In the previous part of this article An Isolated Potentiometer, Part 1, the current sources used in the isopot - both Howland and op-amp V/I - were explained. We now get to the heart of the isopot circuits: the multipliers. The two multipliers comprise the LM13700 dual translinear amplifier. Detailed design derivations and application now proceed, followed by prototype build and testing. The isopot circuit diagram is repeated at the end of this second part of the article.
LM13700 Translinear Amplifier
The simplified circuit diagram of the repeated half of the LM13700 is shown below.
Diodes Q11, Q12 match Q1, Q2 and form a translinear gain cell. (BJTs are used as diodes for better junction matching.) It has the property that the ratios of currents in the diodes equal the ratio of BJT currents. This is evident when KVL is applied to the diode loop and the BJT input loop. Let i(Q11) = iD− and i(Q12) = iD+. Then
Also, for a perfect current mirror, i(Q3) = IY = i(Q1) + i(Q2), for α = 1 (β >> 1). The diff-amp output current is differential and is:
Then with this nomenclature, applying KVL to the diode loop,
For matched BJT b-e junctions, Is values match and cancel. Then
Then equating and solving for the current ratios,
It can be shown algebraically that if
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.
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.
Let iX be the differential current,
Translinear circuits are easier to analyze using the variable, x, to represent the fraction of current that is conducted by one side. Here let
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.