Many discrete analog circuits require precisely matched transistor pairs. Logarithmic and antilogarithmic functions, differential amplifiers, and multipliers possess inherent matching features generally not obtainable with discrete components. The most common method of matching discrete components is by using a curve tracer. However, this is a tedious and expensive method. The circuit shown is an inexpensive circuit that can be used to match two discrete transistors based on their saturation current (IS) (Fig. 1).
If two transistors are operated at the same base-emitter voltage, then the ratio of their collector currents is equal to the ratio of their saturation currents:
IC1/IC2 = IS1/IS2
If the collector currents are equal, then the saturation currents will be equal and the pair will be closely matched. This assumes that the VBE, VCE, and VCB of each transistor are identical. In Figure 1, Q1 is connected in the feedback path of U1A. The summing junction of the op amp forces the collector voltage to zero (within the offset voltage specification of the OP-497, or 50 µV).
With the base connected to ground, the collector-base voltage is zero. The collector current of Q1 is set by R1 and the input voltage, which is a unipolar 0- to 10-V triangular, 1-Hz waveform. This sweeps the collector current of Q1 from 0 to 1 mA, independent of Q1’s beta. The output of U1A is the inverted VBE of Q1, so U1B is used to change the sign of this voltage. This VBE is applied to the base of Q2, which is connected as the antilog element of U1C. The noninverting input of U1C is connected to the same VBE, which forces the collector voltage of Q2 to track its base voltage (again, within the offset voltage limits of U1), forcing the collector-base voltage of Q2 to zero.
If the saturation current of Q2 is matched with the saturation current of Q1, the output voltage of U1C will be the source voltage plus the baseemitter voltage of Q1. U1D subtracts this VBE, leaving a voltage proportional to the collector current of Q2. If the saturation currents are equal, the output voltage of U1D will be identical to the source voltage.
Figure 2 shows the waveforms that result when IS2 is 50% smaller than IS1, while Figure 3 depicts the results when IS2 is 50% larger than IS1. Good performance of the circuit depends on close matching of all resistors.
A byproduct of the circuit configuration is the ability to measure or display the IC versus VBE characteristic of Q1. With the source voltage connected to channel 1 and the output of U1B connected to channel 2, set the oscilloscope in the X/Y mode with channel 1 on 5 V per division and channel 2 on 200 mV per division.