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ECD222 Lesson 16: The Gilbert Cell
Full four-quadrant multiplication, without offset, can be achieved using a
Gilbert cell. The Gilbert cell has become one of the most popular modules in
integrated communication systems.
i36
v3
i54
Q3
v1
Q4
Q5
i1
i2
Q1
Q2
IA
Figure 1
Define the differential input signals as
and
v12 ≡ v1 - v2
v34 ≡ v3 - v4
Then the difference in output currents is given by:
iOD = i36 − i54
v
= I A tanh  34
 2VT

v
 tanh  12

 2VT
Exercise
Derive the preceding expression for iOD.
If v12 , v34 << 2VT then
Q6



v4
v2
ECD222 Lesson 16: The Gilbert Cell
 I
iOD ≅  A 2
 4VT
16-2

 v 34 v 12


This circuit is a fully balanced or 4-quadrant multiplier. Note that the inputs
are differential signals and they are small.
Exercise
If v12 = A 12 sin ω12t and v34 = A 34 sin ω34t, then find an expression for the
output current iOD. Write this expression as a sum of spectral components.
Exercise
A Taylor series expansion of the tanh function up to third-order terms is given
by
tanh(x)
≅
x - x3 /3
Using this approximation, how large can v12 get before the linear
approximation is in error by about 10% ?
Exercise
If v34 is constant, then the output differential current of the Gilbert cell is given
by
v12
iOD
=
K tanh( ----- )
2VT
where K is a constant. Now assume an input signal given by
v12 = A(sin ω1t + sin ω2 t)
Find expressions for iOD, as a sum of spectral components, using
(a)
(b)
the linear approximation for the tanh function;
the cubic approximation for the tanh function, as given in the previous
exercise.
In the last exercise, the output signal includes harmonic combinations of the
original frequency components. These are called intermodulation products
and they contribute to distortion of the information signal. The third-order
products that appear in this exercise are particularly troublesome because
they can appear in the same frequency band as the original frequencies and
so are difficult to remove by filtering.
ECD222 Lesson 16: The Gilbert Cell
16-3
The Gilbert cell can also be used as a two-level current switch. Assume
that the differential signals are binary-valued, with levels H and L, where H
>>0 and L << 0. Then the cell has four possible states:
v12
L
L
H
H
v34
L
H
L
H
i54
0
IA
IA
0
In this case the current i 54 is an exclusive-OR (XOR) function of the binary
input signals.
ECD222 Lesson 16: The Gilbert Cell
16-4
Balanced Mixers
In radio frequency (RF) applications, a mixer multiplies two signals together.
(In audio applications, a mixer is a circuit that adds two signals together.) As
we have already seen, a mixer is used to shift a signal’s spectrum from one
frequency band to another. For example, consider a sinusoidal signal
vs = A s cos ωs t multiplied by a carrier sinusoid vc = Ac cos ωc t:
vo(t)
=
As cos ωs t . Ac cos ωc t
=
As Ac
-------- [cos (ωs + ωc )t + cos (ωs – ωc )t ]
2
The signal, with a spectral component at ω = ωs , is shifted to spectral
locations ω = ωs + ωc and ω = ωs - ωc . This is the response of a balanced
mixer. Notice that this response is similar to the am-modulator’s response,
except that there is no separate carrier component.
The Gilbert cell can be used as a multiplier for this application. For example,
v12 can be used to input the information signal, and v34 can be used to supply
the carrier signal:
 I 
iOD =  A 2 v 34 v 12
 4VT 
 I 
=  A 2  A s A c (cos ( ù c + ù s )t + cos ( ù c − ù s )t )
 8VT 
The following SPICE simulation demonstrates the Gilbert cell as a fully
balanced mixer. Simple voltage sources are used to bias this circuit. These
voltage sources can be replaced by more practical bias circuitry at a later
stage in the design process. The output of the balanced mixer should be
compared with the results of the am modulator. Note in particular that the
envelope of the output waveform no longer represents the information signal
v1. The input signals are sinusoids with frequencies of 1kHz and 10kHz. The
output waveform contains spectral components at 9kHz and 11kHz - there is
no carrier component at 10kHz as there was for am-modulation.
ECD222 Lesson 16: The Gilbert Cell
16-5
SPICE Simulation - Gilbert Cell as a Balanced Mixer
9
R1
1.5k
VCC
6V
R2
1.5k
VOD
∆
2
3
V3
Q3
7
Q4
Q5
6
Q6
VB4
3.5V
V3
4
5
V1
10
Q2
Q1
1
VB2
1.5V
V1
8
IA
2mA
Figure 2
C:\Teaching_2001\mixer.cir Setup1
.TRAN 0.1us 2ms 0 0.1us
.PRINT TRAN V3
.PRINT TRAN VOD
.PRINT TRAN V1
Q3
2
7
4
_BNPN
Q4
3
6
4
_BNPN
Q5
3
7
5
_BNPN
Q6
2
6
5
_BNPN
Q1
4
10
8
_BNPN
Q2
5
1
8
_BNPN
IA
8
0
DC=2mA
R1
9
2
1.5k
R2
9
3
1.5k
VB2 1
0
DC=1.5V
VB4 6
0
DC=3.5V
V3
7
0
DC=3.5V
SIN 3.5V 0.01V 10kHz
V1
10
0
DC=1.5V
SIN 1.5 0.01V 1kHz
VCC 9
0
DC=6V
.MODEL _BNPN
NPN
.END
ECD222 Lesson 16: The Gilbert Cell
16-6
V2
1.540
700.0M
1
1.500
500.0M
VOD in Volts
V1 in Volts
1.520
V1
2
300.0M
VOD
1.480
100.0M
3
1.460
-100.00M
200.0U
600.0U
1.000M
1.400M
WFM.3 VOD vs. TIME in Secs
Gilbert Cell as a Balanced Mixer
Figure 3
1.800M
ECD222 Lesson 16: The Gilbert Cell
16-7
The carrier signal can be a square wave with levels that cause the top crosscoupled pairs to act as current switches. Assume that v34 is binary valued
with levels H and L. When v34 = H then transistors Q3 and Q5 are ON, and Q4
and Q6 are OFF; on the other hand, when v34 = L then Q3 and Q5 are OFF,
and Q4 and Q6 are ON. Therefore,
When v34 = H:
iOD
=
=
=
=
≅
(i 3 + i 6) – (i 4 + i 5)
i3 - i5
i1 - i2
IO tanh(v12/2VT)
(IO/2VT) v12
Or, when v34 = L
iOD
=
=
=
≅
i6 – i 4
i2 - i1
-IO tanh(v12/2VT)
-(IO/2VT) v12
iOD
≅
(IO/2VT) a(t) v12
That is,
where a(t) = 1 or –1. In this case, the circuit behaves as a linear timevarying system. The system is linear because the output is proportional to
the input, and time-varying because the constant of proportionality changes
with time. For a square carrier signal, a(t) periodically shifts between 1 and –1
at the carrier frequency, as shown in fig.4.
a(t)
+1
t
-1
Figure 4
The mixing action of this circuit can be seen by expanding a(t) as a Fourier
series:
a(t)
=
4
--- [ cos ωc t
π
-
cos 3ωc t
----------3
+
cos5ωc t
------------ 5
……. ]
ECD222 Lesson 16: The Gilbert Cell
16-8
Now
iOD
=
=
2IOAs
------- [ cos ωs t cos ωc t - (1/3) cos ωs t cos3ωc t + ……. ]
πVT
IOAs
------ [ cos(ωc + ωs )t + cos(ωc - ωs )t
πVT
- (1/3) [cos(3ωc +ωs )t + cos(3ωc -ωs )t] + ….]
(1)
The frequencies (ωc + ωs ) and |ωc – ωs | are the fundamental frequencies of the
mixer. Usually only one of these frequencies is wanted, the other is removed
by a filter. The filter can also be used to remove the harmonic terms. For
example, if the upper fundamental is kept, then the output of the mixer plus
filter combination will be of the form
iOD
=
K. A s cos(ωc + ωs )t
(2)
In an RF receiver, the incoming signal with frequency ωRF is mixed with a
local oscillator signal with fundamental frequency ωLO , as shown in fig.5.
ωRF
ωIF
ωLO
Figure 5
ωLO is adjusted until the mixer output produces a strong signal at a frequency
called the intermediate frequency ω IF. For example, we may choose
ωIF
=
ωLO - ω RF
A following IF filter would remove the unwanted (ωLO + ωRF) component.
However, a second incoming signal could also be shifted to the intermediate
frequency. This signal has a frequency ωimage such that
ωIF
∴
=
ωimage =
ωLO
+
ωimage
ωIF
-
ωLO
ECD222 Lesson 16: The Gilbert Cell
16-9
ωimage is called the image frequency . It is important for a preceding filter to
remove the image frequency before it gets to the mixer.
Unfortunately, other image frequencies are created by the harmonics of the
mixer, and some of these frequencies may not be so easily removed by the
prefilter. The problems are compounded if these image frequencies
correspond to the frequencies of strong local transmitters (from radio stations
for example).
Exercise
Assume a receiver system that is required to shift an RF signal of frequency
7MHz down to 4.9MHz. Compute the image frequencies for the fundamental
component the third harmonic terms and the fifth harmonic terms.
[Note: The difference frequencies in eq.(1) are always positive so, for
example, if ωs is large such that (3ωLO - ωs ) < 0 then we simply take the
absolute value, which is (ωs – 3ωLO ) .]
ECD222 Lesson 16: The Gilbert Cell
16-10
SPICE SIMULATION: The Gilbert Cell as a Mixer
6
C4
33nF
R1
1.5k
VO2
volts
L1
470uF
1
R7
50
15
2
C1
1uF
VLO
Q12
CA3046
Q8
CA3046
Q10
17
8
R6
1.5k
7
R8
50
18
R5
1.5k
4
13
C2
1uF
16
VRF
CA3046
Q7
CA3046
3
14
VB3
3.5V
VCC
6V
R2
1.5k VO
volts
Q9
CA3046
9
Q11
CA3046
R4
1.5k
11
R3
1.5k
5
VB1
1.5V
VB4
3.5V
10
12
IA
2mA
VB2
1.5V
C:\Teaching_2001\mixer.cir Setup1
*#save V(1) V(2)
*#alias vo v(2)
*#view tran vo
*#alias vo2 v(1)
*#view tran vo2
.TRAN 0.1us 500us 0 0.1us UIC
.OPTIONS reltol=0.001
.PRINT TRAN VO
.PRINT TRAN VO2
C5 7 0 1uF IC=3.5V
IA 5 0 DC=2mA
R1 6 1 1.5k
R2 6 2 1.5k
R3 10 12 1.5k
R4 9 11 1.5k
VB1 11 0 DC=1.5V
VB2 12 0 DC=1.5V
R5 7 13 1.5k
VB4 13 0 DC=3.5V
R6 8 14 1.5k
VB3 14 0 DC=3.5V
VCC 6 0 DC=6V
C1 17 8 1uF IC=-3.5V
VLO 15 0 DC=0 PULSE -0.05 0.05 0 10ns 10ns 10us 20us
C2 9 16 1uF IC=1.5V
VRF 18 0 DC=0 SIN 0 0.05 10kHz
R7 15 17 50
C3
1uF
C5
1uF
ECD222 Lesson 16: The Gilbert Cell
16-11
6.500
8.000
5.500
7.000
4.500
3.500
VO2 in Volts
VO in Volts
R8 16 18 50
Q7 1 8 3 CA3046
.MODEL CA3046 NPN BF=145.76 BR=.1001 CJC=991.79E-15
+ CJE=1.0260E-12 IKF=46.747E-3 IKR=10.010E-3 IS=10.000E-15
+ ISC=10.000E-15 ISE=114.23E-15 ITF=1.7597 MJC=.33333
+ MJE=.33333 NE=1.4830 RC=10 TF=277.09E-12 TR=10.000E-9
+ VAF=100 VAR=100 VTF=16.364 XTF=309.38
Q8 2 8 4 CA3046
Q9 3 9 5 CA3046
Q10 2 7 3 CA3046
Q11 4 10 5 CA3046
Q12 1 7 4 CA3046
C3 10 0 1uF IC=1.5V
L1 6 2 470uF
C4 6 2 33nF
.END
2
6.000
5.000
1
2.500
4.000
410.0U
430.0U
450.0U
470.0U
WFM.1 VO2 vs. TIME in Secs
Mixer Output with Bandpass Filter
490.0U