AD8137YCP-R2 datasheet(22/24 Pages) AD

22 / 24 page

AD8137

Rev. B | Page 22 of 24

This example shows that when RF and RG are large compared

to RT, the gain reduction produced by the increase in RG is

essentially cancelled by the increase in the Thevenin voltage

caused by RT being greater than the output resistance of the

signal source. In general, as RF and RG become smaller in termi-

nated applications, RF needs to be increased to compensate for

the increase in RG.

AD8137

–

2V p-p

52.3

Ω

–

VOCM

Ω

1.02k

Ω

0.1

µF

0.1

µF

+5V

–5V

VIN

SIGNAL

SOURCE

Ω

When generating the typical performance characteristics data,

the measurements were calibrated to take the effects of the

terminations on closed-loop gain into account.

Power Down

The AD8137 features a PD pin that can be used to minimize the

quiescent current consumed when the device is not being used.

PD is asserted by applying a low logic level to Pin 7. The thresh-

old between high and low logic levels is nominally 1.1 V above

the negative supply rail. See the Specification tables (Table 1 to

Table 3) for the threshold limits.

Figure 66. AD8137 with Terminated Input

The 52.3 Ω termination resistor, RT, in parallel with the 1 kΩ

input resistance of the AD8137 circuit, yields an overall input

resistance of 50 Ω that is seen by the signal source. In order to

have matched feedback loops, each loop must have the same RG

if it has the same RF. In the input (upper) loop, RG is equal to the

1 kΩ resistor in series with the (+) input plus the parallel

combination of RT and the source resistance of 50 Ω. In the

upper loop, RG is therefore equal to 1.03 kΩ. The closest

standard value is 1.02 kΩ and is used for RG in the lower loop.

DRIVING AN ADC WITH GREATER THAN 12-BIT

PERFORMANCE

Since the AD8137 is suitable for 12-bit systems, it is desirable to

measure the performance of the amplifier in a system with

greater than 12-bit linearity. In particular, the effective number

of bits, ENOB, is most interesting. The AD7687, 16-bit,

250 KSPS ADC’s performance makes it an ideal candidate for

showcasing the 12-bit performance of the AD8137.

Things become more complicated when it comes to

determining the feedback resistor values. The amplitude of the

signal source generator VIN is two times the amplitude of its

output signal when terminated in 50 Ω. Therefore, a 2 V p-p

terminated amplitude is produced by a 4 V p-p amplitude from

VS. The Thevenin equivalent circuit of the signal source and RT

must be used when calculating the closed-loop gain because RG

in the upper loop is split between the 1 kΩ resistor and the

Thevenin resistance looking back toward the source. The

Thevenin voltage of the signal source is greater than the signal

source output voltage when terminated in 50 Ω because RT

must always be greater than 50 Ω. In this case, RT is 52.3 Ω and

the Thevenin voltage and resistance are 2.04 V p-p and 25.6 Ω,

respectively.

For this application, the AD8137 is set in a gain of two and

driven single-ended through a 20 kHz band-pass filter, while the

output is taken differentially to the input of the AD7687

(see Figure 67). This circuit has mismatched RG impedances and,

therefore, has a dc offset at the differential output. It is included

as a test circuit to illustrate the performance of the AD8137.

Actual application circuits should have matched feedback

networks.

For an AD7687 input range up to −1.82 dBFS, the AD8137 power

supply is a single 5 V applied to VS+ with VS− tied to ground. To

increase the AD7687 input range to −0.45 dBFS, the AD8137

supplies are increased to +6 V and −1 V. In both cases, the VOCM

pin is biased with 2.5 V and the PD pin is left floating. All voltage

supplies are decoupled with 0.1 µF capacitors. Figure 68 and

Figure 69 show the performance of the −1.82 dBFS setup and the

−0.45 dBFS setup, respectively.

Now the upper input branch can be viewed as a 2.04 V p-p

source in series with 1.03 kΩ. Since this is to be a unity-gain

application, a 2 V p-p differential output is required, and RF

must therefore be 1.03 kΩ × (2/2.04) = 1.01 kΩ ≈ 1 kΩ.

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