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I'm following a paper, using ADS, to find the input impedance of a shunt Schottky diode. The paper's schematic is Fig. 1, and my schematic is Fig. 2. The point is that, I could not get the same result as the paper reported. Here is their steps and my corresponding results:

  1. Their method: Shorterning the Impedance Matching tool, varying Pin. It was said the rectifier operates optimally when the diode's peak reverse voltage, Vr_max reaches Bv(6.4V as in the model). So, they found Vr_max = 6.4V @ Pin = 12.320 dBm (Fig. 3).

  2. My results: Vr_max = peak(Vd). With Pin in the range (-5,15), my Vr_max could not reach 6.4V. At Pin = 12.320 dBm, my Vr_max = 4.509V while Vout = 2.599V (Fig.4). Therefore Zin (input impedance) is totally different than the paper's findings. Could anyone tell me what I did wrong? enter image description here enter image description here enter image description here enter image description here

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  • \$\begingroup\$ There is a wide tolerance on Vbr and Infineon's is Vbr= 4V min @ 100uA yet the typical log curve indicates 5uA at 4V with a shift in the log curve at 2.5V then again at 3.5V \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 5 at 3:40
  • \$\begingroup\$ Thank you very much Tony EE. Could you please explain this in more details? Did you mean Vbr to be Bv, the breakdown voltage of the diode? \$\endgroup\$ – Minh Lam Jan 5 at 3:50
  • \$\begingroup\$ Vbr is the threshold at 100uA similar to Vt threshold for zeners and higher is Vzt for rated at 5mA \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 5 at 4:04
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enter image description here

There is a wide tolerance on Vbr and Infineon's is Vbr= 4V min @ 100uA yet the typical log curve indicates 5uA at 4V with a slight shift in the log curve at 2.5V then again at 3.5V

Your signal shows a slight saturation above 3V and almost 10% compression at 4.5V enter image description here

For Infineon the Absolute maximum Voltage rating is Vr=4V

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  • \$\begingroup\$ Thank you very much Tony. Are you saying that although Bv is said to be 6.4V as mentioned in the paper, it is actually 4V? So, what I need to do is to find the Pin correspding to Vmax=4V instead of 6.4V, if I follow the paper's concept? \$\endgroup\$ – Minh Lam Jan 7 at 7:00
  • \$\begingroup\$ Some Paper's are not worth reading , and I haven't read this one. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 7 at 9:05

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