Does AD9277 accept bipolar voltage input？ （-1V ~ +1V）
AD9277接受双极性电压输入吗？ （-1V ~ +1V）
It diode pair will help clip the input signal but the LNA will not be damaged if presented with +-1V. It will compress the input and operate in a nonlinear region. We specify the .1 dB compression point for that reason. If your useful signal is above 733 mVpp then you need to use a voltage divider to scale it down and operate in the linear region of the LNA.
Thank you for your interest in AD9277. If I understand your question correctly, no the analog input cannot go negative with respect to ground.
As can be seen in Table 1 of the datasheet, the input common mode voltage is 1V. This means that the input signal will swing above and below 1V by the peak-to-peak input-swing divided-by-2. The maximum amplitude of the swing depends on the gain setting, but in no case does it go negative with respect to ground.
Thank you very much for your reply!
I noticed Table 1 of AD9277's datasheet(Page 5), but I still didn't understand quitely. In other words, if I design the circuit like the picture below(red rectangle)， could VIN be -1V ~ +1V ，or must VIN be within (Vcommon - Vswing/2) ~ (Vcommon + Vswing/2)?
In my opinion, this circuit will adjust the input voltage to 1V-common-voltage because the inputs are capacitively coupled. But I am not quite sure.
If you apply VIN at the point shown in your circuit picture, it is OK to swing above and below ground, but I believe that -1V to +1V will be too great an amplitude (2Vp-p single ended). If we assume that Cs does not attenuate your signal much, most of that 2Vp-p will make it to the input. As shown below from Table 1 of the AD9277 datasheet, the Input Voltage Range is much less than 2Vp-p.
Hi Dougl, Thank you very much for your reply!
【1】It's a real pity that 2Vp-p is too great an amplitude. How could I design the circuit if the backscattered signal can't ensure the safe swing volatage? The transmit/receive(T/R)-switch chip which is between ultrasonic probe and AD9277 has an output of -1V ~ +1V. There are two output clamping diodes shown in picture below in this chip. Is it necessary to add another clamping diodes between T/R chip and AD9277, or add a pull-down resistor?
【2】I check the User Guides of evaluation board EVAL-AD9277 provided by Analog Devices (http://www.analog.com/media/cn/technical-documentation/user-guides/UG-016.pdf). It offers Analog Input Circuits of AD9277 shown in the picture below(Page11-12). The diodes D101(red rectangle) is MMBD4148SE whose forward voltage is 1V(http://www.mouser.com/ds/2/149/MMBD4148SE-889741.pdf). It means that the input voltage within -1V ~ +1V would pass the diodes. Will it damage AD9277?
Thank you. Shangey
I'll ask the application engineer for AD9277 to look at your question. Hopefully he will be able to help.
If the AD9277 is overloaded due to the T/R SW leakage then it will not damage the chip.. The LNA output will saturate and clip but it will recover quickly. The LNA input levels were designed to handle the medical ultrasound signal levels from different probes, which is up to 733 mVpp. If your probe returns higher signal then it will clip into the LNA unless you use voltage divider circuit to stay in linear range. You can design the feedback resistors to give you 50 ohm equivalent input resistance or lower to reduce input voltage range.
Thank you very much for your reply! Do you mean: if I design the circuit as it shown in the User Guides of evaluation board EVAL-AD9277 (the picture below, Page11-12), the input signal will clip into 733 mVpp. The information out of range would be lost. And If the input voltage is -1V~+1V, it could not damage AD9277 (?).
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