这是一个试探性的答案:每个33120A的接地电压为4000 pF,阻抗为50欧姆。
因此,通过50欧姆+ 4000pF流到地的电流将设置电压降。
这应该确定几MHz范围内的频率截止频率。
如果33120A都没有共同点,就不会发生这种情况。
因此,应该: - 浮动两个33120A中的一个? - 绕过AC的直流电源,使用由电感(约1 mH)和电容(约1 muF)组成的滤波器?
以上来自于谷歌翻译
以下为原文
Here is a tentative for an answer:
Each 33120A has 4000 pF to ground and impedance 50 Ohm. Thus, there will be a voltage drop set by the current flowing to ground through 50 Ohm + 4000 pF. This should determine a frequency cutoff in the range of a few MHz. This would not happen if both 33120A had not common grounds.
So, one should :
- float one of the two 33120A ?
- by-pass the DC source for AC, using a filter made of inductance (about 1 mH) and a capacitance (about 1 muF) ?