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ADA4622-1ARJZ-R2 数据表(PDF) 31 Page - Analog Devices |
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ADA4622-1ARJZ-R2 数据表(HTML) 31 Page - Analog Devices |
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31 / 38 page ![]() Data Sheet ADA4622-1/ADA4622-2/ADA4622-4 Rev. E | Page 31 of 38 The following basic transfer function describes the transimpedance gain of the photodiode preamplifier: F F F PHOTO OUT R sC R I V + × = 1 where: IPHOTO is the output current of the photodiode. The parallel combination of RF and CF sets the signal bandwidth (see the I to V gain trace in Figure 96). s refers to the s-plane. Note that RF must be set so the maximum attainable output voltage corresponds to the maximum diode output current, IPHOTO, which allows use of the full output swing. The attainable signal bandwidth with this photodiode preamplifier is a function of RF, the gain bandwidth product (fGBP) of the amplifier, and the total capacitance at the amplifier summing junction, including CS and the amplifier input capacitance, CD and CM. RF and the total capacitance produce a pole with loop frequency (fP). S F P C R f π = 2 1 With the additional pole from the amplifier open-loop response, the two-pole system results in peaking and instability due to an insufficient phase margin (see Figure 95). log f log f G = 1 G = R2C1s OPEN-LOOP GAIN –180° –135° –90° –45° 0° fP fX fGBP Figure 95. Gain and Phase Plot of the Transimpedance Amplifier Design, Without Compensation OPEN-LOOP GAIN f fp G = 1 f fGBP G = 1 + CS/CF fZ fX fN I TO V GAIN –135° –90° –45° 0° 45° 90° G = RFCS(s) Figure 96. Gain and Phase Plot of the Transimpedance Amplifier Design with Compensation Adding CF creates a zero in the loop transmission that compensates for the effect of the input pole, which stabilizes the photodiode preamplifier design because of the increased phase margin. Adding CF also sets the signal bandwidth (see Figure 96). The signal bandwidth and the zero frequency are determined by F F Z C R f π 2 1 = where fZ is the zero frequency. Setting the zero at the fX frequency maximizes the signal bandwidth with a 45° phase margin. Because fX is the geometric mean of fP and fGBP, it can be calculated by GBP P X f f f × = Combining these equations, the CF value that produces fX is GBP F S F f R C C × × π = 2 The frequency response in this case shows approximately 2 dB of peaking and 15% overshoot. Doubling CF and halving the band- width results in a flat frequency response with approximately 5% transient overshoot. |
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