LPV531 datasheet(19/27 Pages) TI

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-1

2CIN

P1,2 =

4 A0CIN

-R2/R1

1 +

CIN R2

VOUT

VIN

(s) =

A0 R1

R1 + R2

CIN

VOUT

VIN

VOUT

VIN

AV = -

= -

LPV531

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SNOSAK5B – MARCH 2006 – REVISED MARCH 2013

INPUT CAPACITANCE AND FEEDBACK CIRCUIT ELEMENTS

The LPV531 has a very low input bias current (50 fA). To obtain this performance a large CMOS input stage is

used, which adds to the input capacitance of the op amp, CIN. Though this does not affect the DC and low

frequency performance, at higher frequencies the input capacitance interacts with the input and the feedback

impedances to create a pole, which results in lower phase margin and gain peaking. The gain peaking can be

reduced by carefully choosing the appropriate feedback resistor, as well as, by using a feedback capacitance,

CF. For example, in the inverting amplifier shown in Figure 43, if CIN and CF are ignored and the open loop gain

of the op amp is considered infinite then the gain of the circuit is

−R2/R1. An op amp, however, usually has a

dominant pole, which causes its gain to drop with frequency. Hence, this gain is only valid for DC and low

frequency. To understand the effect of the input capacitance coupled with the non-ideal gain of the op amp, the

circuit needs to be analyzed in the frequency domain using a Laplace transform.

Figure 43. Inverting Amplifier

For simplicity, the op amp is modeled as an ideal integrator with a unity gain frequency of A0 . Hence, its transfer

function (or gain) in the frequency domain is A0/s. Solving the circuit equations in the frequency domain, ignoring

CF for the moment, results in the following equation for the gain:

(6)

It can be inferred from the denominator of the transfer function that it has two poles, whose expressions can be

obtained by solving for the roots of the denominator:

(7)

Equation 7 shows that as the values of R1 and R2 are increased, the magnitude of the poles is reduced, and

hence the bandwidth of the amplifier is decreased. Furthermore, R1 and R2 are related by the gain of the

amplifier.

AV = −R2/R1, or alternatively

R2 = −AVR1

It is the presence of pairs of poles in Equation 7 that causes gain peaking. In order to eliminate this effect, the

poles should be placed in Butterworth position, since poles in Butterworth position do not cause gain peaking. To

achieve a Butterworth pair, the quantity under the square root in Equation 7 should be set to equal

−1. Using this

fact and the relation between R1 and R2, the optimum value for R1 can be found. This is shown in Equation 8. If

R1 is chosen to be larger than this optimum value, gain peaking will occur.

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