An Op Amp Tutorial

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An Op Amp Tutorial(Based on material in the book Introduction to Electroacoustics and Audio Am-plifier Design, Second Edition - Revised Printing, by W. Marshall Leach, Jr.,published by Kendall/Hunt,°c 2001.)Anopamphas twoinputs andoneoutput. The circuit isdesignedsothat theoutputvoltageis proportional to the di fference between the two input voltages. In general, an op amp canbe modeled as a three-stage circuit as shown in Fig. 1. The non-inverting input is v .TheI1inverting input is v . The input stage is a di fferential amplifier (Q and Q ) with a currentI2 1 2mirrorload(Q −Q ). Thedi ffamptailsupply isthe dccurrentsourceI . Thesecond stage3 5 Qis a high-gain stage having an inverting or negative gain. A capacitor connects the outputof this stage to its input. This capacitor is called the compensating capacitor. Other namesfor it are lag capacitor, Miller capacitor, and pole-splitting capacitor. It sets the bandwidthof the circuit to a value so that the op amp is stable, i.e. so that it does not oscillate. Theoutput stage is a unity-gain stage which provides the current gain to drive the load.Figure 1: Op amp model.If we assume that Q and Q are matched, that Q and Q are matched, that base1 2 3 4currents can be neglected, and that the Early e ffect can be neglected, we can write thefollowing equation for i :O1i =i −i = i −i = i −i (1)O1 C1 C3 C1 C4 C1 C2But i +i =I and i =I /2+i .ThusweobtainC1 C2 Q C1 Q c1i =2i −I =2i (2)O1 C1 Q c11Open-Loop Transfer ...

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An Op Amp Tutorial

(Based on material in the bookIntroduction to Electroacoustics and Audio Am- pliÞer Design, Second Edition - Revised Printing, by W. Marshall Leach, Jr., published by Kendall/Hunt, c °2001.)

An op amp has two inputs and one output. The circuit is designed so that the output voltage is proportional to the difference between the two input voltages. In general, an op amp can be modeled as a three-stage circuit as shown in Fig. 1. The non-inverting input isv I1 . The inverting input isv I2 input stage is a di. The fferential ampliÞer (Q 1 andQ 2 ) with a current mirror load (Q 3 −Q 5 ). The diffamp tail supply is the dc current sourceI Q second stage. The is a high-gain stage having an inverting or negative gain. A capacitor connects the output of this stage to its input. This capacitor is called the compensating capacitor. Other names for it are lag capacitor, Miller capacitor, and pole-splitting capacitor. It sets the bandwidth of the circuit to a value so that the op amp is stable, i.e. so that it does not oscillate. The output stage is a unity-gain stage which provides the current gain to drive the load.

Figure 1: Op amp model.

If we assume thatQ 1 andQ 2 are matched, thatQ 3 andQ 4 are matched, that base currents can be neglected, and that the Early effect can be neglected, we can write the following equation fori O1 :

i O1 =i C1 −i C3 =i C1 −i C4 =i C1 −i C2

Buti C1 +i C2 =I Q andi C1 =I Q /2 +i c1 . Thus we obtain

i O1 = 2i C1 −I Q = 2i c1

1

(1)

(2)

Open-Loop Transfer Function

We wish to solve for the transfer function forV o /V id , whereV id is the difference voltage between the two op amp inputs. First, we solve for the currentI c1 as a function ofV id . For the diffamp, let us assume that the transistors are matched, thatI E1 =I E2 =I Q /2, the Early effect can be neglected, and the base currents are zero. In this case, the small-signal ac emitter equivalent circuit of the diffamp is the circuit given in Fig. 2(a). In this circuit, r e1 andr e2 are the intrinsic emitter resistances given by V T 2V T r e1 =r e2 =r e = =(3) I E I Q Note that the dc tail supplyI Q does not appear in this circuit because it is not an ac source. From the emitter equivalent circuit, it follows that V id I c1 =I e1 =(4) 2 (r+R) e E whereI c1 =I e1 because we have assumed zero base currents.

Figure 2: Circuit for calculatingI e1 . (b) Circuit for calculatingV o .

Figure 2(b) shows the equivalent circuit which we use to calculateV o assume that. We R eq is the effective load resistance on the current2I c1 this case, the current which. In ßows through the compensating capacitorC c is given by V V V o1id o2 I o1 = 2I c1 + =−(5) R eq r e +R E KR eq where we have used Eq. (4) and the relationV o1 =−V o2 /Kvoltage. The V o2 is given by · ¸ I o1 −V o2 V id V o2 1 V o2 =V o1 ++ = −(6) C c s K r e +R E KR eq C c s If we assume that the output stage has a gain that is approximately unity, thenV o 'V o2 . LetG(s) =V o /V id . It follows from Eq. (6) thatG(s)is given by V o V o2 KR eq 1 G(s) ='=×(7) V id V id r e +R E 1 + (1 +K)R eq C c s This is of the form A G(s) =(8) 1 +s/ω 1 whereAandω 1 are given by KR eq 1 A=ω 1 = 2πf 1 =(9) r e +R E (1 +K)R eq C c

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Figure 3: Asymptotic Bode magnitude plots. (a) Without feedback. (b) With feedback.

Gain Bandwidth Product

The asymptotic Bode magnitude plot for|G(jω)|the pole3(a). Above is shown in Fig. frequencyω 1 , the plot has a slope of−1dec/dec or−20frequency at whichdB/dec. The |G(jω)|= 1is called the unity-gain frequency or the gain-bandwidth product. It is labeled ω x in theÞgure and is given by

K1 1 ω x = 2πf x =Aω 1 =' 1 +K r e +R E C c (r e +R E )C c

(10)

where the approximation holds forKÀ1. It follows that an alternate expression forG(s) is A G(s) =(11) 1 +sA/ω x For maximum bandwidth,f x should be as large as possible. However, iff x is too large, the op amp can oscillate. A value of1MHz is typical for general purpose op amps.

Example 1An op amp is to be designed forf x = 4MHz andI Q = 50µA. IfR E = 0, calculate the required value forC c .

Solution.C c = 1/(2πf x r e ) =I Q /(4πf x V T ) = 38.4pF, where we assume thatV T = 0.0259 V.

Slew Rate

The op amp slew rate is the maximum value of the time derivative of its output voltage. In general, the positive and negative slew rates can be diff1simple model of Fig. erent. The predicts that the two are equal so that we can write

dv O −SR≤ ≤+SR dt

whereSRis the slew rate. To solve for it, we use Eqs. (5) and (6) to write µ ¶ I o1 −V o2 −V o2 1 V o2 =V o1 + = + 2I c1 + C s K KR C s c eq c

3

(12)

(13)

This can be rearranged to obtain · µ ¶¸ 1 1 2I c1 sV o2 =1 + 1 + (14) K R C C eq c c If we assume thatKis large and letV o2 'V o , this equation reduces to 2I c1 sV o '(15) C c Thesoperator in a phasor equation becomes thed/dtoperator in a time-domain equation. Thus we can write dv2i o c1 =(16) dt C c It follows that the slew rate is determined by the maximum value ofi c1 total collector. The current inQ 1 is the sum of the dc value plus the small-signal ac value. Thus we can write i C1 =I Q /2 +i c1 current has the limits. This 0≤i C1 ≤I Q . It follows that the small-signal ac component has the limits−I Q /2≤i c1 ≤I Q /2we can write. Thus −I Q dv o +I Q ≤ ≤(17) C c dt C c It follows that the slew rate is given by I Q SR=(18) C c Example 2Calculate the slew rate of the op amp of Example 1. Solution.SR=I Q /C c = 1.30V/µs.

Relations between Slew Rate and Gain-Bandwidth Product

IfC c (10) and (18), we obtain the relationis eliminated between Eqs. µ ¶ I Q R E SR= 2πf x I Q (r e +R E ) = 4πf x V T 1 +(19) 2V T This equation clearly shows that the slew rate isÞxed by the gain-bandwidth product if R E = 0. IfR E >0, the slew rate and gain bandwidth product can be speciÞed independently. Example 3Emitter resistors with the valueR E = 3kΩare added to the input diffamp in the op amp of Example 1. Iff x is to be held constant, calculate the new value of the slew rate and the new value ofC c . Solution.SR= 2πf x I Q (r e +R E ) = 5.07V/µs.C c =I Q /SR= 9.86pF. The slew rate is greater by a factor of3.9andC c is smaller by the same factor. The above example illustrates how the slew rate of an op amp can be increased without changing its gain-bandwidth product. WhenR E is added,ω x decreases. To makeω x equal to its original value,C c It can be seenmust be decreased, and this increases the slew rate. from Eq. (19) that the slew rate can also be increased by increasingI Q this causes. However, ω x to increase. To makeω x equal to its original value,R E must also be increased. Therefore, the general rule for increasing the slew rate is to either decreaseC c , increaseI Q , or both. All of these makeω x bringincrease. To ω x back down to its original value,R E must be increased. The change inR E does not affect the slew rate.

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Figure 4: (a) Ampli Þlter.

er with feedback. (b) AmpliÞer with feedback preceded by a low-pass

Closed-Loop Transfer Function

Figure 4(a) shows the op amp with a two resistor voltage divider connected as a feedback network. The output voltage can be written

V=G(s) (V−V) =G(s) (V−bV) o i f i o

wherebis the gain of the voltage divider given by

R 1 b= R 1 +R F

Note that0≤b≤1. Eq. (20) can be solved forV o /V i to obtain

V o G(s)A f = = V i 1 +bG(s+) 1 s/ω 1f

(20)

(21)

(22)

where Eq. (11) is used forG(s)dc gain. The A f and the pole frequencyω 1f are given by

A1 A f =' 1 +bA f b 1 +bAω x ω 1f = 2πf 1f =ω x ='bω x A A f

(23)

(24)

where thefin the subscript implies “with feedback” and the approximations assume that bAÀ1. It can be seen from these equations that

A f ω 1f =Aω 1 =ω x

(25)

Fig. 3(b) shows the Bode plot for|V o /V i |for two values ofA f . Asbis increased,A f decreases and the bandwidthω 1f increases so that the product of the two remain constant. This illustrates whyω x is called the gain-bandwidth product.

Example 4An op amp has the gain bandwidth productf x = 8the upperMHz. Calculate −3dB frequencyf u if the op amp is operated at a voltage gain of21.

Solution. The upper−3dB frequency is equal to the pole frequency of the closed-loop transfer function. Thusf u =f 1f =f x /A f = 381kHz.

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Figure 5: (a) No slewing step response. input voltage.

Transient Response

(b) Step response with slewing.

(c) Differential

Let the input voltage to the op amp in Fig. 4(a) be a step of amplitudeV 1 . We can write v I (t) =V 1 u(t), whereu(t)The Laplace transform ofis the unit step function. v I (t)is V i (s) =V 1 /s. The Laplace transform of the output voltage is given by

V 1 A f V o (s) = s1 +s/ω 1f

(26)

The time domain output voltage is obtained by taking the inverse Laplace transform to obtain v O (t) =A f V 1 [1−exp (−ω 1f t)]u(t)(27) A plot ofv O (t)5(a).is shown in Fig. The maximum time derivative ofv O (t)occurs att= 0and is given by

dv O d A f =A f V 1 {[1−exp (−ω 1f t)]u(t)}|=V 1 =ω x V 1 ¯ t=0 dt dtω 1f max

(28)

If the derivative exceeds the slew rate of the op amp , the output voltage will be distorted as shown in Fig. 5(b), where the non-slewing response is shown by the dashed line. The maximum value ofV 1 before the op amp slews is given by

SR SR V 1max = = =I Q (r e +R E ) ω2πf x x

Example 5Calculate the maximum value ofV 1 for the op amps of Examples 2 and 3.

(29)

Solution. For Example 1,V 1max =SR/ω x = 51.7mV. For Example 2,V 1max = 202mV. This is greater by about a factor of3.9the same as the ratio of the two slew rates., i.e.

Input Stage Overload

For the step input signal to the op amp with feedback in Fig. voltage is given by

4(a), the differential input

V 1 v I D (t) =v I (t)−bv O (t[1 +) = bA 0 exp (−bω x t)]u(t) 1 +bA

6

(30)

It follows thatv ID (0) =V 1 andv ID (∞) =V 1 /(1 +bA). A plot ofv I D (t)is shown in Fig. 5(c). The peak voltage occurs att= 0. If the op amp is not to slew, the diffamp input stage must not overload with this voltage. If base currents are neglected, the emitter and collector currents inQ 1 andQ 2 can be written µ ¶ µ ¶ v BE1 v B E2 i E1 =i C1 =I S expi E2 =i C2 =I S exp(31) V T V T whereI S is the BJT saturation current. The differential input voltage can be written

v I D = (v B E1 −v B E2 ) + (i E1 −i E2 )R E

With the relationi E2 =I Q −i E1 , these equations can be solved to obtain µ ¶ i C1 v I D =V T (2ln + i C1 −I Q )R E I Q −i C1

(32)

(33)

The same equation holds fori C2 exceptv I D is replaced with−v I D . Bothi C1 andi C2 must satisfy the inequality0≤i≤I Q . At either limit of this inequality, one transistor in the diffamp is cut offus consider the di. Let ff-amp active range to be the range for whichi C1 andi C2 satisfy0.05I Q ≤i≤0.95I Q is the. This 5%to95%range for the currents. When the diffamp is operated in this range, it follows from Eq. (33) thatv I D must satisfy −v I D(max) ≤v I D ≤ −v I D(max) (34) wherev I D(max) is given by v I D(max) =V T ln 19 + 0.9I Q R E (35) Ifv I D lies in this range, neither transistor in the diffamp can cut offand the op amp cannot exhibit slewing.

Example 6Calculatev I D(max) forI Q = 50µA forR E = 0and forR E = 3kΩ.

Solution. ForR E = 0, we havev I D(max) =V T = 76ln 19 .3mV. ForR E = 3000,v I D(max) = −6 76.3mV+0.9×50×10×3000 = 211mV. These values are greater than the values of V 1max calculated in Example 5 because the analysis here is based on the large signal behavior of the BJT.

Full Power Bandwidth

Figure 6 shows the output voltage of an op amp with a sine wave input for two cases, one where the op amp is not slewing and the other where the op amp is driven into full slewing. The full slewing waveform is a triangle wave. The slew-limited peak voltage is given by the slope multiplied by one-fourth the period, i.e.

T SR V Pslew =SR×= 4 4f

(36)

whereT= 1/f. When the op amp is driven into full slewing, an increase in the amplitude of the input signal causes no change in the amplitude of the output signal. If the frequency is doubled, the amplitude of the output signal is halved.

7

Figure 6: Non-slewing and full slewing output voltage waveforms.

Let the input voltage to the op amp be a sine wave. If the op amp does not slew and is not driven into peak clipping, the output voltage can be writtenv O (t) =V P sinωt. The time derivative is given bydv O /dt=ωV P cosωtmaximum value of. The |dv O /dt|occurs at ωt=nπ, wherenis an integer, and is given by|dv O /dt|=ωV P . For a physical op amp max , this cannot exceed the slew rate, i.e.ωV P < SRfollows that the maximum frequency. It that the op amp can put out the sine wave without slewing is given by

SR f max = 2πV P

Conversely, the peak output voltage without slewing is given by

SR V= P(max) 2πf

(37)

(38)

LetV clip If an op amp is drivenbe the op amp clipping voltage at midband frequencies. at this level and the frequency is increased, the op amp will eventually slew and the maxi- mum output voltage will decrease as the frequency is increased. The full power bandwidth frequencyf FPBW is deÞned as the highest frequency at which the op amp can put out a sine wave with a peak voltage equal toV clip is given by. It

SR f FPBW = 2πV clip

(39)

Figure 7 shows the peak output voltage versus frequency for a sine wave input signal. At low frequencies, the peak voltage is limited to the op amp clipping voltageV clip . As frequency is increased, the peak voltage becomes inversely proportional to frequency when the op amp is driven into full slewing and is given bySR/4f. TheÞgure also shows the peak voltage below which the op amp does not slew. It is given bySR/2πf.

Example 7The op amps of Examples 2 and 3 have clipping voltages of±13V. Calculate the full power bandwidth frequency if the op amps are not to slew at maximum output.

6 Solution. For the op amp of Example 2,f FPBW = 1.3×10/(2π13) = 15.9thekHz. For 6 op amp of Example 3,f FPBW = 5.07×10/(2π13) = 62.1kHz.

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