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I'm reading a paper whose result I am trying to replicate. One of the circuit elements is a clipper circuit as shown below.

This is a very straight-forward circuit. The positive value is limited to \$V_1 + V_{diode}\$ and the negative value is limited to \$V_2 - V_{diode}\$, where \$V_1\$ and \$V_2\$ are the two series voltage sources.

The author is trying to use this circuit to product a sigmoidal voltage response.

According to my equations above, this means, assuming \$V_{diode} = 0.7\$, we should set \$V_1 = 0.3\$ and \$V_2 = 0.7\$. Both the chose \$V_1 = 0.3\$ and \$V_2 = -0.7\$.

Here is a simulation with \$V_1 = 0.3\$ and \$V_2 = 0.7\$.

Here is a simulation with \$V_1 = 0.3\$ and \$V_2 = -0.7\$.

The author says the output voltage is linear in the range between 0 and 1V. Using his values for the series voltages source, the output voltage is linear in the range between -1.4 and 1V. I tried using his values to replicate this circuit:

... using the resistance values from the image below, but had vastly different node voltages:

I also tried using \$V_1 = 0.3\$ and \$V_2 = 0.7\$, but also didn't get similar node voltages. Am I missing something or did the author mistyped \$V_2 = -0.7\$ for \$V_2 = 0.7\$.

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  • \$\begingroup\$ Since you don't have $V_1$ or $V_2$ marked on the schematic, it is impossible to know the polarity of "-0.7 V" in the text. \$\endgroup\$
    – Ben Voigt
    Jun 16, 2021 at 15:27
  • \$\begingroup\$ The author didn't specify. I took V1 to be the left voltage source and V2 to be the right voltage source, and their orientation as shown in the figure. \$\endgroup\$
    – Leonhard Euler
    Jun 16, 2021 at 16:14

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The linear region is the region where none of the diodes conduct. If we assume the polarity of the voltage sources are as shown in the schematic then it is obvious. You are right. He probably mistyped it.

It should be noted though that this doesn't seem a practical circuit to implement. Would you care to share the paper's title?

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  • \$\begingroup\$ Here's a link to the paper: arxiv.org/abs/2006.01981 My goal is to just replicate the result in simulation. \$\endgroup\$
    – Leonhard Euler
    Jun 16, 2021 at 19:10
  • \$\begingroup\$ Thanks Chad. I read the paper. Are you trying to simulate it in LTSpice? \$\endgroup\$
    – Grease Monkey
    Jun 17, 2021 at 6:14
  • \$\begingroup\$ Yes, I tried to use his resistor values for the XOR circuit on his paper and drew the circuit as he indicated but got very different results. Here's the schematic I used. It looks more complicated than it needs to be because I tried to make it more realistic. Because I wasn't getting the expected values, I decide to train a circuit myself. Here's a python script I threw together to do just that. It learns for a while, but after that the loss starts increasing again. \$\endgroup\$
    – Leonhard Euler
    Jun 17, 2021 at 6:35
  • \$\begingroup\$ In the spice model, why are R12 and R13 in parallel? \$\endgroup\$
    – Grease Monkey
    Jun 17, 2021 at 7:04
  • \$\begingroup\$ Sorry, that's an old circuit I sent you. I had cleaned things up. This is the circuit I am using. There are a couple of things I have noticed that maybe you can shed some light on: 1. The choice of the series voltage sources for the non-linearity affect the result greatly. 2. While training the resistances seem to change extremely slowly. As a result, the random values that they get initialized almost determines how well the training goes. \$\endgroup\$
    – Leonhard Euler
    Jun 17, 2021 at 7:26

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