MAT02 Figure 1. Log Conformance Test Circuit LOG CONFORMANCE TESTING kT IC1 The log conformance of the MAT02 is tested using the circuit ∆V In BE = + I q I C1 rBE1 – IC2 rBE2 (2) shown above. The circuit employs a dual transdiode logarithmic C2 converter operating at a fixed ratio of collector currents that are A ramp function that sweeps from 1 V to 10 V is converted by swept over a 10:1 range. The output of each transdiode converter the op amps to a collector current ramp through each transistor. is the VBE of the transistor plus an error term which is the prod- Because IC1 is made equal to 10 IC2, and assuming TA = 25°C, uct of the collector current and rBE, the bulk emitter resistance. the previous equation becomes: The difference of the VBE is amplified at a gain of ×100 by the ∆VBE = 59 mV + 0.9 IC1 rBE (∆rBE ~ 0) AMP01 instrumentation amplifier. The differential emitter-base voltage (∆V As viewed on an oscilloscope, the change in ∆VBE for a 10:1 BE) consists of a temperature-dependent dc level plus an ac error voltage, which is the deviation from true log confor- change in IC is then displayed as shown in Figure 2 below: mity as the collector currents vary. The output of the transdiode logarithmic converter comes from the idealized intrinsic transistor equation (for silicon): OBSOLETE kT I V = In C BE (1) q IS where Figure 2. k = Boltzmann’s Constant (1.38062 × 10–23 J/K) With the oscilloscope ac coupled, the temperature dependent term becomes a dc offset and the trace represents the deviation q = Unit Electron Charge (1.60219 × 10–19 °C) from true log conformity. The bulk resistance can be calculated T = Absolute Temperature, K (= °C + 273.2) from the voltage deviation ∆VO and the change in collector I current (9 mA): S = Extrapolated Current for VBE→0 IC = Collector Current ∆VO r × 1 (3) An error term must be added to this equation to allow for the BE = 9mA 100 bulk resistance (rBE) of the transistor. Error due to the op amp This procedure finds r input current is limited by use of the OP15 BiFET-input op BE for Side A. Switching R1 and R2 will provide the r amp. The resulting AMP01 input is: BE for Side B. Differential rBE is found by making R1 = R2. –6– REV. E