Current-Mode Continuous-Time
Filters
Prof Paul Hasler
Basics of current-mode filters
Current-mode filters, for a given corner frequency,
can result in the lowest power dissipation and smallest area
• Lowpass filters are straight-forward in this technology….
• Unfortunately, highpass (and therefore bandpass filters)
are significantly more difficult in practice (highly rely on matching):
• Highpass of x by subtracting x by the lowpass of x:
x
LP Filter
-
S
Highpass
of x
+
Problem is getting
an ideal subtraction
Key assumptions in CurrentMode Continuous-Time Filters
In our circuit analysis, we assume
• All transistors are matched
• All biasing current sources are matched
• Drawn capacitors are the only significant capacitances.
Often, these conditions are not the case, and
one must take care to achieve these conditions.
In all of these circuits, we must be able to tune the
current sources to get the proper frequency response
Basic current-mode integrator
Vdd
Iin
Iout
C
GND
GND
GND
Basic current-mode integrator
Iout = Is0 exp( DV / UT )
Vdd
dIout(t)
dt
Iin
Iout
V
Iout
C
GND
GND
GND
C
= (Iout / UT)
dV(t)
dt
dV(t)
dt
= Iin - Iout
Assume k = 1
Basic current-mode integrator
Iout = Is0 exp( DV / UT )
Assume k = 1
Vdd
dIout(t)
dt
Iin
Iout
V
Iout
C
GND
GND
C
dIout(t)
dt
= (Iout / UT)
dV(t)
dt
dV(t)
dt
= Iin - Iout
= (Iout / C UT) (Iin - Iout )
GND
This equation is not linear….
approximately linear if Iout does not change much
(linear range)
Basic current-mode integrator
dIout(t)
dt
= (Iout / C UT) (Iin - Iout )
Vdd
Approximately linear if Iout does not change much….
Iin
Iout
V
Iout
C
GND
GND
GND
Basic current-mode integrator
dIout(t)
dt
= (Iout / C UT) (Iin - Iout )
Vdd
Approximately linear if Iout does not change much….
Iin
Iout
V
Iout
C
GND
GND
Iout
Iout + It
Iin
Iin + It
GND
t = C UT / It
Basic current-mode integrator
dIout(t)
dt
= (Iout / C UT) (Iin - Iout )
Vdd
Approximately linear if Iout does not change much….
Iin
Iout
V
Iout
C
GND
GND
Iout
Iout + It
Iin
Iin + It
GND
t = C UT / It
t
dIout(t)
dt
= (Iin - Iout ) ( 1 + (Iout/It))
If Iout/It small, then the circuit is linear
Basic current-mode integrator
t
Vdd
dIout(t)
dt
= (Iin - Iout ) ( 1 + (Iout/It))
t = C UT / It
Vdd
If Iout/It small, then the circuit is linear
It
It
Iin
Iout
GND
C
GND
GND
How small is small?
Basic Current-Mode Blocks
Basic Integrator / Low-Pass Filter
Vdd
Vdd
It
Vdd
It
Iin
Iout
GND
C
GND
GND
t
dIout(t)
dt
Signal Inversion
Vdd
It
Iin
Iout
GND
GND
Iout = -Iin
= (Iin - Iout )
t = C UT / It
It
Basic Highpass Filter
Vdd
It
Iin
GND
Input Stage
Basic Highpass Filter
Vdd
Vdd
Vdd
It
It
It
Iin
Iout
Iin
GND
GND
Copy of Input
GND
Copy of Input
And
Subtraction
Basic Highpass Filter
Vdd
Vdd
It
Vdd
Vdd
Vdd
It
It
GND
Iin
Iin
GND
It
It
C
GND
Iout
GND
GND
GND
Integrating Block
Subtraction
Basic Highpass Filter
Vdd
Vdd
It
Vdd
Vdd
Vdd
It
It
GND
Iin
It
C
GND
GND
t
dIout(t)
dt
Iout
GND
Iin
GND
It
GND
+ Iout = t
dIin(t)
dt
t = C UT / It
Building a current amplifier
Norton Amp:
Iout = Av (I+ - I- )
I-
I+
Iout
Building a current amplifier
Norton Amp:
Iout = Av (I+ - I- )
Vdd
It
I-
I+
Vdd
Iout
I+
-I+
GND
• Need to invert one current
It
GND
Building a current amplifier
Norton Amp:
Iout = Av (I+ - I- )
Vdd
It
I-
I+
Vdd
Iout
I+
-I+
GND
• Need to invert one current
• Need to take a difference
of currents
It
GND
I-
Building a current amplifier
Norton Amp:
Iout = Av (I+ - I- )
Vdd
Vdd
It
I-
I+
Vdd
Iout
I+
It
-I+
GND
GND
Vdd
It
Iout
I-
W/L = n
GND
• Need to invert one current
• Need to take a difference
of currents
• Need to amplify the
response
nIt
Current gain = n
GND
Differential Lowpass filter…
Vdd
Vdd
Vdd
Vdd
t
It
I+
It
-I+
GND
GND
It
= ( n(I+ - I-) - Iout )
nIt
Iout
I-
W/L = n
GND
dIout(t)
dt
C
GND
GND
t = C UT / I t
Current gain = n
Need to emphasize the need for cascode transistors in many applications….
A Translinear Circuit
Vdd
Vdd
I2
Vdd
Iout
I1
I3
I2
GND
GND
GND
GND