BANDWIDTH & SLEW-RATEINEL 5207 - SPRING 2011

Frequency Limits

• Many opamps are internally compensated to have a single

dominant pole at a relatively low frequency.

!

+vp

vn

gm

-a2

iO1

1

Req

Ceq

vO

Noninverting Amplifier

• Open-loop gain can be wrien as:

a(s) = a01

1 + s/!p

a0 = d.c. open-loop gain; fp = 12"ReqCeq

= pole freq.

• For the non-inverting amplifier

A =a

1 + a#

where # = R1R1+R2

.

• Using above a(s) and A0 = a01+#a0

,

A(s) =a(s)

1 + a(s)#= A0

!p(1 + #a0)

s + !p(1 + #a0)

• Corner frequency is increased by 1 + !a0. Gain is decreased

by the same factor.

• Gain-bandwidth product remains constant and equal to unity

gain frequency, ft.

GBP = ft

• This is only true for ! constant and compensated opamp

(dominant pole at low freq.)

Gain of n identical noninverting stages

• If fcl = !p(1 + "a0)/2#, then gain magnitude of one stage is

A = A01

!

1 + (f/fcl)2

• Gain of n identical stages is

An = An0

"

1 + (f/fcl)2#!n/2

• At corner frequency f3dB, An/An0 = 1/

"2 (i.e. -3dB). Thus,

f3dB = fcl

!

21/n ! 1 =ft

A0

!

21/n ! 1

• To design an amplifier with bandwidth fbw and gain K, wemust select n such that K = An

0 and fbw # ftA0

!

12n ! 1.

Inverting Amplifier

• Bw: ftR1

R1+R2; same than non-inv with gain 1 + R2/R1

• A = Aideal1

1+1/T; Aideal = !R2

R1; T = a!non!inv

• For the inverting amplifier, the gain-bandwidth product is

equal to

GBP = ftR2

R1 + R2

so the bandwidth is always lower than that of a non-inverting

amplifier with the same gain.

• Equivalently, we can say that fbw ! (1 + R2R1

) is still constant

and equal to ft, but the the amplifier’s gain magnitude of is

only R2R1

.

Input Impedance

• Di!. input impedance rd is typ. few M!, common-mode rc

is in the G! for BJT. FETs are in the 100’s G!.

• Input capacitance for uA741 is about 1 pF . Appears in par-

allel with rd and/or rc.

• For f = 100kHz, Xc = 10M! so it is comparable to rd. At

higher frecuencies, input impedance drops due to the input

SLEW RATE

TRANSIENT RESPONSE

FOLLOWER (NON-INVERTING WITH UNITY GAIN)

FREQUENCY RESPONSE

STEP RESPONSE

RISE TIME (TIME TO GO FROM 10% TO 90% OF VM)

A =1

1 + j 1ft

vO = Vm (1− exp (−t/τ)) τ =1

2πft

tr =0.35ft

CURRENT STARVING

vn vp

VCC

VEE

Q1 Q2

Q3 Q4

+vp

vngm -a2iO1 1

Req

CC

vO

IA

iO1

EXPONENTIAL & SLEW-RATE LIMITED STEP RESPONSE

HIGHER-ORDER POLES WOULD INTRODUCE“RINGING” - I.E. SMALL OSCILLATIONS ON

RESPONSE

t

Volts

Vm

vI

+

+vI

vO

vO(slew-rate limited)vO (exponential rsponse)

tr

t

Volts

S.R.

vO

vI

Volts

t

vO

vIS.R.

• Expected output: vO,expected = AvI

• If dvO,expected

dt> S.R. then dvO

dt= S.R.

• If dvO,expected

dt≤ S.R. then the response is exponential.

For an input step, the critical output size Vom,crit be-yond which the output becomes slew-rate limited, is

vO,expected = Vom (1− exp (−1/τ))dvO,max

dt=

Vom(crit)

τ= S.R.

Vom(crit) = τ × S.R. = S.R.

2πfB

where τ = 12πfB

, fB is the closed-loop bandwidth i.e.

• fB = ft for unity gain,

• fB = βft for non-unity gain, where β is the non-inverting amplifier feedback factor.

• Vom =| A | ×Vim where Vim is the size of the inputstep and A is the (inverting or non-inverting) ampli-fier gain

When the input is a step and Vom > Vom(crit),

• initially vO changes linearly

vO(t) = SR× t ∀ t < t1

• for t > t1 where t1 is defined by

Vom − vO(t1) < Vom(crit)

the output changes back to a linear response and.

VO(t) = Vom − (Vom − vO(t1)) exp −(t−t1)/τ

For an input sinusoid,

• non-slew-rate limited output signal vO = Vom sin 2πft

• vO rate of change would bedvO

dt= 2πfVomcos2πft

• maximum rate of change is at t = 0,�

dvO

dt

�max

= 2πfVom

• for Vom ≥ Vom,crit, the output becomes slew-rate limited,

2πfVom,crit = S.R.

or

Vom(crit) =S.R.

2πf

• Observe that Vom =| A | ×Vim where Vim is the input signal

peak and A is the amplifier gain.

Given desired Vom we find the maximum frequency ofsinusoid with undistorted output:

fmax =SR

2πVom

Full-power bandwidth (FPB)

Maximum frequency for which the opamp output isundistorted sinusoid with the largest possible amplitude.If saturation voltage is ±Vsat

FPB =SR

2πVsat