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AD8139ACP-R2 数据表(PDF) 22 Page - Analog Devices |
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AD8139ACP-R2 数据表(HTML) 22 Page - Analog Devices |
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22 / 24 page ![]() AD8139 Rev. A | Page 22 of 24 Driving a Capacitive Load A purely capacitive load will react with the bondwire and pin inductance of the AD8139, resulting in high frequency ringing in the transient response and loss of phase margin. One way to minimize this effect is to place a small resistor in series with each output to buffer the load capacitance, see Figure 6 and Figure 61. The resistor and load capacitance will form a first- order low-pass filter; therefore, the resistor value should be as small as possible. In some cases, the ADCs require small series resistors to be added on their inputs. –13 –12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 10M 100M 1G 5 FREQUENCY (MHz) RS = 30.1Ω CL = 15pF RS = 60.4Ω CL = 15pF RS = 60.4Ω CL = 5pF RS = 0Ω CL, dm = 0pF RS = 30.1Ω CL = 5pF VS = ±5V VO, dm = 0.1V p-p G = 1 (RF = RG = 200Ω) RL, dm = 1kΩ Figure 61. Frequency Response for Various Capacitive Load and Series Resistance The Typical Performance Characteristics that illustrate transient response versus the capacitive load were generated using series resistors in each output and a differential capacitive load. Layout Considerations Standard high speed PCB layout practices should be adhered to when designing with the AD8139. A solid ground plane is recom- mended and good wideband power supply decoupling networks should be placed as close as possible to the supply pins. To minimize stray capacitance at the summing nodes, the copper in all layers under all traces and pads that connect to the summing nodes should be removed. Small amounts of stray summing-node capacitance will cause peaking in the frequency response, and large amounts can cause instability. If some stray summing-node capacitance is unavoidable, its effects can be compensated for by placing small capacitors across the feedback resistors. Terminating a Single-Ended Input Controlled impedance interconnections are used in most high speed signal applications, and they require at least one line termination. In analog applications, a matched resistive termination is generally placed at the load end of the line. This section deals with how to properly terminate a single-ended input to the AD8139. The input resistance presented by the AD8139 input circuitry is seen in parallel with the termination resistor, and its loading effect must be taken into account. The Thevenin equivalent circuit of the driver, its source resistance, and the termination resistance must all be included in the calculation as well. An exact solution to the problem requires the solution of several simultaneous algebraic equations and is beyond the scope of this data sheet. An iterative solution is also possible and simpler, especially considering the fact that standard 1% resistor values are generally used. Figure 62 shows the AD8139 in a unity-gain configuration driving the AD6645, which is a 14-bit high speed ADC, and with the following discussion, provides a good example of how to provide a proper termination in a 50 Ω environment. The termination resistor, RT, in parallel with the 268 Ω input resistance of the AD8139 circuit (calculated using Equation 19), yields an overall input resistance of 50 Ω that is seen by the signal source. In order to have matched feedback loops, each loop must have the same RG if they have the same RF. In the input (upper) loop, RG is equal to the 200 Ω resistor in series with the (+) input plus the parallel combination of RT and the source resistance of 50 Ω. In the upper loop, RG is therefore equal to 228 Ω. The closest standard 1% value to 228 Ω is 226 Ω and is used for RG in the lower loop. Greater accuracy could be achieved by using two resistors in series to obtain a resistance closer to 228 Ω. Things get more complicated when it comes to determining the feedback resistor values. The amplitude of the signal source generator VS is two times the amplitude of its output signal when terminated in 50 Ω. Thus, a 2 V p-p terminated amplitude is produced by a 4 V p-p amplitude from VS. The Thevenin equivalent circuit of the signal source and RT must be used when calculating the closed-loop gain because in the upper loop RG is split between the 200 Ω resistor and the Thevenin resis- tance looking back toward the source. The Thevenin voltage of the signal source is greater than the signal source output voltage when terminated in 50 Ω because RT must always be greater than 50 Ω. In this case, it is 61.9 Ω and the Thevenin voltage and resistance are 2.2 V p-p and 28 Ω, respectively. Now the upper input branch can be viewed as a 2.2 V p-p source in series with 228 Ω. Since this is a unity-gain application, a 2 V p-p differential output is required, and RF must therefore be 228 × (2/2.2) = 206 Ω. The closest standard value to this is 205 Ω. When generating the Typical Performance Characteristics data, the measurements were calibrated to take the effects of the terminations on closed-loop gain into account. |
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