor

parameters. .


FOUR TYPES OF NEGATIVE FEEDBACK

� Four basic amplifier types


a. A voltage amplifier with gain Av

b. A current amplifier with gain Ai

c. A transconductance amplifier or voltage-to-current converter. The amplification factor, Ag, or

gm = iOUT/vIN, (A/V or conductance).

d. A transresistance amplifier or current-to-voltage converter. The amplification factor, Ar or

rm, = vOUT/iIN (V/A or resistance).


� The four types of negative feedback In Out Circuit zin zout Converts Ratio Symbol Type of Amplifier

V V VCVS ∞∞∞∞ 0 - vo/vi Av Voltage amplifier I V ICVS 0 0 i to v vo/ii rm Trans-resistance amplifier V I VCIS ∞∞∞∞ ∞∞∞∞ v to i io/vi gm Trans-conductance amplifier I I ICIS 0 ∞∞∞∞ - io/ii Ai Current amplifier

vi vo Avvi HIGH

LOW

VCVS

ii

vo riii LOW

LOW

ICVS

io

gmvi HIGH HIGH

VCIS

~ ~

io

Aivi LOW HIGH

ii

ICIS

vi


∗∗∗∗∗∗∗∗∗∗∗∗ VOLTAGE-CONTROLLED VOLTAGE SOURCE (VCVS)

• High input impedance • Low output impedance • Stiff voltage source

•••• Feedback fraction:

•••• Closed loop gain: •••• Loop gain:

• Error between ideal and exact values:

211

out1R

RRR

vv

�++++

========

1

2

1

21

OL

OL

OL

OLCL R

R1R

RR�

1�A

A�A1

AA ++++====++++====≈≈≈≈≈≈≈≈++++

====

vin vout + _

R1

+VCC

-VEE R2

�A1%100Error%OL++++

====

�A1Gain OLCL ++++====


•••• Impedances:

•••• Output voltage: vin=Avout � Negative feedback: Stabilizes voltage gain, Increases input impedance,

Decreases output impedance, Reduces nonlinear distortion of the amplified signal.

a. Gain stability:

• The gain is stabilized because depends only on the external resistances (i.e., can be precision resistors).

• The gain stability depends on having a low percent error between the ideal and the exact closed-loop voltage gains.

• The smaller the percent error, the better the stability. • The worst-case error of closed-loop voltage gain occurs when the open-loop

voltage gain AOL is minimum.

�A1%100errorMaximum%(min)OL++++

====

�A1RZR)�A1(Z

OL

outoutinOLin ++++

====++++====


b. Nonlinear distortion:

• non-linear distortion will occur with large signals. • input/output response becomes non-linear. • Nonlinear also produces harmonics of the input signal.

• Total harmonic distortion:

%100voltagelFundamenta

voltageharmonicTotalTHD ====


• Negative feedback reduces harmonic distortion (closed-loop harmonic distortion):

�A1THDTHD

OL

OLCL ++++

====

• Quantity 1+AOLββββ has a curative effect. When it is large, it reduces the harmonic distortion to negligible levels, (ex.., high-fidelity sound in audio amplifier system).

Example 19-1, 19-2, 19-3, 19-4 (page 667)


*** CURRENT-CONTROLLED VOLTAGE SOURCE (ICVS) ***

• Low input impedance, Low output impedance. • Stiff voltage source from a current input. • Trans-resistance (rm) (i.e., output voltage is proportional to the current by a

resistance).

• R2 can be selected to have different conversion factors (trans-resistances).

•••• Input and output impedances:

Example: inverting amplifier, 19-5, 19-6 (page 674)

2inOL

OL2inout Ri

A1ARiv ====++++

====

OL

out)CL(out

OL

2)CL(in A1

RZA1

Rz++++

====++++

====

iin vout _

+

+VCC

-VEE

R2


∗∗∗∗∗∗∗∗∗∗∗∗ VOLTAGE-CONTROLLED CURRENT SOURCE (VCIS)

• Transconductance, gm, (i.e., 1/R) • Both input and output impedances are high • Stiff current source.

• Input and output impedances:

Example 19-7 (page 677)

1minm

1

inout

OL

211

inout

R1gwherevg

Rvi

A)RR(R

vi

============

++++++++====

1OL)CL(out

inOL)CL(in

R)A1(Z

R)�A1(Z

++++====

++++====

vin iout + _

R1

+VCC

-VEE RL=R2


∗∗∗∗∗∗∗∗∗∗∗∗ CURRENT-CONTROLLED CURRENT SOURCE (ICIS)

• Low input impedance, High output impedance . • Stiff current source. • Current gain factor Ai.

•••• Input and output impedances:

Example 19-8 (page 678)

iin

iout _ +

R1

+VCC

-VEE RL

R2

1RR

RAR)RR(AA

1

2

1OLL

21OLi ++++≅≅≅≅

++++++++====

21

1

OL

2)CL(in RR

R�where

�A1RZ

++++====

++++==== 1OL)CL(out R)A1(Z ++++====


∗∗∗∗∗∗∗∗∗∗∗∗ BANDWIDTH

• What is bandwidth? • Negative feedback increases the bandwidth of an amplifier. • Because of the roll-off in open-loop voltage gain means less voltage is fed

back, which produces more input voltage as a compensation. • Closed-loop cutoff frequency is higher than the open-loop cutoff frequency.

• The closed-loop cutoff frequency:

•••• Gain Bandwidth Product

“Gain bandwidth product is constant”

unity)CL(2CL ffA ====

• GBP = funity = constant for a given op-amp. • Trade off gain to bandwidth in design. • Less gain, more bandwidth (more gain, less bandwidth).

CL

unity)CL(2 A

ff ====

frequencyGainGBP ⋅⋅⋅⋅====

CLCLOLOL fAfA ====


• Unity frequency determines GBP of the op-amp. • Higher unity frequency op-amp may be needed for specific application

(requires both high gain and high bandwidth.)

Example 19-9, 19-10, 19-11, 19-12, 19-13 (page 682)


FEEDBACK LOOP STABILITY

Stability of the negative feedback loop must be examined to verify that unwanted oscillations will not occur.

• Output of a linear system experiences a relative phase shift of –90O if the driving frequency increases beyond one of the poles of the system function.

• System with three or more poles, a frequency will exist at which the phase shift exceeds 180O.


At some frequency, ωωωω180, the –180O phase shift will change an otherwise negative feedback loop into a positive feedback loop.

The response of the feedback loop at ωωωω180: �A)1(1

A)1(v180

180out −−−−++++

−−−−====


• If A180ββββ=1, the denominator becomes 0 and the output becomes infinite (even vin = 0).

• Such a condition is equivalent to an oscillation at the frequency ωωωω180. • Less stringent inequality A180ββββ≥≥≥≥1 also leads to oscillation at ωωωω180.

• Note: in practice, the saturation limits of the op-amp limit the magnitude of oscillation.


I. FEEDBACK LOOP COMPENSATION

� Use frequency compensation to prevent unwanted oscillations (at ωωωω180). � Alter the open loop response so that the stability condition is met: A180ββββ <<<< 1 � Internal design (stability condition is met up to some maximum value of ββββ,

i.e., ββββ=1). � The LM741 is stable under all negative feedback conditions. The value of A180

of LM741 is less than unity.


� Adding external components to the feedback loop: • Evaluate the feedback loop for stability • If the feedback loop is unstable, external compensation must be added. � External compensation is sometimes preferred over internal compensation

because the latter limits the GBP of the feedback loop.


II. EVALUATION OF STABILITY CONDITION

Gain margin and phase margin to determine stability:

�A1�)�j(A1inargMGain 180�180−−−−====−−−−====−−−−

*** Gain margin must be positive (i.e., A180ββββ <<<< 1).

)�Aarg(180)180()�Aarg(inargMPhase)PM()PM( �)�j(

00�)�j( ++++====−−−−−−−−====−−−−

1�A )�j( ==== at ωωωωPM.

*** Negative phase margin: �A )�j( is greater than unity at ωωωω180 (i.e., circuit unstable.)

• One margin passes the stability test, the other will also. • Design a feedback loop with excess gain or phase margin to ensure stability.


Example: Transfer function open-loop frequency response:

where A0=106, ωωωω1=10rad/sec, ωωωω2=ωωωω3=106rad/sec.

Since ωωωω1 <<<<<<<< ωωωω2, the dominant pole of A(jωωωω) ≈≈≈≈ ωωωω1 . a. How large can the feedback parameter ββββ become before instability results? Assume ββββ

is not a function of frequency. b. Design a non-inverting amplifier that meets the stability conditions.

Solution:

Since ββββ is not frequency dependent, ωωωω180 occurs at the frequency where the angle of the transfer function is at –180O, that is where:

0

3

1801

2

1801

1

1801)�j( 180)

�

�(tan)

�

�(tan)

�

�(tan)Aarg( −−−−====−−−−−−−−−−−−==== −−−−−−−−−−−−

solving for ωωωω: ωωωω180 ≅≅≅≅ 106 rad/sec

)�

�j1)(�

�j1)(�

�j1(

AA

321

0)�j(

++++++++++++====


The magnitude of A(jωωωω) at ωωωω180 as follows:

5)2)(10(

10

)10

10j1)(10

10j1)(10

10j1(

10A25

6

6

6

6

66

6180 ========

++++++++++++

====

Stability of the feedback loop requires:

1�A )�j( ≤≤≤≤ �� 2.051

� ====≤≤≤≤

For the circuit to be stable, the closed-loop gain must therefore meet the minimum condition:

5�

1vv

in

out ≥≥≥≥====

vin vout + _

R1

+VCC

-VEE R2=4R1


EXTERNAL COMPENSATION

A compensation network is added to an under-compensated negative feedback loop.

Example:

a. Show that the minimum allowed ββββ that ensures stability is 0.001.

b. To what closed-loop gain does ββββ correspond?

c. Design a compensation network that will stabilize the op-amp in a non-inverting amplifier with a gain of 5 (i.e., 14dB).


Solution:

• f180 = 3.2*106 Hz (from the angle portion of the Bode plot.)

• A180=60dB=1000

• 1�A )�j( ≤≤≤≤ �� ββββ < 10-3.

• A non-inverting amplifier gain of 5 (ββββ=0.2) will be unstable.

• Compensation network adds a pole to the op-amp open-loop response at a frequency below f180,

• The open-loop gain at f180 will be reduced so that the stability condition can be met.

• If the pole is located well below f90, it will add an additional –90O phase shift at f90, bringing the total phase shift at f90 to –180O.

• The original f90 before compensation will become the new f180 after compensation.

• 0�Aor1�A dBdB9090 ====++++====

• The uncompensated op-amp has a gain magnitude of 90dB at f90.

• If an amplifier with closed-loop gain of 5 (i.e., 14dB) is to be made, then ββββ=1/5 (i.e., - 14dB) and A90 must be reduced from 90dB to 14dB (i.e., 76dB).

• This can be achieved with a pole at fc =50.7Hz. How????????


o Above fc, the pole at fc will

increase the roll-off by –20dB/decade.

Hz7.501031.6

Hz102.310ff

decades8.3decade/dB20dB76ff

3

5

8.390

c

c90

====⋅⋅⋅⋅

⋅⋅⋅⋅========

====−−−−

−−−−====−−−−

o The compensation pole at fc introduced by a simple RC filter

o sec/rad319f�2CR1

cc========

THE NEGATIVE-FEEDBACK LOOP - College of Engineering ... · I V ICVS 0 0 i to v v o/i i r m...

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