MAT02 Figure 3. One-Quadrant Multiplier/Divider APPLICATIONS: NONLINEAR FUNCTIONS these effects can be lumped together as a total effective bulk MULTIPLIER/DIVIDER CIRCUIT resistance rBE. The rBEIC term causes departure from the desired The excellent log conformity of the MAT02 over a very wide logarithmic relationship. The rBE term for the MAT02 is less range of collector current makes it ideal for use in log-antilog than 0.5 Ω and ∆rBE between the two sides is negligible. circuits. Such nonlinear functions as multiplying, dividing, Returning to the multiplier/divider circuit of Figure 1 and using squaring and square-rooting are accurately and easily imple- Equation (4): mented with a log antilog circuit using two MAT02 pairs (see V Figure 3). The transistor circuit accepts three input currents (I BE1A + VBE2A – VBE2B – VBE1B + (I1 + I2 – IO – I3) rBE = 0 1, I If the transistor pairs are held to the same temperature, then: 2 and I3) and provides an output current IO according to IO = I1I2/I3. All four currents must be positive in the log antilog circuit, but negative input voltages can be easily accommodated kT I I 1 2 kT I I In = In S1A S2 A + (I1 + I2 – IO – I3) rBE (6) by various offsetting techniques. Protective diodes across each q I I q I I 3 O S B 1 S2 B base-to-emitter junction would normally be needed, but these If all the terms on the right-hand side were zero, then In diodes are built into the MAT02. External protection diodes (I1 I2/I3 IO) would equal zero, which would lead directly to are, therefore, not needed. the desired result: For the circuit shown in Figure 3, the operational amplifiers make I I1I2 1 = VX/R1, I2 = VY/R2, I3 = VZ/R3, and IO = VO/RO. The IO = , where I1, I2, I3, IO > 0 (7) output voltage for this one-quadrant, log-antilog multiplier/ I3 divider is ideally: Note that this relationship is temperature independent. The right-hand side of Equation (6) is near zero and the output R3RO V XVY V current I O = (VX, VY, VZ > 0) (4) O will be approximately I1 I2/I3. To estimate error, R1R2 V Z define ø as the right-hand side terms of Equation (6): If all the resistors (R OBSOLETE O, R1, R2, R3) are made equal, then I I q ø = In S1A S2 A + (I V 1 + I2 – IO – I3) rBE (8) O = VXVY/VZ I I kT S B 1 S2 B Resistor values of 50 kΩ to 100 kΩ are recommended assuming For the MAT02, In (ISA/ISB) and ICrBE are very small. For small an input range of 0.1 V to +10 V. ø, εØ ~ 1 + ø and therefore: I ERROR ANALYSIS 1I2 = 1 + ø The base-to-emitter voltage of the MAT02 in its forward active I3IO operation is: (9) kT IC V In I1I2 BE = + rBEIC, VCB ~ 0 (5) I (1 – ø) q IS O ~ I3 The first term comes from the idealized intrinsic transistor The In (I equation previously discussed (see equation (1)). SA/ISB) terms in ø cause a fixed gain error of less than ±0.6% from each pair when using the MAT02, and this gain Extrinsic resistive terms and the early effect cause departure error is easily trimmed out by varying RO. The IOUT terms are from the ideal logarithmic relationship. For small VCB, all of REV. E –7–
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