4

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

transconductance multiplier of Figure 2, a current source

There are several difficulties with this simple modulator:

linear.

4. The output voltage is not centered around ground.

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.

will decrease by the same amount. Since the collectors are

cross coupled the current through the load resistors remains

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

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).

V+

V-

FIGURE 1. DIFFERENTIAL AMPLIFIER

A

V

V

OUT

V

IN

----------------

R

L

r

E

-------

==

Substituting r

E

1

g

M

-------

kT

qI

E

---------

==

V

OUT

V

IN

R

L

r

E

-------







V

IN

qI

kT

-------------------

×

==

I

D

V

Y

R

Y

--------

≈

2I

=

V

OUT

qR

L

kTR

Y

---------------

V

X

V

Y

×

()

=

V+

V-

-

+

FIGURE 2. TRANSCONDUCTANCE MULTIPLIER

V+

V-

∆I =

FIGURE 3. VOLTAGE TO CURRENT CONVERTER

ICL8013