Isolated potentiometer design (Part 2)
Keywords:isopot? op-amp? LM13700? BJTs? diodes?
Applying this algebraic identity to the above circuit equations,
where iX = iD+ ? iD-. This can also be expressed as a current-gain transfer function;
The translinear cell is a linear differential-input, differential-output current amplifier. Because the static current ratios, IX and IY set the current gain, by varying either of them, the gain is varied. The diff-amp stage output current, through the unity-current-gain mirrors, is the amplifier output current, and the amplifier gain is given as the diff-amp stage gain above. It is inverting because the diodes are common-anode. (An alternative common-cathode connection, with anodes connected to the diff-amp bases, has positive gain.) The LM13700 can be used as a two-quadrant multiplier or VGA. The input circuit is shown below.
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This amplifier inputs a unipolar IY and a unipolar (positive) vI . When speed is a consideration, it is best to make vI the gain control and IY (which becomes iY ) the faster waveform. This implementation has a voltage-source input as shown below. When vI = 0 V, then the two sides of the circuit are symmetrical and iD- = iD+ = IX/2.
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Let iX be the differential current,
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Translinear circuits are easier to analyse using the variable, x, to represent the fraction of current that is conducted by one side. Here let
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By KCL,
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and thus
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The range of x is 1. However, to switch all the current from one diode to the other, an infinite differential voltage is required. Because x is a hyperbolic tangent function of vX, the current only asymptotically approaches a complete switchover. Huge voltages are theoretically required to switch decades of currents at near-infinitesimal values. In practice, only about 150 mV will switch a differential BJT or diode pair.
Translinear circuit design
Translinear circuit design requires that a design decision be made about the full-scale value of x. The zero-scale value is x = 0.5, where iD+ = iD- . For a choice of x(fs) = 0.75, then the ratio of diode currents is
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and ln(3) ? 1.009 ? 1. The output fraction, iO /IY at full-scale is also 0.5, and the full-scale multiplier (or VGA) gain is 0.5. Therefore, IY must be twice as large as the desired iO.
Substituting into the previously derived equation for vI and applying a KVL equation involving the resistors,
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