icl8013 Intersil Corporation, icl8013 Datasheet - Page 4

• Manufacturer Part Number: icl8013
• Description: 1mhz, Four Quadrant Analog Multiplier
• Manufacturer: Intersil Corporation

Application Information
Detailed Circuit Description
The fundamental element of the ICL8013 multiplier is the
bipolar differential amplifier of Figure 1.
The small signal differential voltage gain of this circuit is
given by:
The output voltage is thus proportional to the product of the
input voltage V
transconductance multiplier of Figure 2, a current source
comprising Q
with the drop across D
A
Substituting r
V
I
V
D
V
OU T
OU T
≈
=
------- -
R
V
Y
Y
V
V
--------------- -
=
=
FIGURE 2. TRANSCONDUCTANCE MULTIPLIER
V
IN
OUT
=
R
I N
V
-------------- - V
kTR
L
qR
IN
2I
FIGURE 1. DIFFERENTIAL AMPLIFIER
E
E


=
L
Y
R
------ -
3
r
and
E
=
, D
(
R
------ -
L
lN
r
Q
V-
E



V
L
X
3
-------
g
OUT
1
1
and the emitter current I
=
M
×
, and R
V+
V
V
V
=
IN
2I
IN
Y
R
E
-------- -
qI
1
L
)
kT
, then
×
V
E
D
qI
------------------ -
R
Y
L
+
V
-
E
k T
is used. If V
OUT
V
R
4
OUT
L
R
D
= K (V
V+
Y
V-
I
1
D
2I
V
E
X
Y
R
x V
Y
L
Y
E
is large compared
) =
. In the simple
kTR
qR
L
Y
(V
X
x V
Y
)
ICL8013
There are several difficulties with this simple modulator:
The first problem relates to the method of converting the V
voltage to a current to vary the gain of the V
pair. A better method, Figure 3, uses another differential pair
but with considerable emitter degeneration. In this circuit the
differential input voltage appears across the common emitter
resistor, producing a current which adds or subtracts from
the quiescent current in either collector. This type of voltage
to current converter handles signals from 0V to ±10V with
excellent linearity.
The second problem is called feedthrough; i.e., the product
of zero and some finite Input signal does not produce zero
output voltage. The circuit whose operation is illustrated by
Figures 4A, 4B, and 4C overcomes this problem and forms
the heart of many multiplier circuits in use today.
This circuit is basically two matched differential pairs with
cross coupled collectors. Consider the case shown in Figure
4A of exactly equal current sources basing the two pairs.
With a small positive signal at V
and Q
will decrease by the same amount. Since the collectors are
cross coupled the current through the load resistors remains
unchanged and independent of the V
In Figure 4B, notice that with V
of biasing current sources will produce a common mode
voltage across the load resistors. The differential output
voltage will remain zero. In Figure 4C we apply a differential
input voltage with unbalanced current sources. If I
I
pair Q
collector currents will be unequal and a differential output
voltage will result. By replacing the separate biasing current
sources with the voltage to current converter of Figure 3 we
have a balanced multiplier circuit capable of four quadrant
operation (Figure 5).
E2
1. V
2. Some portion of the signal at V
3. V
4. The output voltage is not centered around ground.
the gain of differential pair Q
unless I
linear.
Y
X
4
3
FIGURE 3. VOLTAGE TO CURRENT CONVERTER
must be positive and greater than V
must be a small signal for the differential pair to be
will increase but the collector currents of Q
and Q
E
= 0.
4
. Therefore, the change in cross coupled
I
E
+ ∆I
V
I
E
IN
∆I =
∆V
OUT
IN
V
lN
R
V-
V+
1
IN
E
= 0 any variation in the ratio
, the collector current of Q
and Q
X
will appear at the output
lN
I
input voltage.
2
E
I
E
is twice the gain of
- ∆I
D
X
.
differential
E1
2
and Q
is twice
Y
3
1