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T&M?? # Isolated potentiometer design (Part 2)

Posted: 22 Feb 2016 ?? ? Print Version ?

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

Applying the transfer function, the fractional output current becomes This equation can be expressed in vI as To determine the required range for ±vI, at full-scale let x = 0.75, IX= 100?A, and R||RI = 49.9 k Ω Then The second term in the above equation accounts for the non-linearity of the diodes in series with resistors. If R||RI is made relatively large, the first term dominates and the output-current fraction becomes approximately a linear function of the input voltage. If x is restrained to be only slightly larger than 0.5, the currents do not deviate much from being balanced and linearity is maintained.

RI can be chosen for a given vI (fs) and R found from the parallel combination. For vI (fs) = 4 V, then and

For the output multiplier (U1B) the full-scale gain from the full-scale Y input of IY = 100?A to the output is IY /2 = 50?A because (2 x x 鈭? 1) at the full-scale x of x(fs) = 0.75 is 0.5. The same full-scale x is used for the multiplier as for the V/I converter (U1A). The vXB inputs are at about The multiplier Y input is from the Howland current source. Its output current is sourced into the IY (pin 16) input of LM13700 U1B and is For a potentiometer VP = 10 V, then IYB= 100?A. The fs gain of the multiplier (with x(fs) = 0.75) is 0.5 and the fs output current is thus 100?A.

The multiplier X input has a current-source drive instead of a vI , RI drive. However, its corresponding equation is similar: IXB = 200?A, twice that of IXA = 100?A. With a current gain of 0.5, iOA(fs) = iIB (fs) 鈮? 50?A out of the XIN+ node of U1B. With an even 100?A split between the X-input diodes, an offset of 50?A makes the X current ratio 150?A/50?A = 3 = x/(1 鈭? x), corresponding to x a full-scale x = 0.75.

Output interface
The IYB = 100?A is multiplied at fs by 鈭?0.5, resulting in an output current of 鈭?50?A (negative current into the IO pin). This current flows through Q5, which has its base at somewhat over the minimum 1.75 V needed for the U1 output to function linearly. Q5 is needed for voltage range extension on the high side, for VP+ 鈮? 50 V. The Q5 common-base stage drives a PNP current mirror with a current gain of about one. The output current is at a voltage about a junction drop down from VP+ . The pot emulation falls short of the full range by this drop. This current then develops a voltage across RO = R8 in the isopot circuit diagram. With 0 V of control voltage in, U1A output current is zero as is U1B. i(Q3) = 0 A, R8 has no voltage drop, and VPW = VP- .

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