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AD8318ACPZ-R2 データシート(PDF) 15 Page - Analog Devices

部品番号 AD8318ACPZ-R2
部品情報  1 MHz to 8 GHz, 70 dB Logarithmic Detector/Controller
PDF  24 Pages
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メーカー  AD [Analog Devices]
ホームページ  http://www.analog.com
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AD8318ACPZ-R2 データシート(HTML) 15 Page - Analog Devices

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Data Sheet
AD8318
USING THE AD8318
analog.com
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



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