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Basics of Transconductance –Capacitance Filters
Dr. Paul Hasler
Operational Amplifiers
OP Amp: Vout = Av (V+ - V- )
Balanced Op Amp: Vout
+ = Av (V+ - V- )Vout
- = Av (V- - V+)
V+
V- V-
V+
VoutVout
+
Vout-
Av is frequency dependent
Transconductance Amplifiers
• Most op-amps can be used as OTAs, most OTAs can be used as op-amps. (depends which application is being optimized)
OTA: I = Gm (V+ - V- )
Balanced OTA: I+ = Gm (V+ - V- )I- = Gm (V- - V+)
V+
V- V-
V+
I I+
I -
Major Issues for OTAs
Major Issues for OTAs: • Linearity• Offsets• Output Resistance (depends upon application)
How to Build OTAs
Basic transistor differential amplifer, Wide-output-range differential amplifier …
Build with cascodes or folded cascode or differential approaches.
Analysis of Diff-Pair
Differential Pair Currents
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.40
0.5
1
1.5
2
2.5
3
Differential input voltage (V)
Out
put c
urre
nt (
nA)
Iout+Iout
-
Basic Differential Amplifier
Transfer Functions
Wide-Output Range Differential Amplifier
Wide-output range amplifier
Simple differential amplifer
"Vmin" effect
Simplest Gm-C Filter
IVoutVin
C1
GND
dVout(t) dtBy KCL: Gm(Vin - Vout ) = C1
Defining τ = C1 / Gm
dVout(t) dtVin = Vout + τ
Also, should mention the high-pass version as well.
We can set Gm and build C sufficiently big enough (slow down the amplifier), or set by C (smallest size to get enough SNR), and change Gm.
If an ideal op-amp, then Vin = Vout
Tuning Frequency using Bias Current
6Hz1/25GΩ1pA
6kHz1/25MΩ1nA
6MHz1/25kΩ1µA
600MHz1/250Ω1mA
Cutoff FrequencyTransconductanceBias Current
This approach is the most power-efficient approach for any filter.
All electronically tunable: Advantage: we can electronically change the cornerDisadvantage: we need a method to set this frequency (tuning)
C = 1pF, W/L of input transistors = 30, Ith ~ 10µA.
A Low-Pass FilterC (dVout/dt) = I
= Ibias tanh(κ(Vin - Vout)/(2 UT) )= ((κ Ibias )/(2 UT) ) (Vin - Vout)
Time-constant changeswith bias
τ = C UT / κ Ibias
τ (dVout/dt) + Vout = Vin
Linear Range
C (dVout/dt) = Ibias tanh(κ(Vin - Vout)/(2 UT) )
This is not linear….“Small-signal analysis: How small is “small”?
Another First-Order Low-Pass Filter
GND
CGND
10nS 1nSVin GND
Vout
Second-Order Behavior
1 + (τ/Q) s + τ2 s2
1Vout
Vin
=
Gain = Q at ω = 1/τ,
Second-Order Behavior
1 + (τ/Q) s + τ2 s2
1Vout
Vin
=
Gain = Q at ω = 1/τ,
High Q = very narrow band(width ∝ 1/Q)
Basic Second-Order Sections
Vout
1kΩ
Vin
Gm1
Vin
Gm1
I
C1
GND
I
C1
GND
Gm2
Diff2 Based Second-Order Filter
1 + (τ/Q) s + τ2 s2
1Vout
Vin
=
Vin
-+ Vout
V1 +-
Vτ2Vτ1
τ = (τ1 τ2)1/2
Q = (τ2/τ1)1/2
Diff2 Responses
Diff2 Second-Order Section
Vout
Vin
Gm1Gm2
I
C1
GND
I
C1
GND
We will see this circuit again for sample and hold elements
Issue of feedback with an outputbuffer for an op-amp.
Tow-ThomasSecond-Order Section
Vout
Vin
Gm1Gm2
I
C1
GND
I
C1
GND
GND
GND
Bandpass Second-Order Section
GND
Vout
Vin
Gm1
C
GND
CV Gm2
Another Second-Order Section
GND
Vout
Vin
Gm1
CL
GNDGNDC1
GND
Cw
Gm2
Gm3
V1
Even Bigger Gm-C Filter
GND
Vin
Gm1
C
GND
C Gm2
GND
Vout
Gm1
C
GND
C Gm2
1
Filter Design Problem
Design example(s): Design a Chebyshev filter with the following specs:
150kHzfsb
100kHzfpb
-25dBTsb
-1dB ( 0.5088)Tpb
We get the following filter:
From our MATLAB functions, we get the following τ’s and Q’s:
Filter Design Example
101 102 103 104 10510
-2
10-1
100
Frequency (Hz)
Mag
nitu
de
From MATLAB code: N = 5
101
102
103
104
105
106
10-3
10-2
10-1
100
101
Q = 5.5561e+000, tau = 1.6009e-006Q = 1.3987e+000, tau = 2.4290e-006
tau = 5.4974e-006