使用HP 8753B 2.4GHz测量空心铆钉电感

我设计了测试方法来确定ADS中使用的SMT库的值（和寄生值），因此我对此有很多经验。
首先，由于两者之间的互感，两个接地铆钉不会是1铆钉的电感的1/2。
将它们放在相反的两侧有帮助，但仍然会有一个共同的感应器。
第二：SMA连接器的边缘可能约为70 fF左右，因此您必须从测量中退出。
你真的在测量SMA边缘帽的平行电路和电感。
现在凌晨1点在中国，为时已晚，我无法启动ADS模拟，但是你得到了图片。
最后，当你使用铆钉时，边缘区域将完全不同。
如果您使用它将信号从电路板的一侧传递到另一侧，则其有效电感将非常强烈地依赖于附近的接地。
事实上，你可以认为它没有任何电感;
相反，认为它是高阻抗的传输线。
这是解释地面效应的一种方法。
另一种是试图模拟并联电容。
所以，如果你可以发布一个你打算如何使用铆钉的图纸，我可以更好地了解它的模型。



     以下为原文

  I designed the test methods to determine the values (and parasitic values ) for the SMT library used in ADS, so I have a lot of experience with this.

First, two rivets to ground won't be 1/2 the inductance of 1 rivet because of mutual inductance between the two. Putting them on opposite sides help but  there will still be a mutual inductacne.

Second: the fringing from an SMA connector is likely to be about 70 fF or so, so you have to back out that from the measurement.  You are really measuring the parrallel circuit of the fringing cap of the SMA and the inducatance.  It's 1 AM in china and too late for me to spin up an ADS simulation, but you get the picture.

Lastly, when you use the rivet, the fringing fields will be entirely different.  If you are using it to pass signals from one side of a board to another, it's effective inductance will depend very strongly on nearby grounds. In fact, you could consider that it hasn't any inductance; rather, consider it an transmission line of high impedance.  That's the one way to account for the ground effects.  The other is to try to model a shunt capacitance.

So, if you can post a drawing of how you intend to use the rivet, I can give a better idea of what to model it as.

我怀疑你是如何使用它的。
我的第一个HP标志是一个公共基极缓冲放大器，其基极电感是工作中的关键因素（用于8753A）。
要在“使用”情况下测量电感，请制作直通传输线（无晶体管），然后通过铆钉将传输线中心接地。
也就是说，用分流接地电感制作传输线。
测量S21和/或S11。
然后，您可以将等效电感建模为L并联50欧姆。
当然，在低频时，没有传输。
我想我



     以下为原文

  I suspected that was how you were using it.  My first dsign for HP was a common base buffer amplifier where the base inductance was THE critical factor in the operation (used in the 8753A).

To measure the inductance in an "as used" case, make a thru transmission line (no transistor) and then ground the transmission line center through the rivet.  That is, make a transmission line with a shunt ground inductor.  Measure S21 and or S11.  You can then model the equivalent inductance as L in parallel with 50 ohms.  Of course at low freq there is no transmission.
I think I

因此，快速的ADS模拟显示，pH值约为14.4。
对您的过渡器接地可能是一个很好的估计。
可能有趣的是看到切割一个或两个顶侧接地的效果，并且还看到通过中心导体到背板接地的单个铆钉是什么样的。
我猜测上层土地给你带来了很多好处。
ShuntL.PNG22.8 KB



     以下为原文

  So a quick ADS simulation shows that to be about 14.4 pH equivalent inductance.  probably a good estimate for your transitor grounding.  Might be interesting to see the effect of cutting one or both top side grounds, and seeing also what a single rivet to ground through the center conductor to backplane looks like.  I'm guessing the topside ground is doing you a lot of good.

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• ShuntL.PNG22.8 KB
  

嗨，Dunsmore博士，非常感谢您的回答和解释，我当然必须朝着正确的方向前进。
我已经买了你的最后一本书，这对我来说可能非常珍贵。
圣诞快乐，为您和您的家人带来新年快乐。
瑞诺



     以下为原文

  Hi Dr Dunsmore , many thanks for your answer and explanations, I must certainly be in the right direction.

I have bought your last book and it probably will be very precious for me .

Merry Xmas and happy new year for you and all your family.
Guerrino