# Clarification on a clipper circuit

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\$$.

• 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. Commented Jun 16, 2021 at 15:27
• 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. Commented Jun 16, 2021 at 16:14