Time Domain Circuits

Passive Filters GND V out R V in C C(dVout/dt) = (V in – V out ) / R  (dV out /dt) = (Vin – Vout)  = R * C Equate currents: GND V out C V in R C ( d(Vin – Vout)/dt) = Vout/R  (d(Vin – Vout)/dt) = Vout

Response of a linear system Behavior is independent of boundary conditions Boundary conditions only scale the result (double the input, double the output) Impulse response of a low-pass filter V = V(0)exp( -t /  ) V=(1)exp( -t/  ) V = (e -1 )exp( -t/  ) V = (e -2 )exp( -t/  )

Analysis of Diff-Pair

Basic Differential Amplifier

Transfer Functions

A Unity-Gain Follower

Amp Gain = A v Solving: V out = V in AvAv 1 + A v

A Unity-Gain Follower Amp Gain = A v Solving: V out ~ V in

A Unity-Gain Follower Amp Gain = A v Solving: V out ~ V in

A Low-Pass Filter C (dV out /dt) = I bias (  ) (Vin - Vout) /(2 UT) I bias tanh(  (Vin - Vout)/(2 UT) )

A Low-Pass Filter Time-constant changes with bias  = C U T /  I bias  (dV out /dt) + V out = V in C (dV out /dt) = I bias (  ) (Vin - Vout) /(2 UT)

C (dV out /dt) = I bias tanh(  (V in - V out )/(2 U T ) ) This is not linear…. “Small-signal analysis: How small is “small”? A Low-Pass Filter |V in - V out | < 2U T

Follower Integrator ( Simplest Gm-C Filter) I V out V in C1C1 GND dV out (t) dt By KCL: G m (V in - V out ) = C 1 Defining  = C 1 / G m dV out (t) dt V in = V out +  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 V in = V out U T / k I bias

C (dV out /dt) = I bias tanh(  (V in - V out )/(2 U T ) ) This is not linear…. “Small-signal analysis: How small is “small”? A Low-Pass Filter |V in - V out | < 2U T |V in - V out | > 10U T tanh(X) = +/- 1

A Low-Pass Filter C (dV out /dt) =I bias I bias tanh(  (Vin - Vout)/(2 UT) ) Slew-Rate changes with bias Slew Rate = C / I bias (dV out /dt) = 1 / 

Tuning Frequency using Bias Current Bias CurrentTransconductanceCutoff Frequency 1mA 1/250  600MHz 1A1A1/25k  6MHz 1nA 1/25M  6kHz 1pA 1/25G  6Hz This approach is the most power-efficient approach for any filter. All electronically tunable: Advantage: we can electronically change the corner Disadvantage: we need a method to set this frequency (tuning) C = 1pF, W/L of input transistors = 30, I th ~ 10  A.

Comparing Low-Pass Filters GND V out R V in C Linear Filtering Requires no additional power Transistor Capacitor design Time-constant is electrically tunable No input current load