AD8244

Data Sheet

APPLICATIONS INFORMATION

ELECTROCARDIOGRAM (ECG)

In an ECG system, mismatches between the source impedance

of different leads, working against the input impedance of the

front-end amplifier, can create unbalanced voltage dividers that

reduce the system CMRR. When presented to a moderately

high input impedance amplifier, the combined impedance of the

skin, electrolyte, electrodes, and the protection resistors can be

enough to cause power line noise pickup, current noise issues, and

signal division. Dry electrode systems, which are becoming

increasingly common and have significantly higher source

impedance, are especially sensitive to these errors. Typically, a high

input impedance, low bias current, FET input op amp is used to

buffer the electrode signal before it is presented to an

instrumentation amplifier. This buffer solves the majority of

these problems; however, when an instrument is in the field, it

can be subject to dust pickup and humidity. If the op amp input

is not guarded, these environmental factors can create unwanted

leakage currents that bring back the aforementioned issues from

insufficient input impedance. The AD8244 pinout is configured to

make it simple to guard the inputs from parasitic resistance and

capacitance while it also drives the instrumentation amplifier

inputs, creating a more robust design, while saving power and

board space. The CMRR of the AD8244 driving an instrumentation

amplifier initially depends on the gain matching for the chosen

supplies and voltage range, as well as the instrumentation

amplifier used, but it can be improved with design techniques

such as right leg drive (RLD) or digital filtering.

FILTERING

In filtering applications, it is generally recommended to use

capacitors such as C0G or NP0 ceramics for distortion and

dielectric absorption performance. These types of capacitors

do not have a high volumetric efficiency and are only available

in values less than a few tens of nanofarads, depending on the

case size and voltage rating. For a given cutoff frequency, using

smaller capacitors requires larger resistor values. At low

frequencies where the resistor values become very large, the bias

current of a typical op amp can introduce significant offsets and

additional noise. The subpicoampere bias current of the

AD8244 allows resistor values in the tens of megaohms with no

additional error while providing an excellent low power, small

footprint solution for filter design. Between the four channels of

the AD8244, a filter with more than eight poles can be

implemented while using less space than the same filter with a

quad op amp.

Sallen-Key Low-Pass Filter

1/4

AD8244

VOUT

C1

2nF

VIN

C2

1nF

R1

R2

NOTES

1. R1 = R2 = R

2. R = 112.5MΩ/fC, Q = 0.707

Figure 42. Sallen-Key Low-Pass Filter

The following equations describe the corner frequency, fC, and

quality factor, Q, for the low-pass filter case of the Sallen-Key

topology, shown in Figure 42:

fC = 1/(2π

C2

C1

R2

R1

×

×

×

)

Q = (

C2

C1

R2

R1

×

×

×

)/(C2 × (R1 + R2))

For an example of a design with this topology, choose a filter

where Q = 0.707 and R1 = R2 = R. This requires that C1 = 2 × C2.

The corner frequency equation can now be simplified to

fC = 1/(2π × R × C2 × √2)

If an available capacitor, such as 1 nF, is chosen for C2, R can be

written in terms of the desired cutoff frequency:

R = 1/(2√2 × π × 1 nF × fC) = 112.5 MΩ/fc (that is,

R = 750 kΩ for fC = 150 Hz)

Sallen-Key High-Pass Filter

1/4

AD8244

VOUT

C1

22nF

VIN

C2

22nF

R1

R2

NOTES

1. R2 = R, R1 = R/2

2. R = 10.2MΩ/fC, Q = 0.707

Figure 43. Sallen-Key High-Pass Filter

The high-pass filter case of the Sallen-Key topology has the

same corner frequency equation as the low-pass filter. However,

the equation for Q changes to

Q = (

C2

C1

R2

R1

×

×

×

)/(R1 × (C1 + C2))

In this case, a Q of 0.707 is achieved with C1 = C2 = C, and 2 ×

R1 = R2 = R, which is a symmetrical result to the low-pass filter

case.

The corner frequency then simplifies to

fC = 1/(√2 × π × R × C)

For a low corner frequency, a larger available capacitor such as

22 nF can be chosen, yielding the following expression for R:

R = 10.2 MΩ/fc (that is, a 0.5 Hz filter requires

R1 = 10 MΩ and R2 = 20 MΩ)

Rev. A | Page 16 of 20