Data Sheet

AD8318

USING THE AD8318

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Rev. E | 15 of 24

When X = 1, the typical output voltage swing is 0.5 V to 2.1 V. The

output voltage swing is modeled using Equation 5 to Equation 7

and restricted by Equation 8:

VOUT(MIN) < VOUT < VOUT(MAX)

(8)

When X = 4 and VPOS = 5 V,

(X × VOFFSET) < VOUT < (VPOS − 400 mV)

(4 × 0.5 V) < VOUT < (2.1 V × 4)

2 V < VOUT < 4.6 V

For X = 4, slope = −100 mV/dB; VOUT can swing 2.6 V, and the

usable dynamic range is reduced to 26 dB from 0 dBm to –26 dBm.

The slope is very stable vs. process and temperature variation.

When base-10 logarithms are used, VSLOPE/DECADE represents the

output voltage per decade of input power. One decade is equal

to 20 dB; VSLOPE/DEC/20 = VSLOPE/dB represents the output voltage

slope in V/dB.

As noted in Equation 3, the VOUT voltage has a negative slope.

This is the correct slope polarity to control the gain of many power

amplifiers and other VGAs in a negative feedback configuration.

Because both the slope and intercept vary slightly with frequency,

refer to Table 1 for application-specific values for the slope and

intercept.

Although demodulating log amps respond to input signal voltage,

not input signal power, it is customary to discuss the amplitude

of high frequency signals in terms of power. In this case, the

characteristic impedance of the system, Z0, must be known to

convert voltages to corresponding power levels. Beginning with the

definitions of dBm and dBV,

P (dBm) = 10 × log10(Vrms2/(Z0 × 1 mW))

(9)

V (dBV) = 20 × log10(Vrms/1 Vrms)

(10)

When Equation 9 is expanded

P (dBm) = 20 × log10(Vrms) − 10 × log10(Z0 × 1 mW)

(11)

and given Equation 10, Equation 11 can be rewritten as

P (dBm) = V (dBV) − 10 × log10(Z0 × 1 mW)

(12)

For example, PINTERCEPT for a sinusoidal input signal, expressed in

terms of dBm (decibels referred to 1 mW), in a 50 Ω system is

PINTERCEPT (dBm) = VINTERCEPT (dBV)

− 10 × log10(Z0 × 1 mW) =

(13)

7 dBV − 10 × log10(50 × 10−3) = 20 dBm

For further information on the intercept variation dependence upon

waveform, refer to the AD8313 and AD8307 data sheets.

DEVICE CALIBRATION AND ERROR

CALCULATION

The measured transfer function of the AD8318 at 2.2 GHz is shown

in Figure 32. The figure shows plots of both output voltage vs. input

power and calculated log conformance error vs. input power.

As the input power varies from −65 dBm to 0 dBm, the output

voltage varies from 2 V to about 0.5 V.

Figure 32. Transfer Function at 2.2 GHz

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

level calibration is performed to achieve high accuracy.

The equation can be rewritten for output voltage, from the Measure-

ment Mode section, using an intercept expressed in dBm.

VOUT = Slope × (PIN – Intercept)

(14)

In general, the calibration is performed by applying two known

signal levels to the AD8318 input and measuring the corresponding

output voltages. The calibration points are generally chosen to be

within the linear-in-dB operating range of the device (see Figure

32). Calculation of the slope and intercept is done by:

Slope = (VOUT1 − VOUT2)/(PIN1 − PIN2)

(15)

Intercept = PIN1 − VOUT1/Slope

(16)

Once the slope and intercept are calculated, an equation can be

written to allow calculation of an (unknown) input power based on

the output voltage of the detector.

PIN(unknown) = VOUT (measured)/Slope + Intercept

(17)

Using the equation for the ideal output voltage (see Equation 13) as

a reference, the log conformance error of the measured data can

be calculated as

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

(18)

Figure 32 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. This

is because the log amp does not perfectly follow the ideal VOUT