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AD8318ACPZ-WP Datasheet(PDF) 15 Page - Analog Devices

Part # AD8318ACPZ-WP
Description  Logarithmic Detector/Controller
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Manufacturer  AD [Analog Devices]
Direct Link  http://www.analog.com
Logo AD - Analog Devices

AD8318ACPZ-WP Datasheet(HTML) 15 Page - Analog Devices

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Data Sheet
AD8318
Rev. D | Page 15 of 24
The maximum output voltage is 2.1 V × X, and cannot exceed
400 mV below the positive supply.
VOUT(MAX) = (2.1 V × X) when X < (VPOS − 400 mV)/(2.1 V)
(6)
VOUT(MAX) = (VPOS − 400 mV) when X ≥ (VPOS − 400 mV)/ (2.1 V)
(7)
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.
2.2
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
2.5
2.0
1.5
1.0
0.5
0
–0.5
–1.0
–1.5
–2.0
–65 –60 –55
–45 –40 –35 –30 –25 –20 –15
–5
0
5
PIN (dBm)
INTERCEPT
PIN1
PIN2
VOUT +25°C
VOUT –40°C
VOUT +85°C
ERROR +25°C
ERROR –40°C
ERROR +85°C
VOUT2
VOUT1
VOUTIDEAL = SLOPE × (PIN – INTERCEPT)
SLOPE = (VOUT1 – VOUT2)/(PIN1 – PIN2)
INTERCEPT = PIN1 – (VOUT1/SLOPE)
ERROR (dB) = (VOUT × VOUTIDEAL)/SLOPE
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
Measurement 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 corre-
sponding 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)


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