ADL5519ACPZ-R2 AD [Analog Devices], ADL5519ACPZ-R2 Datasheet - Page 16

• Manufacturer Part Number: ADL5519ACPZ-R2
• Description: 1 MHz to 10 GHz, 50 dB Dual Log Detector/Controller
• Manufacturer: AD [Analog Devices]

ADL5519
In this configuration, all four measurements, OUT[A, B, P,
N], are made available simultaneously. A differential
output can be taken from OUTP − OUTN, and VLVL can
be used to adjust the common-mode level for an ADC
connection.
MEASUREMENT MODE
The ADL5519 requires a single supply of 3.0 V to 5 V. The
supply is connected to the three supply pins, VPSA, VPSB,
and VPSR. Each pin should be decoupled using the two
capacitors with values equal or similar to those shown in
Figure 19. These capacitors must provide a low impedance
over the full frequency range of the input, and they should
be placed as close as possible to the positive supply pins.
Two different capacitors are used in parallel to provide a
broadband ac short to ground.
The device is placed in measurement mode by connecting
OUTA and/or OUTB to VSTA and/or VSTB, respectively. As
seen in Figure 18, the ADL5519 has an offset voltage, a
negative slope, and a V
high end of its input signal range.
The output voltage vs. input signal voltage of the ADL5519
is linear-in-dB over a multidecade range. The equation for
this function is of the form
where:
X is the feedback factor in V
V
V
of the V
V
An offset voltage, V
the detector signal, so that the minimum value for V
X × V
SLOPE/DEC
INTERCEPT
INTERCEPT
Figure 18. Typical Output Voltage vs. Input Signal, Single Channel
V
X × V
OUT
OFFSET
OUT
= X × V
SLOPE/dB
is nominally –440 mV/decade or −22 mV/dB.
is the x-axis intercept of the linear-in-dB portion
is +2 dBV for a sinusoidal input signal.
. So for X = 1, minimum V
vs. V
× 20 × log
SLOPE/DEC
IN
curve (Figure 18).
OFFSET
OUT[A,B]
× log
, of 0.35 V is internally added to
10
(V
SET
10
measurement intercept at the
IN
(V
= V
/V
IN
INTERCEPT
/V
OUT
INTERCEPT
OUT
/X.
is 0.35 V.
)
) =
OUT
is
Rev. PrB | Page 16 of 27
(1)
(2)
The slope is very stable vs. process and temperature variation.
When base-10 logarithms are used, V
volts/decade. A decade corresponds to 20 dB; V
V
As noted in Equation 1 and Equation 2, the V
negative slope. This is also the correct slope polarity to control
the gain of many power amplifiers in a negative feedback
configuration. Because both the slope and intercept vary slightly
with frequency, it is recommended to refer to the Specifications
section for application-specific values for 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 charac-
teristic impedance of the system, Z
voltages to their corresponding power levels. The following
equations are used to perform this conversion:
For example, P
terms of dBm (decibels referred to 1 mW), in a 50 Ω system is
For a square wave input signal in a 200 Ω system,
Further information on the intercept variation dependence upon
waveform can be found in the AD8313 and AD8307 data sheets.
As the input signal to Channel A and Channel B are swept over
their nominal input dynamic range of +10 dBm to −50 dBm,
the output swings from 0.5 V to 1.75 V. The voltages OUTA and
OUTB are also internally applied to a difference amplifier with
a gain of two. So as the dB difference between INA and INB
ranges from approximately −30 dB to +30 dB, the difference
voltage on OUTP and OUTN swings from 0.5 V to 1.75 V.
Input differences larger than ±30 dB can be measured as long as
the absolute input level at INA and INB are within their nominal
ranges of +10 dBm to −50 dBm. However, measurement of large
differences between INA and INB are affected by on-chip signal
leakage. The common-mode level of OUTP and OUTN is set by
the voltage applied to VLVL. These output can be easily biased
up to a common-mode voltage of 2.5 V by connecting VREF to
VLVL. As the gain range is swept, OUTP swings from
approximately 0.5 V to 1.75 V and OUTN swings from 1.75 V
to 0.5 V.
SLOPE/dB
P(dBm) = 10 × log
P(dBV) = 20 × log
P(dBm) = P(dBV) − 10 × log
P
1 mW/1 V
+2 dBV − 10 × log
P
+6 dBm
INTERCEPT
INTERCEPT
represents the slope in volts/dB.
(dBm) = P
= −1 dBV − 10 × log
rms
INTERCEPT
2
) =
10
10
10
INTERCEPT
(V
(50×10
(V
for a sinusoidal input signal expressed in
Preliminary Technical Data
rms
rms
/1 V
2
/(Z
-3
(dBV) – 10 × log
) = +15 dBm
10
0
rms
10
[(200 Ω × 1 mW/1V
× 1 mW))
(Z
)
0
, must be known to convert
0
× 1 mW/1 V
SLOPE/DECADE
SLOPE/DECADE
OUT
10
represents the
(Z
voltage has a
rms
0
2
rms
)
×
2
/20 =
)] =
(3)
(4)
(5)
(6)