Current-Mode Continuous-Time Filters

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Current-Mode Continuous-Time Filters

• Prof Paul Hasler

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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
Highpass of x
S

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Key assumptions in Current-Mode 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
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Basic current-mode integrator
Vdd
Iin
Iout
C
GND
GND
GND
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Basic current-mode integrator
Iout Is0 exp( DV / UT )
Assume k 1
Vdd
(Iout / UT)
Iin
Iout
C Iin - Iout
V
Iout
C
GND
GND
GND
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Basic current-mode integrator
Iout Is0 exp( DV / UT )
Assume k 1
Vdd
(Iout / UT)
Iin
Iout
C Iin - Iout
V
Iout
C
(Iout / C UT) (Iin - Iout )
GND
GND
GND
This equation is not linear. approximately
linear if Iout does not change much
(linear range)
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Basic current-mode integrator
(Iout / C UT) (Iin - Iout )
Vdd
Approximately linear if Iout does not change
much.
Iin
Iout
V
Iout
C
GND
GND
GND
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Basic current-mode integrator
(Iout / C UT) (Iin - Iout )
Vdd
Approximately linear if Iout does not change
much.
Iin
Iout
Iout It
V
Iout
Iout
C
Iin
Iin It
GND
GND
GND
t C UT / It
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Basic current-mode integrator
(Iout / C UT) (Iin - Iout )
Vdd
Approximately linear if Iout does not change
much.
Iin
Iout
Iout It
V
Iout
Iout
C
Iin
Iin It
GND
GND
GND
t C UT / It
t (Iin - Iout ) ( 1 (Iout/It))
If Iout/It small, then the circuit is linear
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Basic current-mode integrator
t (Iin - Iout ) ( 1 (Iout/It))
t C UT / It
Vdd
Vdd
If Iout/It small, then the circuit is linear
It
It
How small is small?
Iin
Iout
C
GND
GND
GND
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Basic Current-Mode Blocks
Basic Integrator / Low-Pass Filter
Signal Inversion
Vdd
Vdd
Vdd
Vdd
It
It
It
It
Iin
Iout
Iin
Iout
C
GND
GND
GND
GND
GND
Iout -Iin
t (Iin - Iout )
t C UT / It
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Basic Highpass Filter
Vdd
It
Iin
GND
Input Stage
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Basic Highpass Filter
Vdd
Vdd
Vdd
It
It
It
Iin
Iout
Iin
GND
GND
GND
Copy of Input
Copy of Input And Subtraction
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Basic Highpass Filter
Vdd
Vdd
It
It
Vdd
Vdd
Vdd
It
It
It
C
GND
GND
Iin
Iout
GND
Iin
GND
GND
GND
Integrating Block
Subtraction
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Basic Highpass Filter
Vdd
Vdd
It
It
Vdd
Vdd
Vdd
It
It
It
C
GND
GND
Iin
Iout
GND
Iin
GND
GND
GND
t Iout t
t C UT / It
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Building a current amplifier
Norton Amp Iout Av (I - I- )
I-
Iout
I
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Building a current amplifier
Norton Amp Iout Av (I - I- )
Vdd
Vdd
It
It
I-
I
Iout
-I
I
GND
GND

• Need to invert one current

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Building a current amplifier
Norton Amp Iout Av (I - I- )
Vdd
Vdd
It
It
I-
I
Iout
-I
I
I-
GND
GND

• Need to invert one current
• Need to take a difference
• of currents

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Building a current amplifier
Norton Amp Iout Av (I - I- )
Vdd
Vdd
Vdd
Vdd
It
It
It
nIt
I-
I
Iout
Iout
-I
I
I-
W/L n
GND
GND
GND
GND

• Need to invert one current
• Need to take a difference
• of currents
• Need to amplify the
• response

Current gain n
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Differential Lowpass filter
Vdd
Vdd
Vdd
Vdd
t ( n(I - I-) - Iout )
It
It
It
nIt
t C UT / It
I
Iout
-I
I-
W/L n
Current gain n
C
GND
GND
GND
GND
GND
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Need to emphasize the need for cascode
transistors in many applications.
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A Translinear Circuit
Vdd
I2
Vdd
Vdd
Iout
I1
I3
I2