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AD8363

Rev. 0 | Page 22 of 36

DEVICE CALIBRATION AND ERROR CALCULATION

The measured transfer function of the AD8363 at 2.14 GHz is

shown in Figure 55. It shows plots of both output voltage vs.

input amplitude (power) and calculated error vs. input amplitude

(power). As the input power varies from −50 dBm to 0 dBm,

the output voltage varies from 0.25 V to about 2.8 V.

5.0

4.5

4.0

3.5

3.0

2.5

2.0

1.5

1.0

0.5

2.5

1.5

0.5

–0.5

–1.5

–2.5

–60

–50

–40

–30

–20

–10

PIN, INHI (dBm)

+25°C

–40°C

+85°C

Figure 55. 2.14 GHz Transfer Function Using Two-Point Calibration

Because slope and intercept vary from device to device, board-

level calibration must be performed to achieve high accuracy.

The equation for output voltage can be written as

VOUT = Slope × (PIN − Intercept)

(14)

where:

Slope is the change in output voltage divided by the change in

power (dB).

Intercept is the calculated input power level at which the output

voltage would be 0 V. (Note that Intercept is a theoretical value;

the output voltage can never achieve 0 V).

In general, calibration is performed by applying two (or more)

known signal levels into the input of the AD8363 and by measuring

the corresponding output voltages. The calibration points are

generally within the linear-in-dB operating range of the device

(see the Specifications section for more details).

The slope and intercept are calculated as follows:

Slope = (VOUT1 − VOUT2)/(PIN1 − PIN2)

(15)

Intercept = PIN1 − (VOUT1/Slope)

(16)

The previous formula for intercept is a shorthand formula based

upon Equation 14 and the assumption that the AD8363 is

operating within the linear-in-dB operating range. When the

slope and intercept are calculated, an equation can be written

that allows the calculation of the ideal input power based on the

output voltage of the detector.

PIN (unknown) = (VOUT1(MEASURED)/Slope) + Intercept

(17)

The log conformance error is the deviation of the detector from

the ideal calculated power and is given by

Error (dB) = (VOUT(MEASURED) − VOUT(IDEAL))/Slope

(18)

Figure 56 includes a plot of the error at 25°C, the temperature at

which the log amp is calibrated. Note that the error is not zero

because the detector does not perfectly follow the ideal straight

line. The error at the calibration points (in this case, −40 dBm

and −21 dBm) are, however, equal to zero by definition. Note that

Figure 55 is slightly different from those found in the Typical

Performance Characteristics section; its slope and intercept are

calculated using a two-point calculation and not based on multiple

points, as was used for the Typical Performance Characteristics.

Figure 55 also includes error plots for the output voltage at −40°C

and +85°C. These error plots are calculated using the slope and

intercept at 25°C. Another way of saying this is that the hot and

cold temperatures are calculated with respect to the output voltage

at ambient, and by definition, the error at ambient becomes

equal to 0. This is consistent with calibration in a mass production

environment, where calibration at temperature is not practical.

5.0

4.5

4.0

3.5

3.0

2.5

2.0

1.5

1.0

0.5

2.5

1.5

0.5

–0.5

–1.5

–2.5

–60

–50

–40

–30

–20

–10

PIN, INHI (dBm)

+25°C

–40°C

+85°C

Figure 56. 2.14 GHz Transfer Function Using a Three-Point Calibration

SELECTING AND INCREASING CALIBRATION

POINTS TO IMPROVE ACCURACY OVER A

REDUCED RANGE

Choose the amount and location of the calibration points carefully

because they can optimize the performance of the detector. In

some applications, increasing the dynamic range of the AD8363

may be desirable; however, in others, very high accuracy is required

at one power level or over a reduced input range. For example,

in a wireless transmitter, the accuracy of the high power amplifier

(HPA) is most critical at or close to full power. These objectives

can be achieved by the proper selection of the amount and

location of the calibration points.

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