ADA4310-1 数据手册DataSheet 下载
... Twisted pair line drivers Power line communications Video line drivers ARB line drivers I/Q channel amplifiers ...
... Twisted pair line drivers Power line communications Video line drivers ARB line drivers I/Q channel amplifiers ...
AD8350 数据手册DataSheet 下载
... Low Noise Figure: 5.9 dB @ 250 MHz Two Gain Versions: AD8350-15: 15 dB AD8350-20: 20 dB –3 dB Bandwidth: 1.0 GHz Single Supply Operation: 5 V to 10 V ...
... Low Noise Figure: 5.9 dB @ 250 MHz Two Gain Versions: AD8350-15: 15 dB AD8350-20: 20 dB –3 dB Bandwidth: 1.0 GHz Single Supply Operation: 5 V to 10 V ...
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LTC1046 - Inductorless 5V to -5V Converter
... approximately two times higher than output current, and losses will occur on both the charge and discharge cycle. This means that using a capacitor with 1Ω of ESR for CIN will have the same effect as increasing the output impedance of the LTC1046 by 4Ω. This represents a significant increase in the ...
... approximately two times higher than output current, and losses will occur on both the charge and discharge cycle. This means that using a capacitor with 1Ω of ESR for CIN will have the same effect as increasing the output impedance of the LTC1046 by 4Ω. This represents a significant increase in the ...
basics of electrical circuits laboratory
... physical quantities will be presented, then, a simple electrical circuit will be set up for the experiment and voltage-current values on this circuit will be measured. It will be checked that whether the measured values satisfy Kirchhoff’s current and voltage laws or not. In addition, how to utilize ...
... physical quantities will be presented, then, a simple electrical circuit will be set up for the experiment and voltage-current values on this circuit will be measured. It will be checked that whether the measured values satisfy Kirchhoff’s current and voltage laws or not. In addition, how to utilize ...
AD7400A 数据手册DataSheet下载
... on-chip digital isolation based on Analog Devices, Inc., iCoupler® technology. The AD7400A operates from a 5 V power supply and accepts a differential input signal of ±250 mV (±320 mV full scale). The analog input is sampled continuously by the analog modulator, eliminating the need for external sam ...
... on-chip digital isolation based on Analog Devices, Inc., iCoupler® technology. The AD7400A operates from a 5 V power supply and accepts a differential input signal of ±250 mV (±320 mV full scale). The analog input is sampled continuously by the analog modulator, eliminating the need for external sam ...
15 Single-phase series ac circuits
... The phasor diagram is shown in Figure 15.7(b). (Note that in a.c. circuits, the supply voltage is not the arithmetic sum of the p.d.’s across components but the phasor sum) Problem 12. A coil of inductance 159.2 mH and resistance 20 is connected in series with a 60 resistor to a 240 V, 50 Hz sup ...
... The phasor diagram is shown in Figure 15.7(b). (Note that in a.c. circuits, the supply voltage is not the arithmetic sum of the p.d.’s across components but the phasor sum) Problem 12. A coil of inductance 159.2 mH and resistance 20 is connected in series with a 60 resistor to a 240 V, 50 Hz sup ...
raman amplifier
... less constant at a level much higher than the noise floor in the system. In case of EDFA the signal level reaches close to the noise floor just before the amplifier, and just after the amplifier it increases to such a high value that the non-linear effects like SPM, XPM, FWM become effective. In pre ...
... less constant at a level much higher than the noise floor in the system. In case of EDFA the signal level reaches close to the noise floor just before the amplifier, and just after the amplifier it increases to such a high value that the non-linear effects like SPM, XPM, FWM become effective. In pre ...
AC Circuits and Resonance Conclusion
... A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. ...
... A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. ...
Parallel Resistance, Series/Parallel Circuit Combinations, and
... parallel circuits. Reading this appendix before doing Problem 1 will make this question considerably easier. 1.) In this problem you will derive from basic principles the equivalent resistance formula for parallel circuits. (See Lab Appendix) a.) Write an equation relating the power supply voltage ( ...
... parallel circuits. Reading this appendix before doing Problem 1 will make this question considerably easier. 1.) In this problem you will derive from basic principles the equivalent resistance formula for parallel circuits. (See Lab Appendix) a.) Write an equation relating the power supply voltage ( ...
TDA7200 ASK/FSK Single Conversion Receiver Version 1.0
... To demodulate frequency shift keyed (FSK) signals a PLL circuit is used that is contained fully on chip. The Limiter output differential signal is fed to the linear phase detector as is the output of the 10.7 MHz center frequency VCO. The demodulator gain is typically 200µV/kHz. The passive loop fil ...
... To demodulate frequency shift keyed (FSK) signals a PLL circuit is used that is contained fully on chip. The Limiter output differential signal is fed to the linear phase detector as is the output of the 10.7 MHz center frequency VCO. The demodulator gain is typically 200µV/kHz. The passive loop fil ...
Ver 9024 E1.1 Analysis of Circuits (2016) E1.1 Circuit Analysis
... 26. (a) Despite the current IB , it is still the case that V+ = 0 and so, because of the negative feedback, 0−Y P = 0 also. KCL @ P gives 0−U 10 +IB + 40 = 0 from which −4U +40IB −Y = 0 or Y = −4U +40IB . Substituting IB = 0.0001 mA gives Y = −4U + 0.004 so there is an output error of 4 mV. (b) This ...
... 26. (a) Despite the current IB , it is still the case that V+ = 0 and so, because of the negative feedback, 0−Y P = 0 also. KCL @ P gives 0−U 10 +IB + 40 = 0 from which −4U +40IB −Y = 0 or Y = −4U +40IB . Substituting IB = 0.0001 mA gives Y = −4U + 0.004 so there is an output error of 4 mV. (b) This ...
