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PRECISION INTEGRATED CIRCUIT RESISTANCE MIRROR

IP.com Disclosure Number: IPCOM000005976D
Original Publication Date: 1990-Oct-01
Included in the Prior Art Database: 2001-Nov-21
Document File: 3 page(s) / 137K

Publishing Venue

Motorola

Related People

Jeffrey P. Kotowski: AUTHOR [+2]

Abstract

Treating Q, and R, as a black box impedance Z,, the average impedance (Z,& as seen from Node 1 will be V,/IRIAVC1. Substituting IRIAVO with K(V,/R,) from above yields ZIAMj = RI/K. Since Ksl, then Z,A\~; rR1.

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MO7VROLA Technical Developments Volume 11 October 1990

PRECISION INTEGRATED CIRCUIT RESISTANCE MIRROR

by Jeffrey f? Kotowski and Jeffery R. Spitzer

Treating Q, and R, as a black box impedance Z,, the average impedance (Z,& as seen from Node 1 will be V,/IRIAVC1. Substituting IRIAVO with K(V,/R,) from above yields ZIAMj = RI/K. Since Ksl, then Z,A\~; rR1.

The current through the capacitor, C, in Figure 2 is io(t) = CT. Integrating both sides yieldsfi&)df = CldV(t).

Treating dt and dVas small finite changes n t and n v respectively yields i(f)nf = C/Iv. Solving for n t yields n t = C(Av/i,(t)). Since the Schmitt Trigger has designed (voltage) switching points, n vwill be constant. Since C and A v are both constant, they can be treated as one constant K, = CAv

Let A t, be the part of the cycle when Cl, switches and remains off. Using the equation for a f above, n r, can be defined as nf, = K,/i,(t). Since 0, is off, i&j = I,. Therefore n f, = KIN,.

   Let n t2 be the part of the cycle when Cl, is switching and remaining on. Using the equation for n f above, A t can be defined as n fz = K,/i,(r) Since Q, is on, i&J = I R, - I, (summing the currents at Node 1). Substituting (InI - I,) for L(t) yields n f2 = K& - I,). The duty cycle, K, defined in terms of nf, and n f, is K = D t,/(nf, + n tzl. Substituting the previously determined values for a f, and af, and reducing yields K = II/In,. Since I, = VI/RA and IR1 = &/RI, K can be rewritten in terms of RI and RI: K = R,IRn. Remembering ZIAvr, = R,lK and substituting RllRa for K yields Ztwo = R,/(RI/Ra) = R& therefore, a single resistor, Rn, external to an IC, can be duplicated (mirrored) anywhere inside the IC.

Note that since ZlauG = Ra and ZIAvG 2 RI, Rn has to be greater than or equal to RI (and Rz since the two resistors are identical) for proper operation.

Also note the duty cycle of the switching circuit is self adjusting. The integrator/Schmitt Trigger circuit in Figure 2 is using the duty cycle to achieve and maintain the desired impedance VI/l1 = Rn at Node 1.

The analogous circuit of Qz and R1 works exactly the same as Q, and RI. Both transistors and resistors are iden- tical and both circuits are controlled by the same switching circuit.

When Q2 is on, IR2 = VIN/R2. When Q2 is off, I ~2 = 0. The average current through R2 will also be proportional to the duty cycle of Q2 (or Q,); therefore, I- = KVINIR~.

Treating Q2 and R2 as a black box impedance Z2. the average impedance of Z2 as seen at Node 2 will be: z2n, = VIN/Im. Substituting Imm with KVIN/RP (see above) yields .&,+I = RdK.

   The black box average impedance of Q1 and R, will be ZqnE = VI/l RIAXj = VI/l, = RI. ZmAE was previously deter- mined to be: Z,nv~ = RI/K. Since RI = R2 and RzlK = 4nKi, Z,*% = Zsw = Ra. Therefore, an impedance can be created at Node 2 equal to Rn. In fact, many impedances similar to the series impedance of Q2 and Rz, ZZAVC, can be created and controlled by the same switching circuit. By taking advan...