具有41 V/W光转换增益的10 GHz带宽平衡光接收器

R. Costanzo, Zhanyu Yang, Nicolas Raduazo, A. Beling, S. Bowers
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引用次数: 7

摘要

演示了一种10 GHz带宽的光电接收机。光电接收器由两个平衡配置的光电二极管(PD)组成,为跨阻放大器(TIA)产生单端输入。在设计的偏置下,pd的响应率为0.48 a /W,带宽为15 GHz,可驱动50 D的负载。可调节级联码TIA在130 nm RF CMOS工艺中实现。TIA实现了39 dBD的跨阻增益,完整的光电接收器在10 GHz带宽上实现了41 V/W的转换增益,其中TIA从2 V电源消耗56 mW, pd从+/ - 5 V电源消耗1.2 mA。由于86 pWA/Hz的低噪声等效功率(NEP),实现了高灵敏度。
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A 10 GHz bandwidth balanced photoreceiver with 41 V/W optical conversion gain
A 10 GHz bandwidth photoreceiver is demonstrated. The photoreceiver consists of two photodiodes (PD) in a balanced configuration to generate a single-ended input to the transimpedance amplifier (TIA). The PDs achieves a responsivity of 0.48 A/W, and a 15 GHz bandwidth while driving a 50 D load under the designed biasing. The regulated cascode TIA is implemented in a 130 nm RF CMOS process. The TIA achieves a transimpedance gain of 39 dBD The complete photoreceiver achieves a conversion gain of 41 V/W across a 10 GHz bandwidth with the TIA consuming 56 mW from a 2 V supply, and the PDs drawing 1.2 mA from +/−5 V supplies. High sensitivity is achieved due to a low noise equivalent power (NEP) of 86 pWA/Hz.
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