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    • 4. 发明公开
    • OPERATIONAL AMPLIFIER CIRCUIT WITH DC OFFSET SUPPRESSION
    • 操作流程麻省理工学院UNTERDRÜCKUNGDES DC-OFFSETS
    • EP3096452A1
    • 2016-11-23
    • EP15168460.2
    • 2015-05-20
    • ALi Corporation
    • Berner, Raphael
    • H03F3/45
    • H03F3/45475H03F3/45183H03F3/45659H03F3/45748H03F3/45946H03F3/45973H03F2203/45424H03F2203/45441H03F2203/45521H03F2203/45681H04B1/30
    • An Op-Amp circuit (100), a low pass filter and a RF receiver using the Op-Amp circuit are provided. The Op-Amp circuit includes a first operational amplifier (110) and a feedback circuit (120; 130). The feedback circuit is coupled to an output stage (113) of the first operational amplifier and generates a feedback signal (fb1; fb2) according to a reference voltage (ref) and one of a pair of differential output signals out_p; out_n). The feedback circuit outputs the feedback signal to the output stage (113) to adjust a bias current of the output stage corresponding to the one of the pair of differential output signals, so that a DC voltage level of the one of the pair of differential output signals is close to a voltage level of the reference voltage (ref) and a DC offset of the one of the pair of differential output signals is suppressed.
    • 提供了运算放大器电路(100),低通滤波器和使用运算放大器电路的RF接收器。 运算放大器电路包括第一运算放大器(110)和反馈电路(120; 130)。 所述反馈电路耦合到所述第一运算放大器的输出级(113),并且根据参考电压(ref)产生反馈信号(fb1; fb2),并且一对差分输出信号out_p中的一个; OUT_N)。 反馈电路将反馈信号输出到输出级(113),以调整与一对差分输出信号中的一个对应的输出级的偏置电流,使得该对差分输出中的一个的直流电压电平 信号接近参考电压(ref)的电压电平,并且抑制该对差分输出信号之一的DC偏移。
    • 8. 发明公开
    • Level shift circuit for differential signals
    • PegelschiebeschaltungfürDifferenzsignale。
    • EP0265044A1
    • 1988-04-27
    • EP87307134.4
    • 1987-08-12
    • TEKTRONIX INC.
    • Diller, Calvin DeanGladden, Donald D.
    • H03F3/45G01R13/22
    • H03F3/26H03F3/45475H03F3/45946
    • A level shift circuit for shifting the common mode level of the output signal provided by a differential amplifier has first and second input terminals for receiving the output signal from the differential amplifier, which has a Thevenin source impedance R s , and also has first and second output terminals that are connected to a reference voltage level through a load impedance R o . The circuit comprises a differential transconductance amplifier having two output terminals that are connected respectively to the first and second output terminals of the circuit. The amplifier has the property that it responds to an input voltage E e between its input terminals by providing a current equal to E e g m1 /(1+τs) (where g m1 is the transconductance of the amplifier, τ is the response time constant of the amplifier and s is the Laplace transform operator) at its output terminals. Two equal-valued capacitors are connected respectively between the first input terminal of the circuit and the first output terminal of the circuit and between the second input terminal of the circuit and the second output terminal of the circuit, the capacitance C c of each capacitor being such that R s C c is much greater than τ. The circuit also comprises two equal-valued resistors connected in series between the first input terminal of the circuit and the second output terminal of the circuit and having their connection point connected to one of the input terminals of the amplifier, and two more equal-valued resistors connected in series between the first output terminal of the circuit and the second input terminal of the circuit and having their connection point connected to the other of the two input terminals of the amplifier. The value of g m1 is selected to be equal to (1/2R s + 1/R).
    • 用于移动由差分放大器提供的输出信号的共模电平的电平移位电路具有用于接收来自差分放大器的输出信号的第一和第二输入端,差分放大器具有戴维南源阻抗R s,并且还具有第一和第二 输出端子,通过负载阻抗R o连接到参考电压电平。 电路包括差分跨导放大器,其具有分别连接到电路的第一和第二输出端子的两个输出端子。 放大器具有通过提供等于E的电流来响应输入端之间的输入电压E e的特性,例如m1 /(1+ tau s)(其中g m1是放大器的跨导,τ是响应时间 放大器的常数和s是拉普拉斯变换算子)。 电路的第一输入端子与电路的第一输出端子和电路的第二输入端子和电路的第二输出端子之间分别连接两个等价电容器,每个电容器的电容C c为 使得R sC c比tau大得多。 该电路还包括串联连接在电路的第一输入端和电路的第二输出端之间并具有连接点连接到放大器的一个输入端的两个等值电阻,以及两个等值的 电路串联连接在电路的第一输出端和电路的第二输入端之间,并将它们的连接点连接到放大器的两个输入端中的另一个。 选择g m1的值等于(1 / 2Rs + 1 / R)。