Reading into the authors explanation of the transconductance value (Kn) it was discovered that this value is not the same transconductance value PSpice uses (KP). The author wastes a paragraph of ink explaining that the value Kn used in the book is *commonly* known as the transconduction parameter, however the reader should refer to the value as the conduction parameter. The author defines the conduction parameter as:

1: Kn = (W*un*Cox)/(2L) and Cox = eox/tox

**THEN** at the bottom of the page it is written that *"we can rewrite the conduction parameter in the form"*:

2: Kn = (k'n/2)(W/L)

The following page defines k'n as the process conduction parameter:

3: k'n = un*Cox

By default Pspice defaults the width and length to .1uM. Therefor equation two becomes:

4: Kn = k'n/2

Based on the evidence that the PSpice output was off by a factor of 2, **k'n** represents the value PSpice considers to be the *transconductance parameter*. The author of the book considers this value to be the *process conduction parameter*. This result was confirmed with other examples from the book.

I think I am more confused now than before. How do I know what value to use when I look at a spec sheet!?!?

You can determine the transfer function \$H(s)\$ of the circuit reasoning on the following circuit:

and thinking of \$V_1\$ and \$V_2\$ as two independent inputs. Since the circuit is linear superimposition applies, and the output (in the s-domain) of the circuit when \$V_2\$ is off is simply that of an inverting amplifier (\$R_3\$ shorts the non inverting input to ground, assuming an ideal op-amp):

\$ V_{out1} = - \dfrac{R_2}{R_1} V_1 \$

Analogously, when \$V_1\$ is off, the circuit acts as a non-inverting amplifier whose input is filtered by the series \$C-R_3\$. Thus applying the non-inverting amp gain formula and the voltage divider formula you get:

\$ V_{out2} = \left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}} V_2 \$

The full response is the sum of the two above:

\$ V_{out} = V_{out1} + V_{out2} =
- \dfrac{R_2}{R_1} V_1 +
\left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}} V_2 \$

Your circuit is like the one I posted, but with \$V_1 = V_2\$, therefore the full response becomes:

\$ V_{out} = V_{in} \cdot \left[
- \dfrac{R_2}{R_1} + \left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}}
\right] \$

from which you get:

\$ H(s) = \dfrac{V_{out}}{V_{in}} =
- \dfrac{R_2}{R_1} + \left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}} \$

This simplifies, after a bit of algebra, into:

\$H(s) = \dfrac{s - \frac{R_2}{R_1 R_3 C}}{s + \frac{1}{R_3 C}} \$

Which shows that the circuit acts as an active filter with a 1st order frequency response.

Such a topology is used, for example, to create all-pass filters if \$R_2 = R_1\$.

**EDIT**

The derivation of the final form of H(s) follows:

\$ H(s)
= - \dfrac{R_2}{R_1} + \left(1 + \dfrac{R_2}{R_1} \right)\dfrac{R_3}{R_3 + \frac{1}{C s}}
= - \dfrac{R_2}{R_1} + \dfrac{R_1 + R_2}{R_1} \dfrac{R_3 C s}{R_3 C s + 1} = \$

\$
= - \dfrac{R_2}{R_1} + \dfrac{(R_1 + R_2)R_3 C s}{R_1(R_3 C s + 1)}
= \dfrac{-R_2(R_3 C s + 1) + (R_1 + R_2)R_3 C s}{R_1(R_3 C s + 1)}
\$

\$
= \dfrac{-R_2 R_3 C s - R_2 + R_1 R_3 C s + R_2 R_3 C s}{R_1(R_3 C s + 1)}
= \dfrac{- R_2 + R_1 R_3 C s }{R_1(R_3 C s + 1)}
= \dfrac{R_1 R_3 C s - R_2 }{R_1 R_3 C s + R_1}
\$

dividing numerator and denominator by \$R_1 R_3 C \$ we get:

\$ H(s)
= \dfrac{s - \frac{R_2}{R_1 R_3 C}}{s + \frac{R_1}{R_1 R_3 C}}
= \dfrac{s - \frac{R_2}{R_1 R_3 C}}{s + \frac{1}{R_3 C}}
\$

## Best Answer

There might be a better way to do it (look at the VPWL and VPWL_ENH sources), but to keep this simple, try putting two sources in series, each producing half the waveform.