**T&M??**

# Isolated potentiometer design (Part 2)

**Keywords:isopot?
op-amp?
LM13700?
BJTs?
diodes?
**

Applying this algebraic identity to the above circuit equations,

where *i _{X}* =

*i*?

_{D+}*i*. This can also be expressed as a current-gain transfer function;

_{D-}The translinear cell is a linear differential-input, differential-output current amplifier. Because the static current ratios, *I _{X}* and

*I*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.

_{Y}This amplifier inputs a unipolar *I _{Y}* and a unipolar (positive)

*v*. When speed is a consideration, it is best to make

_{I}*v*the gain control and

_{I}*I*(which becomes

_{Y}*i*) the faster waveform. This implementation has a voltage-source input as shown below. When

_{Y}*v*= 0 V, then the two sides of the circuit are symmetrical and

_{I}*i*=

_{D-}*i*=

_{D+}*I*/2.

_{X}Let *i _{X}* be the differential current,

Translinear circuits are easier to analyse using the variable, *x*, to represent the fraction of current that is conducted by one side. Here let

By KCL,

and thus

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 *v _{X}*, 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 *i _{D+}* =

*i*. For a choice of

_{D-}*x*(fs) = 0.75, then the ratio of diode currents is

and ln(3) ? 1.009 ? 1. The output fraction, *i _{O}* /

*I*at full-scale is also 0.5, and the full-scale multiplier (or VGA) gain is 0.5. Therefore,

_{Y}*I*must be twice as large as the desired

_{Y}*i*.

_{O}Substituting into the previously derived equation for *v _{I}* and applying a KVL equation involving the resistors,

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