the negative-feedback loop - u of s engineering 323 -feedback and stability page 16 of 27 ***...

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EE 323 -Feedback and stability Page 1 of 27 FEEDBACK AND STABILITY THE NEGATIVE-FEEDBACK LOOP x S = input signal x OUT = Ax IN x F = “feedback” signal x IN = x S – x F linear feedback: x F = βx OUT β=feedback factor ( constant) x OUT = Ax IN = A(x S – x F ) x OUT = A(x S βx OUT ) x IN X OUT A Σ A β x S x IN x OUT x F + _ Feedback network Output Signal source Open loop Closed loop

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EE 323 -Feedback and stability Page 1 of 27

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

EE 323 -Feedback and stability Page 2 of 27

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

xOUT(1+Aβ) =AxS or A1A

xx

AS

OUTfb +

==

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

1AAAfb =≈

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

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

EE 323 -Feedback and stability Page 3 of 27

• 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

EE 323 -Feedback and stability Page 4 of 27

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:

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

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

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

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

EE 323 -Feedback and stability Page 5 of 27

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

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

parameters. .

EE 323 -Feedback and stability Page 6 of 27

FOUR TYPES OF NEGATIVE FEEDBACK

Ø Four basic amplifier types

EE 323 -Feedback and stability Page 7 of 27

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).

EE 323 -Feedback and stability Page 8 of 27

Ø 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

EE 323 -Feedback and stability Page 9 of 27

∗∗∗ 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

+==

12

121

OL

OL

OL

OLCL R

R1R

RR1AA

A1AA +=+=≈≈

+=

vin vout + _

R1

+VCC

-VEE R2

A1%100Error%OL+

=

A1Gain OLCL +=

EE 323 -Feedback and stability Page 10 of 27

• Impedances:

• Output voltage: vin=Avout v 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 +

=+=

EE 323 -Feedback and stability Page 11 of 27

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 =

EE 323 -Feedback and stability Page 12 of 27

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

A1THDTHD

OL

OLCL +

=

• Quantity 1+AOLβ 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)

EE 323 -Feedback and stability Page 13 of 27

∗∗∗ 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

EE 323 -Feedback and stability Page 14 of 27

∗∗∗ 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

EE 323 -Feedback and stability Page 15 of 27

∗∗∗ 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

RwhereA1RZ

+=

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

EE 323 -Feedback and stability Page 16 of 27

∗∗∗ 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). • Unity frequency determines GBP of the op-amp.

CL

unity)CL(2 A

ff =

frequencyGainGBP ⋅=

CLCLOLOL fAfA =

EE 323 -Feedback and stability Page 17 of 27

• 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)

EE 323 -Feedback and stability Page 18 of 27

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.

EE 323 -Feedback and stability Page 19 of 27

At some frequency, ω180, the –180O phase shift will change an otherwise negative feedback loop into a positive feedback loop.

The response of the feedback loop at ω180: A)1(1

A)1(v180

180out −+

−=

EE 323 -Feedback and stability Page 20 of 27

• If A180β=1, the denominator becomes 0 and the output becomes infinite (even vin = 0).

• Such a condition is equivalent to an oscillation at the frequency ω 180. • Less stringent inequality A180β≥1 also leads to oscillation at ω180.

• 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 ω180). § Alter the open loop response so that the stability condition is met: A180β < 1 § Internal design (stability condition is met up to some maximum value of β,

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

of LM741 is less than unity.

EE 323 -Feedback and stability Page 21 of 27

§ 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.

EE 323 -Feedback and stability Page 22 of 27

II. EVALUATION OF STABILITY CONDITION

Gain margin and phase margin to determine stability:

A1)j(A1inargMGain 180180−=−=−

*** Gain margin must be positive (i.e., A180β < 1).

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

00)j( +=−−=−

1A )j( = at ωPM.

*** Negative phase margin: A )j( is greater than unity at ω180 (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.

EE 323 -Feedback and stability Page 23 of 27

Example: Transfer function open-loop frequency response:

where A0=106, ω1=10rad/sec, ω2=ω3=106rad/sec.

Since ω1 << ω2, the dominant pole of A(jω) ≈ ω1 . a. How large can the feedback parameter β become before instability results? Assume β

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

Solution:

Since β is not frequency dependent, ω180 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 ω: ω180 ≅ 106 rad/sec

)j1)(j1)(j1(

AA

321

0)j(

+++=

EE 323 -Feedback and stability Page 24 of 27

The magnitude of A(jω) at ω180 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:

1A )j( ≤ ⇒ 2.051 =≤

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

51vv

in

out ≥=

vin vout + _

R1

+VCC

-VEE R2=4R1

EE 323 -Feedback and stability Page 25 of 27

EXTERNAL COMPENSATION

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

Example:

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

b. To what closed-loop gain does β 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).

EE 323 -Feedback and stability Page 26 of 27

Solution:

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

• A180=60dB=1000

• 1A )j( ≤ ⇒ β < 10-3.

• A non-inverting amplifier gain of 5 (β=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.

• 0Aor1A 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 β=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????????

EE 323 -Feedback and stability Page 27 of 27

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/rad319f2CR1

cc==