Memristor Emulator Lab Report

In section 2, mathematical equations of memristor model are briefly explained. The proposed memristor emulator is discussed along with mathematical equations and operation in section 3. The results of experiments and simulations of the proposed circuit are discussed in section 4. A short conclusion is given in section 5.
Mathematical Model of Memristor
Figure 2 is the schematic of the two terminal memristor [1]. A thin titanium dioxide (TiO2) is sandwiched between two metal wires (platinum). The first layer is oxygen deficient titanium oxide i.e., TiO2-x which has low resistance (RON) and acts as a conductor. The second layer is pure titanium oxide, acts as an insulator which has high resistance (ROFF). From figure 2, w is the width of the …show more content…
The circuit is designed to implement the features of the memristor using current feedback operational amplifier and analog multiplier.
C
Figure 4: Memristor emulator
In figure 4, voltage at the input terminal is
Vin(t)=Iin(t)Rx+V03(t) (6) where Iin is the input current, Rx is the resistance at the inverting terminal and V03 is the output of a analog multiplier feedback to the non-inverting terminal of CFOA1.
To get the V03, three devices- a resistor, a capacitor and voltage multiplier are used. The output voltage across the resistor (Ry) and capacitor (C) are multiplied using a analog multiplier AD633. According to the port characteristics of AD844,
Vx=Vy; iy=0; iz=ix; Vw=Vz
In figure 4, stage 1 is a current to voltage converter, by analyzing the circuit,
Iint=V01(t)Ry (7)
The output voltage across the resistor Ry is expressed as
V01(t)=RyIin(t) …show more content…
The output of the 1st stage, V01, is given as an input to the non-inverting terminal of stage 2.
Since V01=Vin2, ict=Vin2(t)Rz=V01(t)Rz (10) where Rz is the resistance at the inverting input terminal of the integral circuit.
By substituting eq. (10) in eq. (9),
V02t=1CV01(t)Rz dt (11)
The output of first and second stage is given to the voltage multiplier (AD633) as an input. The output of the analog multiplier is,
Vw=(Vx1-Vx2)(Vy1-Vy2)10+ Vz (12)
Since Vx1=V01, Vy1=V02; then eq. (12) can be expressed as
V03(t)= V01(t)10 V02(t) (13)
By inserting eq. (13) into eq. (6), we get
Vin=RxIin+V01V0210 (14)
Substituting eq. (8) and eq. (11) in eq. (14), then
Vint=Iint[Rx+Ry210RzC qt] (15)
So the memristance is given