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ICL8013BCTX

1MHz, Four Quadrant Analog Multiplier

Intersil 

ICL8013

Application Information

Detailed Circuit Description

The fundamental element of the ICL8013 multiplier is the

bipolar differential amplifier of Figure 1.

V+

RL

RL

VOUT

VIN

2IE

V-

FIGURE 1. DIFFERENTIAL AMPLIFIER

The small signal differential voltage gain of this circuit is

given by:

AV = V----V-O---I-U-N---T-- = -Rr---E-L-

Substituting rE = g---1-M--- = q--k--I-T-E--

VOUT

=



VI N 

R-r---E-L- 

=

VIN × q----I--E-k---T--R-----L-

The output voltage is thus proportional to the product of the

input voltage VlN and the emitter current IE. In the simple

transconductance multiplier of Figure 2, a current source

comprising Q3, D1, and RY is used. If VY is large compared

with the drop across D1, then

ID ≈ R-V----YY-- = 2IE and

VOUT = k---q-T--R--R---L--Y--(VX × VY)

V+

RL

VIN

VOUT

RL

qRL

VOUT = K (VX x VY) =

(VX x VY)

kTRY

2IE

RY

Q3

+

ID VY

VD - D1

V-

FIGURE 2. TRANSCONDUCTANCE MULTIPLIER

There are several difficulties with this simple modulator:

1. VY must be positive and greater than VD.

2. Some portion of the signal at VX will appear at the output

unless IE = 0.

3. VX must be a small signal for the differential pair to be

linear.

4. The output voltage is not centered around ground.

The first problem relates to the method of converting the VY

voltage to a current to vary the gain of the VX differential 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.

V+

IE + ∆I

VIN

∆VOUT

VIN

∆I =

RE

IE - ∆I

IE

IE

V-

FIGURE 3. VOLTAGE TO CURRENT CONVERTER

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 VlN, the collector current of Q1

and Q4 will increase but the collector currents of Q2 and Q3

will decrease by the same amount. Since the collectors are

cross coupled the current through the load resistors remains

unchanged and independent of the VlN input voltage.

In Figure 4B, notice that with VIN = 0 any variation in the ratio

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 IE1 is twice

IE2 the gain of differential pair Q1 and Q2 is twice the gain of

pair Q3 and Q4. Therefore, the change in cross coupled

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).

4

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