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S. RossEECS 40 Spring 2003 Lecture 27 Today we will Put a twist on our normally linear operational amplifier circuits to make them perform nonlinear computations.

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Presentation on theme: "S. RossEECS 40 Spring 2003 Lecture 27 Today we will Put a twist on our normally linear operational amplifier circuits to make them perform nonlinear computations."— Presentation transcript:

1 S. RossEECS 40 Spring 2003 Lecture 27 Today we will Put a twist on our normally linear operational amplifier circuits to make them perform nonlinear computations Make a linear circuit model for the nonlinear NMOS transistor (Preview of EE 105) Next time we will Show how we can design a pipelined computer datapath at the transistor level Use a relay to design an analog circuit that counts LECTURE 27 Trying to expose you to various complicated circuits/topics to use the tools you’ve developed and prepare you for final exam…

2 S. RossEECS 40 Spring 2003 Lecture 27 NONLINEAR OPERATIONAL AMPLIFIERS When I put a nonlinear device in an operational amplifier circuit, I can compute a nonlinear function. Consider the following circuit using the realistic diode model: +  R V IN V OUT

3 S. RossEECS 40 Spring 2003 Lecture 27 +  R V IN V OUT NONLINEAR OPERATIONAL AMPLIFIERS Computes an exponential function of V IN ! IDID

4 S. RossEECS 40 Spring 2003 Lecture 27 +  R V IN V OUT NONLINEAR OPERATIONAL AMPLIFIERS Computes a natural logarithm function of V IN ! IDID What if I switch the positions of the resistor and the diode (and make sure V IN ≥ 0 V)? Changing the position of the elements inverted the function performed!

5 S. RossEECS 40 Spring 2003 Lecture 27 G D S I DSAT +_+_ +_+_ V DD V GS This circuit acts like a constant current source, as long as the transistor remains in saturation mode. I DSAT TRANSISTOR AS CURRENT SOURCE Load But this hides the fact that I DSAT depends on V GS ; it is really a voltage-dependent current source! If V GS is not constant, the model fails. What if V GS changes? What if there is noise in the circuit?

6 S. RossEECS 40 Spring 2003 Lecture 27 V DS IDID V GS = 2 V THE EFFECT OF A SMALL SIGNAL V GS = 2.1 V V GS = 1.9 V If V GS changes a little bit, so does I D.

7 S. RossEECS 40 Spring 2003 Lecture 27 THE SMALL-SIGNAL MODEL Let’s include the effect of noise in V GS. Suppose we have tried to set V GS to some value V GS,DC with a fixed voltage source, but some noise  V GS gets added in. V GS = V GS,DC +  V GS + V GS - S D G  I DSAT = g(  V GS ) We get the predicted I DSAT,DC plus a change due to noise,  I DSAT. No current flows into or out of the gate because of the opening. I DSAT,DC = f(V GS,DC )

8 S. RossEECS 40 Spring 2003 Lecture 27 THE SMALL-SIGNAL MODEL To be even more accurate, we could add in the effect of. When is nonzero, I D increases linearly with V DS in saturation. We can model this with a resistor from drain to source: The resistor will make more current flow from drain to source as V DS increases. roro + V GS - S D G  I DSAT = g(  V GS ) I DSAT,DC = f(V GS,DC )

9 S. RossEECS 40 Spring 2003 Lecture 27 THE SMALL-SIGNAL MODEL How do we find the values for the model? I DSAT,DC = ½ W/L  N C OX (V GS,DC – V TH ) 2 This is a constant depending on V GS,DC. This is a first-order Taylor series approximation which works out to  I DSAT = W/L  N C OX (V GS,DC – V TH )  V GS We often refer to W/L  N C OX (V GS,DC – V TH ) as g m, so  I DSAT = g m  V GS.

10 S. RossEECS 40 Spring 2003 Lecture 27 THE SMALL-SIGNAL MODEL Including the effect of via r o, the added current contributed by the resistor is I r0 = ½ W/L  N C OX (V GS – V TH ) 2 V DS To make things much easier, since the effect is small anyway, we neglect the effect of  V GS in the resistance, so the current is I r0 ≈ ½ W/L  N C OX (V GS,DC – V TH ) 2 V DS = I DSAT,DC V DS This leads to r 0 = V DS / I r0 = ( I DSAT,DC ) -1

11 S. RossEECS 40 Spring 2003 Lecture 27 G D S IDID + V GS _ + V DS _ +_+_ +_+_ 4 V 3 V 1.5 k  EXAMPLE Revisit the example of Lecture 20, but now the 3 V source can have noise up to ± 0.1 V. Find the range of variation for I D. We figured out that saturation is the correct mode in Lecture 20. The “noiseless” value of I D is I DSAT,DC = ½ W/L  N C OX (V GS,DC – V TH ) 2 = 250  A/V 2 (3 V – 1 V) 2 = 1 mA V TH(N) = 1 V, W/L  n C OX = 500  A/V 2, = 0 V -1.

12 S. RossEECS 40 Spring 2003 Lecture 27 G D S IDID + V GS _ + V DS _ +_+_ +_+_ 4 V 3 V 1.5 k  EXAMPLE The variation in I D due to noise:  I DSAT = W/L  N C OX (V GS,DC – V TH )  V GS = 500  A/V 2 (3 V – 1 V) 0.1 V = 100  A So I D could vary between 1.1 mA and 0.9 mA. Will saturation mode be maintained? Yes.

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