Datasheet LT6604-10 (Analog Devices) - 10
| 制造商 | Analog Devices |
| 描述 | Dual Very Low Noise, Differential Amplifier and 10MHz Lowpass Filter |
| 页数 / 页 | 16 / 10 — APPLICATIONS INFORMATION. Evaluating the LT6604-10. Figure 3. Figure 4 |
| 文件格式/大小 | PDF / 208 Kb |
| 文件语言 | 英语 |
APPLICATIONS INFORMATION. Evaluating the LT6604-10. Figure 3. Figure 4
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LT6604-10
APPLICATIONS INFORMATION
In Figure 3 the LT6604-10 is providing 12dB of gain. The The voltage at VMID, for VS = 3.3V, is 1.65V. The voltage at gain resistor has an optional 62pF in parallel to improve the DAC pins is given by: the passband fl atness near 10MHz. The common mode R1 R1•R2 output voltage is set to 2V. VDAC = VMID • +I R1+R2+ 402 IN • R1+R2 Use Figure 4 to determine the interface between the LT6604-10 and a current output DAC. The gain, or “tran- =103mV +I simpedance,” is defi ned as A = V IN • 43.6Ω OUT/IIN. To compute the transimpedance, use the following equation: I + – IN is IIN or IIN . The transimpedance in this example is 50.4Ω. A = 402 •R1(Ω) (R1+R2)
Evaluating the LT6604-10
By setting R1 + R2 = 402Ω, the gain equation reduces to The low impedance levels and high frequency operation A = R1 (Ω). of the LT6604-10 require some attention to the matching The voltage at the pins of the DAC is determined by R1, R2, networks between the LT6604-10 and other devices. The the voltage on V + previous examples assume an ideal (0Ω) source impedance MID and the DAC output current (IIN or I – and a large (1k) load resistance. Among practical examples IN ). Consider Figure 4 with R1 = 49.9Ω and R2 = 348Ω. where impedance must be considered is the evaluation of the LT6604-10 with a network analyzer. 62pF 5V 0.1μF V V 100Ω 25 – 4 3 3 VIN – + + 34 1/2 + 27 VOUT VOUT 2 LT6604-10 2 6 0.01μF – – VOUT 500mVP-P (DIFF) + 2 – VOUT 1 1 V 29 IN + V + 100Ω IN 7 0 t – + 0 t V 660410 F03 IN – 2V 62pF
Figure 3
CURRENT 3.3V OUTPUT 0.1μF DAC – 25 IIN R2 4 – 27 34 + V + OUT R1 1/2 0.01μF 6 LT6604-10 I + R2 – IN 2 – + VOUT 29 7 R1 V + – OUT – VOUT 402 • R1 = I + – I – R1 + R2 660410 F04 IN IN
Figure 4
660410fb 10
APPLICATIONS INFORMATION
In Figure 3 the LT6604-10 is providing 12dB of gain. The The voltage at VMID, for VS = 3.3V, is 1.65V. The voltage at gain resistor has an optional 62pF in parallel to improve the DAC pins is given by: the passband fl atness near 10MHz. The common mode R1 R1•R2 output voltage is set to 2V. VDAC = VMID • +I R1+R2+ 402 IN • R1+R2 Use Figure 4 to determine the interface between the LT6604-10 and a current output DAC. The gain, or “tran- =103mV +I simpedance,” is defi ned as A = V IN • 43.6Ω OUT/IIN. To compute the transimpedance, use the following equation: I + – IN is IIN or IIN . The transimpedance in this example is 50.4Ω. A = 402 •R1(Ω) (R1+R2)
Evaluating the LT6604-10
By setting R1 + R2 = 402Ω, the gain equation reduces to The low impedance levels and high frequency operation A = R1 (Ω). of the LT6604-10 require some attention to the matching The voltage at the pins of the DAC is determined by R1, R2, networks between the LT6604-10 and other devices. The the voltage on V + previous examples assume an ideal (0Ω) source impedance MID and the DAC output current (IIN or I – and a large (1k) load resistance. Among practical examples IN ). Consider Figure 4 with R1 = 49.9Ω and R2 = 348Ω. where impedance must be considered is the evaluation of the LT6604-10 with a network analyzer. 62pF 5V 0.1μF V V 100Ω 25 – 4 3 3 VIN – + + 34 1/2 + 27 VOUT VOUT 2 LT6604-10 2 6 0.01μF – – VOUT 500mVP-P (DIFF) + 2 – VOUT 1 1 V 29 IN + V + 100Ω IN 7 0 t – + 0 t V 660410 F03 IN – 2V 62pF
Figure 3
CURRENT 3.3V OUTPUT 0.1μF DAC – 25 IIN R2 4 – 27 34 + V + OUT R1 1/2 0.01μF 6 LT6604-10 I + R2 – IN 2 – + VOUT 29 7 R1 V + – OUT – VOUT 402 • R1 = I + – I – R1 + R2 660410 F04 IN IN
Figure 4
660410fb 10