AD8315ARM datasheet(15/24 Pages) AD

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AD8315

Rev. C | Page 15 of 24

Example

Assume that the gain magnitude of the power amplifier runs

from a minimum value of ×0.316 (−10 dB) at VAPC = 0 to ×100

(40 dB) at VAPC = 2.5 V. Applying Equation 9, GO = 0.316 and

VGBC = 1 V. Using a coupling factor of k = 0.0316 (that is, a

30 dB directional coupler) and recalling that the nominal value

of VSLP is 480 mV and VZ = 316 μV for the AD8315, first calculate

the range of values needed for VSET to control an output range of

+33 dBm to −17 dBm. This can be found by noting that, in the

steady state, the numerator of Equation 7 must be 0, that is:

VSET = VSLP log10 (kVPA/VZ)

(12)

where VIN is expanded to kVPA, the fractional voltage sample of

the power amplifier output. For 33 dBm, VPA = 10 V rms, which

evaluates to

VSET (max) = 0.48 log10 (316 mV/316 μV) = 1.44 V

(13)

For a delivered power of −17 dBm, VPA = 31.6 mV rms

VSET (min) = 0.48 log10 (1 mV/316 μV) = 0.24 V

(14)

Check that 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.

Now, the value of VAPC is of interest, although it is a dependent

parameter, inside the loop. It depends on the characteristics of

the power amplifier, and the value of the carrier amplitude VCW.

Using the control values previously derived, that is, GO = 0.316

and VGBC = 1 V, and assuming the applied power is fixed at

−7 dBm (so VCW = 100 mV rms), the following is true using

Equation 11

VAPC(max) = (VSETVGBC)/VSLP − log10 kGOVCW/VZ

= (1.44 × 1)/0.48 − log10(0.0316 × 0.316 × 0.1/316 μV)

= 3.0 − 0.5 = 2.5 V

(15)

VAPC(min) = (VSETVGBC)/VSLP − log10 kGOVCW/VZ

= (0.24 × 1)/0.48 − log10(0.0316 × 0.316 × 0.1/316 μV)

= 0.5 − 0.5 = 0

(16)

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.

RF PA

DIRECTIONAL COUPLER

RF DRIVE: UP

TO 2.5GHz

AD8315

VRF

VCW

VIN = kVRF

VSET

VAPC

CFLT

RESPONSE-SHAPING

OF OVERALL CONTROL-

LOOP (EXTERNAL CAP)

Figure 34. 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

TO = (VGBC/VSLP) T

= (1/0.48)3.07 μs × 2 (nF) = 12.8 μs

(17)

PRACTICAL LOOP

At 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)

varies with the setpoint because 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 vs. the APC voltage, draw a tangent at the

point on this curve where the slope is highest (see Figure 35).

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 vs. 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

VGBC = 20 (V2 − V1)/(P2 − P1)

(18)

This should be used to calculate the filter capacitance. The

response time at high and low power levels (on the shoulders

of the curve shown in Figure 35) is slower. Note also that it is

sometimes useful to add a 0 in the closed-loop response by

placing a resistor in series with CFLT. For more information on

this, see the Transient Response section.

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