AD8315
Check: The power range is 50 dB, which should correspond
to a voltage change in VSET of 50 dB ¥ 24 mV/dB = 1.2 V,
which agrees.
of the curve shown in Figure 5) will be slower. Note also that it
is sometimes useful to add a zero in the closed-loop response by
placing a resistor in series with CFLT. For more about these
matters, refer to the Applications section.
Now, the value of VAPC is of interest, although it is a dependent
parameter, inside the loop. It depends on the characteristics of
V , P
2
2
the power amplifier, and the value of the carrier amplitude VCW
Using the control values derived above, that is, GO = 0.316 and
.
33
23
13
V
GBC = 1 V, and assuming the applied power is fixed at –7 dBm
(so VCW = 100 mV rms), the following is true using Equation 11:
V
APC(max) = (VSETVGBC )/VSLP – log10 kGOVCW /VZ
= (1.44 ¥ 1)/0.48 – log10 (0.0316 ¥ 0.316 ¥ 0.1/316 mV )
= 3.0 – 0.5 = 2.5V
(15)
(16)
V
APC(min) = (VSETVGBC )/VSLP – log10 kGOVCW /VZ
3
= (0.24 ¥ 1)/0.48 – log10 (0.0316 ¥ 0.316 ¥ 0.1/316 mV )
= 0.5 – 0.5 = zero
both of which results are consistent with the assumptions made
about the amplifier control function. Note that the second term
is independent of the delivered power and a fixed function of
the drive power.
–7
0
0.5
1.0
1
1.5
2.0
2.5
V , P
1
V
– V
APC
Figure 5. Typical Power-Control Curve
A Note About Power Equivalency
In using the AD8315, it must be understood that log amps do not
fundamentally respond to power. It is for this reason that dBV
(decibels above 1 V rms) are used rather than the commonly used
metric of dBm. The dBV scaling is fixed, independent of termi-
nation impedance, while the corresponding power level is not.
For example, 224 mV rms is always –13 dBV (with one further
condition of an assumed sinusoidal waveform; see the AD640
data sheet for more information about the effect of waveform on
logarithmic intercept), and this corresponds to a power of 0 dBm
when the net impedance at the input is 50 W. When this impedance
is altered to 200 W, however, the same voltage corresponds to a
power level that is four times smaller (P = V2/R) or –6 dBm. A
dBV level may be converted to dBm in the special case of a 50 W
system and a sinusoidal signal by simply adding 13 dB (0 dBV
is then, and only then, equivalent to 13 dBm).
V
V
CW
DIRECTIONAL COUPLER
RF
RF PA
RF DRIVE: UP
TO 2.5GHz
V
= kV
V
IN
RF
V
AD8315
APC
SET
RESPONSE-SHAPING
OF OVERALL CONTROL-
LOOP (EXTERNAL CAP)
C
FLT
Figure 4. Idealized Control Loop for Analysis
Finally, using the loop time constant for these parameters and
an illustrative value of 2 nF for the filter capacitor CFLT
:
Therefore, the external termination added ahead of the AD8315
determines the effective power scaling. This will often take the
form of a simple resistor (52.3 W will provide a net 50 W input),
but more elaborate matching networks may be used. The choice
of impedance determines the logarithmic intercept, that is, the
input power for which the VSET versus PIN function would
cross the baseline if that relationship were continuous for all
values of VIN. This is never the case for a practical log amp; the
intercept (so many dBV) refers to the value obtained by the
minimum error straight line fit to the actual graph of VSET versus
PIN (more generally, VIN). Where the modulation is complex, as
in CDMA, the calibration of the power response needs to be
adjusted; the intercept will remain stable for any given arbitrary
waveform. When a true power (waveform independent) response is
needed, a mean-responding detector, such as the AD8361,
should be considered.
TO = (VGBC /VSLP )T
(17)
= (1/0.48) 3.07 ms ¥ 2(nF) = 12.8 ms
Practical Loop
At the present time, power amplifiers, or VGAs preceding such
amplifiers, do not provide an exponential gain characteristic. It
follows that the loop dynamics (the effective time constant) will
vary with the setpoint, since the exponential function is unique
in providing constant dynamics. The procedure must, therefore,
be as follows. Beginning with the curve usually provided for the
power output versus the APC voltage, draw a tangent at the
point on this curve where the slope is highest (see Figure 5).
Using this line, calculate the effective minimum value of the
variable VGBC and use it in Equation 17 to determine the time
constant. Note that the minimum in VGBC corresponds to the
maximum rate of change in the output power versus VAPC
.
For example, suppose it is found that, for a given drive power,
the amplifier generates an output power of P1 at VAPC = V1 and
P2 at VAPC = V2. Then, it is readily shown that:
The logarithmic slope, VSLP in Equation 1, which is the amount
by which the setpoint voltage needs to be changed for each decibel
of input change (voltage or power), is, in principle, independent
of waveform or termination impedance. In practice, it usually
falls off somewhat at higher frequencies, due to the declining
gain of the amplifier stages and other effects in the detector
cells (see TPC 13).
(18)
VGBC = 20(V2 –V )/(P –P1)
1
2
This should be used to calculate the filter capacitance. The
response time at high and low power levels (on the “shoulders”
REV. B
–11–
ADI [ ADI ]
