7AnalogMultipliers(4p)

```5.1
Dr. Yuri Panarin, DT021/4, Electronics
Introduction
Nonlinear operations on continuous-valued analog signals are often required in
instrumentation, communication, and control-system design.
These operations include
• rectification,
• Modulation - demodulation,
• frequency translation,
7. Gilbert cell &
Analog Multipliers
• multiplication, and division.
In this chapter we analyze the most commonly used techniques for performing
multiplication and division within a monolithic integrated circuit
In analog-signal processing the need often arises for a circuit that takes two
Such circuits are termed analog multipliers.
In the following sections we examine several analog multipliers that depend on
the exponential transfer function of bipolar transistors .
Recommended Text: Gray, P.R. & Meyer. R.G.,
Edition), Wiley (1992) pp. 667-681
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DT021/4 Electronics – 7 Analog Multipliers
The Emitter-Coupled Pair
The emitter-coupled pair, was shown in to
produce output currents that were related to
the differential input voltage by :
The Emitter-Coupled Pair
Q1
Q2
+
Vi1
The emitter-coupled pair, was shown in to
produce output currents that were related to
the differential input voltage by :
I c1 = I c 2 exp(Vid / VT ) = ( I EE − I c1 ) exp(Vid / VT )
Ic2
Ic1
Vi1 − Vbe1 + Vbe 2 − Vi 2 = 0
+
IEE
-
-
Vi2
 V −V 
V
I c1 I S1
exp be1 be 2  = exp id
=
I c2 I S 2
VT


 VT



+
Vid
I EE = −( I e1 + I e 2 ) = ( I c1 + I c 2 ) / α F ≈ I c1 + I c 2
Q1
Q1
I EE exp(Vid / VT )
I EE
=
1 + exp(Vid / VT ) 1 + exp(−Vid / VT )
+
IEE
-
+
Vid
I c1 =
IEE
I EE
1 + exp(−Vid / VT )
Ic2 =
I EE
1 + exp(Vid / VT )
-
Vi2
Ic2
Ic1
Q2
-
Q2
+
Vi1
I c1 =
Ic2
Ic1
Ic2
Ic1
I c1 (1 + exp(Vid / VT )) = I EE exp(Vid / VT )
Vbe1 = VT ln(I c1 / I S1 ) I c1 = I S1 exp(Vbe1 / VT )
Vbe 2 = VT ln (I c 2 / I S 2 ) I c 2 = I S 2 exp(Vbe 2 / VT )
2
DT021/4 Electronics – 7 Analog Multipliers
Q1
Q2
IEE
I c1 = I c 2 exp(Vid / VT ) = ( I EE − I c1 ) exp(Vid / VT )
DT021/4 Electronics – 7 Analog Multipliers
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5.2
Dr. Yuri Panarin, DT021/4, Electronics
Notes
The Emitter-Coupled Pair
The emitter-coupled pair, was shown in to
produce output currents that were related to
the differential input voltage by :
Q1
+
I EE
I EE
I c1 =
Ic2 =
1 + exp(−Vid / VT )
1 + exp(Vid / VT )
Vid
Ic2
Ic1
Q2
IEE
∆I c = I c1 − I c 2 = I EE tanh(Vid / 2VT )
This relationship is plotted => and
shows that the emitter-coupled pair
by itself can be used as a primitive
multiplier.
tanh(x) = x + x 3 / 3 + .... ≅ x
or assuming (Vid / 2VT ) << 1, ⇒ ∆I c = I EE (Vid / 2VT )
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DT021/4 Electronics – 7 Analog Multipliers
DT021/4 Electronics – 7 Analog Multipliers
6
Notes
Simple Multiplier
The current IEE is actually the bias current
for the emitter-coupled pair.
With the addition of more circuitry, we
can make IEE proportional to a second
input signal.
Thus we have
Ic2
Ic1
+
Vid
-
I EE ≅ K o (Vi 2 − VBE ( on ) )
Q1
Q2
IEE
R
+
The differential output current of the V
i2
emitter-coupled pair can be calculated to
give
∆I c ≅
Q3
Q4
K oVid (Vi 2 − VBE ( on ) )
2VT
DT021/4 Electronics – 7 Analog Multipliers
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5.3
Dr. Yuri Panarin, DT021/4, Electronics
Notes
Thus we have produced a circuit that functions as a multiplier
under the assumption that Vid is small, and that Vi2 is greater
than VBE(on).
The latter restriction means that the multiplier functions in
only two quadrants of the Vid - Vi2 plane, and this type of
circuit is termed a two-quadrant multiplier.
The restriction to two quadrants of operation is a severe one
for many communications applications, and most practical
The Gilbert multiplier cell, shown, is a modification of the
emitter-coupled
cell,
which
allows
multiplication.
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DT021/4 Electronics – 7 Analog Multipliers
DT021/4 Electronics – 7 Analog Multipliers
Notes
Gilbert multiplier cell
The Gilbert multiplier cell is
the basis for most integratedcircuit balanced multiplier
systems.
The series connection of an
emitter-coupled pair with two
cross-coupled, emitter-coupled
pairs produces a particularly
useful transfer characteristic,.
IO =I35 - I46
I35
I3
I4
Q3
Q4
I c5 =
I c1
I c1
I =
1 + exp(−V1 / VT ) c 4 1 + exp(V1 / VT )
Ic2
1 + exp(V1 / VT )
Ic6 =
I6
Q5
I1
Q6
I2
Q1
I c3 =
I46
I5
V1
10
Q2
V2
IEE
Ic2
1 + exp(−V1 / VT )
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5.4
Dr. Yuri Panarin, DT021/4, Electronics
Gilbert cell - DC Analysis
The two currents Ic1 and Ic2 are related to V2
I3
I4
Q3
Q4
I EE
[1 + exp(−V1 / VT )][1 + exp(−V2 / VT )]
I c4 =
[1 + exp(V1 / VT )][1 + exp(−V2 / VT )]
I c5 =
[1 + exp(V1 / VT )][1 + exp(V2 / VT )]
I c6 =
[1 + exp(−V1 / VT )][1 + exp(V2 / VT )]
I46
I5
Substituting Ic1 and Ic2 in expressions for
Ic3 , Ic4, Ic5 and Ic6 get :
V1
I c3 =
IO =I35 - I46
I35
I EE
I EE
I c1 =
I c2 =
1 + exp(−V2 / VT )
1 + exp(V2 / VT )
Notes
I6
Q5
Q6
I1
I EE
I2
Q1
Q2
V2
I EE
IEE
I EE
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DT021/4 Electronics – 7 Analog Multipliers
DT021/4 Electronics – 7 Analog Multipliers
Gilbert cell
First show that
1
1
−
= tanh( x / 2)
1 + e−x 1 + e x
14
Notes
e− x / 2e x / 2 = 1
e x / 2e x / 2 = e x
1
1
1 + e x −1 − e− x
−
=
=
1 + e− x 1 + e x 1 + e− x 1 + e x
(
=
e
(e
x/2
)
−x
e −e
=
+ e− x / 2 e x / 2 e− x / 2 + e x / 2
x
−x / 2
)(
e − x / 2e − x / 2 = e − x
) (
)
a 2 − b 2 = ( a + b)(a − b)
e x – e -x = (e x/ 2 + e -x/ 2 ) ⋅ (e x/ 2 - e -x/ 2 )
=
(e
(e
)(
)(
)
)
+ e−x / 2 e x / 2 − e−x / 2
= tanh( x / 2)
x/2
+ e−x / 2 e−x / 2 + e x / 2
x/2
DT021/4 Electronics – 7 Analog Multipliers
tanh (x) =
e x - e -x
e x + e -x
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5.5
Dr. Yuri Panarin, DT021/4, Electronics
Gilbert cell
Notes
The differential output current is then given by
∆I = I c 3−5 − I c 4−6 = I c 3 + I c 5 − (I c 4 + I c 6 ) = (I c 3 − I c 6 ) − (I c 4 − I c 5 )
I c3 − I c 6 =
(1 + e
I EE
=
1 + e −V1 / VT
(
I EE
I EE
−
=
1 + e −V2 / VT
1 + e −V1 / VT 1 + eV2 / VT
−V1 / VT
)(
) (
)(

1
1

−
−V2 / VT
1 + eV2 / VT
 1+ e
)(
) (
)

I EE
 =
tanh(V2 / 2VT )
−V1 / VT
 1+ e
) (
)
1
1
−
= tanh( x / 2)
1 + e−x 1 + e x
Similar:
Ic 4 − Ic4 =
(1 + e
V1 / VT
I EE
I EE
I EE
−
=
tanh(V2 / 2VT )
1 + e −V2 / VT
1 + eV1 / VT 1 + eV2 / VT
1 + eV1 / VT
)(
) (
)(
) (
)
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DT021/4 Electronics – 7 Analog Multipliers
DT021/4 Electronics – 7 Analog Multipliers
Gilbert cell
Notes
The differential output current is then given by
∆I = I c 3−5 − I c 4−6 = I c 3 + I c 5 − (I c 4 + I c 6 ) = (I c 3 − I c 6 ) − (I c 4 − I c 5 )
Where
I c3 − I c6 =
I EE
(1 + e
−V1 / VT
∆I = (I c 3 − I c 6 ) − (I c 4 − I c 5 ) =
) tanh(V
2
I EE
(1 + e
−V1 / VT
I c 4 − I c5 =
/ 2VT )
) tanh(V
2
/ 2VT ) −
18
I EE
(1 + e
V1 / VT
I EE
(1 + e
V1 / VT
) tanh(V
2
) tanh(V
2
/ 2VT )
/ 2VT ) =

I EE
I EE 
=
−
tanh(V2 / 2VT ) = I EE tanh(V1 / 2VT ) tanh(V2 / 2VT )
−V1 / VT
1
+
e
1
+
eV1 /VT 

(
Finally
) (
)
∆I = I EE tanh(V1 / 2VT ) tanh(V2 / 2VT )
The dc transfer characteristic, then, is the product of the hyperbolic
tangent of the two input voltages. The are three main application of
Gilbert cell depending of the V1 an V2 range:
tanh(V1, 2 / 2VT ) ≅ V1, 2 / 2VT
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5.6
Dr. Yuri Panarin, DT021/4, Electronics
Gilbert cell Applications
Notes
∆I = I EE tanh(V1 / 2VT ) tanh(V2 / 2VT )
(1) If V1 < VT and V2 < VT then :
and it woks as multiplier
(2) If one of the inputs of a signal that is large compared to VT, this
effectively multiplies the applied small signal by a square wave,
and acts as a modulator.
(3) If both inputs are large compared to VT, and all six transistors in
the circuit behave as nonsaturating switches. This is useful for the
detection of phase differences between two amplitude-limited
signals, as is required in phase-locked loops, and is sometimes
called the phase-detector mode.
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DT021/4 Electronics – 7 Analog Multipliers
Gilbert cell as Multiplier
22
Notes
(1) If V1 < VT and V2 < VT then : tanh(x) = x + x 3 / 3 + .... ≅ x
Thus for small-amplitude signals,
the circuit performs an analog
multiplication. Unfortunately, the
amplitudes of the input signals are
often much larger than VT,
An alternate approach is to
introduce a nonlinearity that
predictors the input signals to
compensate for the hyperbolic
tangent transfer characteristic of
the basic cell.
The required nonlinearity is an
inverse
hyperbolic
tangent
characteristic
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5.7
Dr. Yuri Panarin, DT021/4, Electronics
Pre-warping circuit inverse hyperbolic tangent
Notes
We assume for the time being that the circuitry within the box develops a
differential output current that is linearly related to the input voltage 7i. Thus
I1 = I o1 + K1V1 and I 2 = I o1 − K1V1
Here Io1 is the dc current that flows in each output lead if V1 is equal to zero, and
K1 is the transconductance of the voltage-to-current converter
 I + K1V1 
 I − K1V1 
 I + K1V1 

 = VT ln o1
 - VT ln o1
∆V = VT ln o1
Is
Is
 I o1 − K1V1 




The differential voltage developed across the
two diode-connected transistors is
tanh -1x = ln((1 + x)/(1 - x) ) /2
Using the identity:
We get
And finally
KV 
∆V = 2VT tanh −1  1 1 
 I o1 
 KV  K V 
∆I = I EE  1 1   2 2 
 I o1   I o 2 
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DT021/4 Electronics – 7 Analog Multipliers
Complete Analog Multiplier
Vout = I EE K 3
26
Notes
K1 K 2
V1V2 = 0.1V1V2
I o1 I o 2
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5.8
Dr. Yuri Panarin, DT021/4, Electronics
Notes
Balanced Modulator
In communications systems, the need frequently arises for the multiplication of a
continuously varying signal by a square wave.
This is easily accomplished with the multiplier circuit by applying a sufficiently
large signal directly to the cross-coupled pair.
Vm (t ) = Vm cos ω mt
∞
 nπ  nπ
Vc (t ) = ∑ An cos nω c t , where An = sin 
/
 2  4
n =1
∞
Vo (t ) = K [Vc (t )Vm (t )] = K ∑ AnVm cos ω mt cos nω c t =
n =1
∞
= K∑
n =1
AnVm
cos(nω c t − ω nt ) cos(nω c t + ω nt )
2
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DT021/4 Electronics – 7 Analog Multipliers
Spectra for balanced modulator
Notes
The spectrum has components located at frequencies ωm above and below each
of the harmonics of ωc, but no component at the carrier frequency ωc or its
harmonics. The spectrum of the input signals and the resulting output signal is
shown below.
The lack of an output component at the carrier frequency is a very useful
property of balanced modulators. The signal is usually filtered following the
modulation process so that only the components near ωc. are retained
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5.9
Dr. Yuri Panarin, DT021/4, Electronics
Phase Detector
Notes
If unmodulated signals of identical frequency coo are applied to the two inputs, the circuit
behaves as a phase detector and produces an output whose dc component is proportional
to the phase difference between the two inputs.
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DT021/4 Electronics – 7 Analog Multipliers
34
Notes
The output waveform that results is shown in Fig. and consists of a dc component and a
component at twice the incoming frequency. The dc component is given by:
Vaverage =
1
2π
∫
2π
0
Vo (t ) d (ω ot ) =
−1
[A1 − A2 ]
π
where areas A1 and A2 are as indicated. Thus
π −ϕ
ϕ
 2ϕ 

Vaverage = −  I EE RC
− I EE RC  = I EE RC 
− 1
π
π
π


If input signals are comparable to or
smaller than VT, the circuit still acts as
a phase detector.
However, the output voltage then
depends both on the phase difference
and on the amplitude of the two input
waveforms
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5.10
Dr. Yuri Panarin, DT021/4, Electronics
Notes
Figure shows the complete multiplier AD534.
Four-quadrant operation is achieved by using two transconductance pairs with the bases
driven in antiphase and the emitters driven by a second V-I converter.
Z1 − Z 2 = K ( X 1 − X 2 )(Y1 − Y2 )
K=
Rz
R y Rx I x
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DT021/4 Electronics – 7 Analog Multipliers
38
Notes
The basic connection for four-quadrant multiplication, is used in
• amplitude modulation,
• voltage-controlled amplification, and
• instantaneous power measurements.
When one of the inputs is zero,
the output should also be zero,
regardless of the signal at the other input.
In practice, a small fraction of the other input will feed through to the output,
causing an error.
This can be minimized by applying an external voltage to the X2 or Y2 input.
This basic configuration has a number of useful variations.
• For instance, tying the inputs together yields the squaring function.
• Deriving Z1 from Vo via a voltage divider allows for scale factors other than 1/(10
V).
• Applying a signal to the Z1 terminal will cause it to be summed to the output
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5.11
Dr. Yuri Panarin, DT021/4, Electronics
Notes
Z − Z = K ( X − X )(Y − Y )
1
2
1
2
1
2
− Vz = (1 / 10) × Vx (−Vo )
Vo = 10× Vz / Vx
− Vz = (1 / 10) × Vo (−Vo )
Vo = 10 × Vz
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DT021/4 Electronics – 7 Analog Multipliers
Test
Show that
(
)
Vo = Vx2 − V y2 / 10
X 1 = V X and Y1 =
V X + VY
2
Notes
Z1 − Z 2 = K ( X 1 − X 2 )(Y1 − Y2 )
Z1 =
42
10 kΩ
V
VO = O
10 kΩ + 30 kΩ
4
VO
V + VY  VX + VY
V − VY VX + VY
V 2 − VY2

= 0.1× X
×
= 0.1× X
= 0.1×  VX − X
×
4
2 
2
2
2
4

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