User Guide

Chapter 9 AC analyses
240
Using a DC source
Consider the circuit shown here. At the DC bias point,
PSpice calculates the partial derivatives which determine
the linear response of the multiplier as follows:
For this circuit, this equation reduces to:
This means that the multiplier acts as an amplifier of the
AC input with a gain that is set by the DC input.
Caution: multiplying AC sources
Suppose that you replace the 2 volt DC source in this
example with an AC source with amplitude 1 and no DC
value (DC=0). When PSpice computes the bias point, there
are no DC sources in the circuit, so all nodes are at 0 volts
at the bias point. The linear equivalent of the multiplier
block is a block with gain 0, which means that there is no
output voltage at the fundamental frequency.
VOut()VIn1()
VOut()∂
VIn1()∂
----------------------
⋅ VIn2()
VOut()∂
VIn2()∂
----------------------
⋅+=
VIn1()= VIn2()⋅ VIn2()VIn1()⋅+
VOut()VIn1()2VIn2()0⋅+⋅=
T
h
is is exact
l
y
h
ow a
d
ou
bl
e-
b
a
l
ance
d
mixer behaves. In practice, this is a simple
multiplier.
Note
A double-balanced mixer with inputs
at the same frequency would produce
outputs at DC at twice the input frequency,
but these terms cannot be seen with a
linear, small-signal analysis.
Pspug.book Page 240 Wednesday, November 11, 1998 1:14 PM