Datasheet AD8350 (Analog Devices) - 8

ManufacturerAnalog Devices
DescriptionLow Distortion 1.0 GHz Differential Amplifier
Pages / Page14 / 8 — AD8350. APPLICATIONS. S/2. LS/2. Using the AD8350. RLOAD. RS/2. CAC. 0.1. …
RevisionC
File Format / SizePDF / 247 Kb
Document LanguageEnglish

AD8350. APPLICATIONS. S/2. LS/2. Using the AD8350. RLOAD. RS/2. CAC. 0.1. ENBL (5V). +VS (5V TO 10V). 0.001. SOURCE. LOAD. Z = 100. Z = 200

AD8350 APPLICATIONS S/2 LS/2 Using the AD8350 RLOAD RS/2 CAC 0.1 ENBL (5V) +VS (5V TO 10V) 0.001 SOURCE LOAD Z = 100 Z = 200

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AD8350 APPLICATIONS C L AC C S/2 AC LS/2 Using the AD8350
Figure 1 shows the basic connections for operating the AD8350.
8 7 6 5 R
A single supply in the range 5 V to 10 V is required. The power
S/2 AD8350
supply pin should be decoupled using a 0.1 μF capacitor. The
C P CP
ENBL pin is tied to the positive supply or to 5 V (when V
VS RLOAD
CC =
+
10 V) for normal operation and should be pulled to ground to put the device in sleep mode. Both the inputs and the outputs
RS/2 1 2 3 4
have dc bias levels at midsupply and should be ac-coupled. Also shown in Figure 1 are the impedance balancing requirements,
LS/2 C LS/2 AC CAC
either resistive or reactive, of the input and output. With an
0.1 F
input and output impedance of 200 Ω, the AD8350 should be
ENBL (5V)
driven by a 200 Ω source and loaded by a 200 Ω impedance. A
+VS (5V TO 10V)
reactive match can also be implemented. Figure 3. Reactively Matching the Input and Output
C2 C4 0.001 SOURCE F 0.001 F LOAD L CAC CAC L Z = 100 S S 8 7 6 5 AD8350 8 7 6 5 RS AD8350 Z = 200 + C C V P P S RLOAD + 1 2 3 4 Z = 100 1 2 3 4 C1 C3 0.001 F 0.001 F C5 C C 0.1 F AC AC ENBL (5V) 0.1 F +V ENBL (5V) S (5V TO 10V)
Figure 1. Basic Connections for Differential Drive
+VS (5V TO 10V)
Figure 4. Single-Ended Equivalent Circuit Figure 2 shows how the AD8350 can be driven by a single- ended source. The unused input should be ac-coupled to ground. When the source impedance is smaller than the load impedance, When driven single-endedly, there will be a slight imbalance in a step-up matching network is required. A typical step-up network the differential output voltages. This will cause an increase in is shown on the input of the AD8350 in Figure 3. For purely the second order harmonic distortion (at 50 MHz, with V resistive source and load impedances the resonant approach may CC = 10 V and V be used. The input and output impedance of the AD8350 can be OUT = 1 V p-p, –59 dBc was measured for the second harmonic on AD8350-15). modeled as a real 200 Ω resistance for operating frequencies less than 100 MHz. For signal frequencies exceeding 100 MHz, classi-
LOAD
cal Smith Chart matching techniques should be invoked in order to deal with the complex impedance relationships. Detailed S
C2 C4 8 7 6 5 0.001 F 0.001 F
parameter data measured differentially in a 200 Ω system can be
AD8350
found in Tables II and III.
Z = 200
For the input matching network the source resistance is less
+
than the input resistance of the AD8350. The AD8350 has a nominal 200 Ω input resistance from Pins 1 to 8. The reactance
1 2 3 4
of the ac-coupling capacitors, CAC, should be negligible if 100 nF
SOURCE
capacitors are used and the lowest signal frequency is greater
Z = 200 C1 C3
than 1 MHz. If the series reactance of the matching network
0.001 F 0.001 F C5
inductor is defined to be XS = 2 π f LS, and the shunt reactance
ENBL (5V) 0.1 F
of the matching capacitor to be XP = (2 π f CP)–1, then:
+VS (5V TO 10V)
RS RLOAD RS Figure 2. Basic Connections for Single-Ended Drive X = × where = × S XP RLOAD (1) XP RLOAD – RS
Reactive Matching
In practical applications, the AD8350 will most likely be matched For a 70 MHz application with a 50 Ω source resistance, and using reactive matching components as shown in Figure 3. assuming the input impedance is 200 Ω, or RLOAD = RIN = 200 Ω, Matching components can be calculated using a Smith Chart or then XP = 115.5 Ω and XS = 86.6 Ω, which results in the follow- by using a resonant approach to determine the matching network ing component values: that results in a complex conjugate match. In either situation, CP = (2 π × 70 × 106 × 115.5)–1 = 19.7 pF and the circuit can be analyzed as a single-ended equivalent circuit to ease calculations as shown in Figure 4. LS = 86.6 × (2 π × 70 × 106)–1 = 197 nH –8– REV. C