Download Single Transistor Mixers and Large Signal Distortion

Berkeley
Single Transistor Mixers and Large Signal
Distortion
Prof. Ali M. Niknejad
U.C. Berkeley
Copyright c 2015 by Ali M. Niknejad
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BJT with Large Sine Drive
IC
vi
VA
vi = V̂i cos !t
IC = IS e
vBE
Vt
vBE = VA + V̂i cos !t
Consider a bipolar device driven with a large sine signal.
This occurs in many types of non-linear circuits, such as
oscillators, frequency multipliers, mixers and class C amplifiers.
2 / 35
BJT Collector Current
The collector current can be factored into a DC bias term and
a periodic signal
Vˆi
I C = I S e V A V t e Vt
cos !t
IC = IS e a e b cos !t
Where the normalized bias is a = Va /Vt and the normalized
drive signal is b = V̂i /Vt .
Since IC is a periodic function, we can expand it into a Fourier
Series. Note that the Fourier coefficients of e b cos !t are
modified Bessel functions In (b)
e b cos !t = I0 (b) + 2I1 (b) cos !t + 2I2 (b) cos 2!t + · · ·
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BJT DC Current
Assume that the bias current of the amplifier is stabilized.
Then
✓
◆
2I1 (b)
2I2 (b)
a
IC = IS e I0 (b) 1 +
cos !t +
cos 2!t + · · ·
| {z }
I0 (b)
I0 (b)
IQ
6
5
IC = IS e a e b cos !t
IQ b cos !t
=
e
I0 (b)
4
IC
IQ
3
2
1
1
2
3
4
5
4 / 35
Collector Current Waveform
8
b = 10
7
6
IC
IQ
5
4
3
b=4
2
b = .5
1
-0.4
-0.2
0
0.2
0.4
With increasing input drive, the current waveform becomes
“peaky”. The peak value can exceed the DC bias by a large
factor.
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Harmonic Current Amplitudes
1
n=1
0.8
In (b)
I0 (b)
n=2
n=3
0.6
n=4
0.4
n=5
n=6
0.2
n=7
2.5
5
7.5
10
12.5
15
17.5
20
b
The BJT output spectrum is rich in harmonics.
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Small-Signal Region
0.01
0.008
n=1
In (b)
I0 (b)
0.006
0.004
n=2
0.002
n=3
0.05
0.1
b
0.15
0.2
If we zoom in on the curves to small b values, we enter the
small-signal regime, and the weakly non-linear behavior is
predicted by our power series analysis.
7 / 35
BJT with Stable Bias
Neglecting base current (
given by
1), the voltage at the base is
R1
VA
+
vi
R2
VA0 =
RE
CE
R2
VCC
R1 + R2
VE = VA0
VBE
V 0 VBE
VE
= A
RE
RE
We see that the bias is fixed since VBE does not vary too
much. Typically VA0 is a few volts.
IQ =
8 / 35
Di↵erential Pair with Sine Drive
The large signal equation for IC 1 is given by
IC 1 + IC 2 = IEE
VBE 1
VBE 2 = Vi = Vt ln
✓
IC 1
IC 2
IC 1 =
+
Vi
2
Vi
2
+
IEE
◆
IEE
1 + e vi /Vt
vi = V̂i cos !t
IC 1 =
IEE
1 + e b cos !t
9 / 35
Di↵ Pair Waveforms
1
0.8
IC1
IEE
0.6
0.4
b =1
0.2
b =3
b =10
1
2
3
4
5
6
For large b, the waveform approaches a square wave
✓
◆
IC
1 2
1
1
= +
cos !t
cos 3!t + cos 5!t + · · ·
IEE
2 ⇡
3
5
10 / 35
Di↵ Pair Harmonic Currents
Normalized Harmonic Currents
2
⇡
n=1
0.6
0.5
0.4
0.3
n=3
0.2
(negative)
n=5
0.1
n=7
2.5
5
7.5
10
12.5
(negative)
15
17.5
20
b
As expected, the ideal di↵erential pair does not produce any
even harmonics.
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Mixer Analysis
As we have seen, a mixer has three ports, the LO, RF, and IF
port.
Assume that a circuit is “pumped” with a periodic large signal
at the LO port with frequency !0 .
From the RF port, though, assume we apply a small signal at
frequency !s .
Since the RF input is small, the circuit response should be
linear (or weakly non-linear). But since the LO port changes
the operating point of the circuit periodically, we expect the
overall response to the RF port to be a linear time-varying
response
io (t) = g (t)vin
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Mixer Assumptions
gm (t)
The transconductance varies periodically and can be expanded
in a Fourier series
g (t) = g0 + g1 cos !0 t + g2 cos 2!0 t + · · ·
Applying the input vin = Vˆ1 cos !s t
io (t) = (g0 + g1 cos !0 t + g2 cos 2!0 t + · · · ) ⇥ Vˆ1 cos !s t
13 / 35
Mixer Output Signal
Expanding the product, we have
g1
g2
io (t) = g0 Vˆ1 cos !s t+ Vˆ1 cos(!0 ±!s )t+ Vˆ1 cos(2!0 ±!s )t+· · ·
2
2
The first term is just the input amplified. The other terms are
all due to the mixing action of the linear time-varying periodic
circuit.
Let’s say the desired output is the IF at !0
conversion gain is therefore defined as
gconv =
!s . The
|IF output current|
g1
=
|RF input signal voltage—
2
14 / 35
A BJT Mixer
C3
LO
L3
IF
L1
C1
C
RF
C
C2
L2
The transformer is used to sum the LO and RF signals at the
input. The winding inductance is used to form resonant tanks
at the LO and RF frequencies.
The output tank is tuned to the IF frequency.
Large capacitors are used to form AC grounds.
15 / 35
AC Eq. Circuit
IF
LO
L1
C3
L3
C1
RF
C2
L2
The AC equivalent circuit is shown above.
16 / 35
BJT Mixer Analysis
When we apply the LO alone, the collector current of the
mixer is given by
✓
◆
2I1 (b)
2I2 (b)
IC = IQ 1 +
cos !t +
cos 2!t + · · ·
I0 (b)
I0 (b)
We can therefore define a time-varying gm (t) by
gm (t) =
IC (t)
qIC (t)
=
Vt
kT
The output current when the RF is also applied is therefore
given by iC (t) = gm (t)vs
✓
◆
qIQ
2I1 (b)
2I2 (b)
iC =
1+
cos !t +
cos 2!t + · · · ⇥ Vˆs cos !s t
kT
I0 (b)
I0 (b)
17 / 35
BJT Mixer Analysis (cont)
The output at the IF is therefore given by
qIQ I1 (b)
iC |!IF = Vˆs
cos(!0 !s )t
| {z }
kT I0 (b)
|{z}
!IF
gmQ
The conversion gain is given by
1
0.8
gconv
gmQ
gconv = gmQ
I1 (b)
I0 (b)
0.6
0.4
0.2
2
4
b=
V̂i
V
6
8
10
18 / 35
LO Signal Drive
For now, let’s ignore the small-signal input and determine the
impedance seen by the LO drive. If we examine the collector
current
✓
◆
2I1 (b)
2I2 (b)
IC = IQ 1 +
cos !t +
cos 2!t + · · ·
I0 (b)
I0 (b)
The base current is simply IC / , and so the input impedance
seen by the LO is given by
Zi | ! 0 =
Vˆo
iB,!0
=
=
Vˆo
iC ,!0
=
Vˆo
(b)
IQ 2II01(b)
=
bVt
2I1 (b)
IQ I0 (b)
b
I0 (b)
=
2 gmQ I1 (b)
Gm
19 / 35
RF Signal Drive
The impedance seen by the RF singal source is also the base
current at the !s components. Typically, we have a high-Q
circuit at the input that resonates at RF.
iB (t) =
=
1 qIQ
kT
✓
iC (t)
2I1 (b)
Vˆs cos !s t +
cos(!0 ± !s )t + · · ·
I0 (b)
◆
The input impedance is thus the same as an amplifier
Rin =
Vˆs
=
|component in iB at !s |
kT
=
qIQ
gmQ
20 / 35
Mixer Analysis: General Approach
If we go back to our original equations, our major assumption
was that the mixer is a linear time-varying function relative to
the RF input. Let’s see how that comes about
IC = IS e vBE /Vt
where
vBE = vin + vo + VA
or
Vˆs
IC = IS e VA /Vt ⇥ e b cos !0 t ⇥ e Vt
cos !s t
If we assume that the RF signal is weak, then we can
approximate e x ⇡ 1 + x
21 / 35
General Approach (cont)
Now the output current can be expanded into
✓
◆
2I1 (b)
2I2 (b)
IC = IQ 1 +
cos !t +
cos 2!t + · · ·
I0 (b)
I0 (b)
!
Vˆs
⇥ 1+
cos !s t
Vt
In other words, the output can be written as
= BIAS + LO + Conversion Products
In general we would filter the output of the mixer and so the
LO terms can be minimized. Likewise, the RF terms are
undesired and filtered from the output.
22 / 35
Distortion in Mixers
Using the same formulation, we can now insert a signal with
two tones
vin = Vˆs1 cos !s1 t + Vˆs2 cos !s2
IC = IS e VA /Vt ⇥ e b cos !0 t ⇥ e
Vˆs1
Vt
cos !s1 t+
Vˆs2
Vt
cos !s2 t
The final term can be expanded into a Taylor series
IC = IS e VA /Vt ⇥ e b cos !0 t ⇥
1 + Vs1 cos !s 1t + Vs2 cos !s 2t + ( )2 + ( )3 + · · ·
The square and cubic terms produce IM products as before,
but now these products are frequency translated to the IF
frequency
23 / 35
Harmonic Mixer
RF
Second
Harmonic
Mixer
IF
LO
We can use a harmonic of the LO to build a mixer.
Example, let LO = 500MHz, RF = 900MHz, and
IF = 100MHz.
Note that IF = 2LO
RF = 1000
900 = 100
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Harmonic Mixer Analysis
The nth harmonic conversion tranconductance is given by
gconv ,n =
|IF current out|
gn
=
|input signal voltage|
2
For a BJT, we have
gconv ,n = gmQ
In (b)
I0 (b)
The advantage of a harmonic mixer is the use of a lower
frequency LO and the separation between LO and RF.
The disadvantage is the lower conversion gain and higher
noise.
25 / 35
FET Large Signal Drive
ID
vo
VQ
Consider the output current of a FET driven by a large LO
signal
µCox W
ID =
(VGS VT )2 (1 + VDS )
2 L
where VGS = VA + vLO = VA + Vo cos !0 t. Here we implicitly
assume that Vo is small enough such that it does not take the
device into cuto↵.
26 / 35
FET Large Signal Drive (cont)
0.5
1
1.5
2
2.5
3
That means that VA + V0 cos !0 t > VT , or VA V0 > VT , or
equivalently V0 < VA VT . Under such a case we expand the
current
ID / (VA
VT )2 + V02 cos2 !0 t + 2(VA
VT )V0 cos !0 t
27 / 35
FET Current Components
The cos2 term can be further expanded into a DC and second
harmonic term.
Identifying the quiescent operating point
IQ =
ID = IDQ + µCox
µCox W
(VA
2 L
0
W B
B
L @
VT )2 (1 + VDS )
1 2
V
4 0
|{z}
bias point shift
1
+ (VA
|
VT )Vo cos !0 t +
{z
}
LO modulation
C
V02
cos(2!0 t)C
A (1 + VDS )
| 4 {z
}
LO 2nd harmonic
28 / 35
FET Time-Varying Transconductance
The transconductance of a FET is given by (assuming strong
inversion operation)
g (t) =
@ID
W
= µCox (VGS
@VGS
L
VT )(1 + VDS )
VGS (t) = VA + V0 cos !0 t
W
(VA
L
✓
g (t) = gmQ 1 +
g (t) = µCox
VT + V0 cos !0 t)(1 + VDS )
◆
V0
cos !0 t (1 + VDS )
VA VT
This is an almost ideal mixer in that there is no harmonic
components in the transconductance.
29 / 35
MOS Mixer
vs
We see that we can build a
mixer by simply injecting an
LO + RF signal at the gate of
the FET (ignore output
resistance)
IF Filter
vo
VQ
i0 = g (t)vs = gmQ
i0 |IF =
✓
1+
V0
VA
VT
◆
cos !0 t Vs cos !s t
gmQ
V0
cos(!0 ± !s )tVs
2 VA VT
gc =
i0 |IF
gmQ
V0
=
Vs
2 VA VT
30 / 35
MOS Mixer Summary
But gmQ = µCox W
L (VA
VT )
gc =
µCox W
V0
2 L
which means that gc is independent of bias VA . The gain is
controlled by the LO amplitude V0 and by the device aspect
ratio.
Keep in mind, though, that the transistor must remain in
forward active region in the entire cycle for the above
assumptions to hold.
In practice, a real FET is not square law and the above
analysis should be verified with extensive simulation.
Sub-threshold conduction and output conductance complicate
the picture.
31 / 35
“Dual Gate” Mixer
io
M2
+
vo
VA
G2
+
vs
M1
G1
D2
shared junction
(no contact)
S1
VA
The “dual gate” mixer, or more commonly a cascode
amplifier, can be turned into a mixer by applying the LO at
the gate of M2 and the RF signal at the gate of M1. Using
two transistors in place of one transistor results in area savings
since the signals do not need to be combined with a
transformer or capacitively .
32 / 35
Dual Gate Mixer Operation
tri
o
de
saturation
0.5
LO
1
1.5
2
2.5
3
Without the LO signal, this is simply a cascode amplifier. But
the LO signal is large enough to push M1 into triode during
part of the operating cycle.
The transconductance of M1 is therefore modulated
periodically
gm |sat = µCox
W
(VGS
L
gm |triode = µCox
VT )
W
VDS
L
33 / 35
Dual Gate Waveforms
vLO,1
vLO,2
Active Region
Triode Region
gm (t)
gm,max
VGS2 is roughly constant since M1 acts like a current source.
VD1 = vLO
g (t) =
⇢
VGS2 = VA2 + V0 cos !0 t
VGS2
µCox W
VT )
VD1 > VGS
L (VGS1
µCox (VA2 VGS2 |V0 cos !0 t|) VD1 < VGS
VT
VT
34 / 35
Realistic Waveforms
A more sophisticated analysis would take sub-threshold
operation into account and the resulting g (t) curve would be
smoother. A Fourier decomposition of the waveform would
yield the conversion gain coefficient as the first harmonic
amplitude.
35 / 35