Current Loop Has 5-kV Isolation

Jan. 7, 1993
By using an AD7245A DACPORT, a linear optoisolator, and a discrete V-I converter to control loop current, a 4-to-20-mA isolated current loop may be digitally controlled across a 5-kV barrier (

By using an AD7245A DACPORT, a linear optoisolator, and a discrete V-I converter to control loop current, a 4-to-20-mA isolated current loop may be digitally controlled across a 5-kV barrier (see the figure).

The complete circuit exhibits excellent linearity between the digital input word and the isolated analog output current. Up to 11-bit performance is achievable with 12-bit monotonicity. The AD7245A contains a 12-bit digital-to-analog converter, an output amplifier, and a +5-V reference. Although the reference is used internally for the DAC, it can also be used for external biasing.

The IL300 optocoupler contains a single LED and two identical but isolated photodiodes. Both photodiodes receive equal irradiation from the LED. Therefore, by biasing each photodiode with a similar voltage, an identical current will flow through each. Because the relationship between both the LED and photodiode currents is nonlinear, it's necessary to use a feedback loop that monitors the current flowing through one of the photodiodes and then servos the LED drive current when needed.

The DACPORT is configured to operate in the 0-to-+5-V output range. This output swing (V\[subscript\]1) is attenuated by R1 and R2 to give 0 to +1.666 V at the noninverting amplifier input (V2). The negative feedback loop, which consists of the LED and one of the photodiodes in the IL300 (PD1) ensures that the voltage (V3) at the inverting terminal of the amplifier (A1) will be at the same potential as the voltage at the noninverting terminal.

The current flowing through the photodiode will be very small, and more importantly, the change in current flow for a 1 LSB change in DAC code will be negligible (12 nA). Therefore, it's important that a highly accurate and stable voltage reference source is used as a bias supply. The +5-V reference output on the AD7245A provides this bias.

A similar current will flow through the second photodiode, generating an equivalent voltage across resistor R\[subscript\]5. Again, it's essential that an accurate and stable voltage reference bias is used so that any current variations are minimized. In this case, the REF02 supplies a stable +5-V bias voltage.

The isolated output voltage (V4) is buffered by A2. V4 controls a V-I converter formed using A3and Q1.

The buffered output voltage (V5) ranges between 0 and +1.666 V, depending on the code applied to the DAC. This is transformed into a loop current by A3 and transistor Q1. The loop current (Iloop) is monitored with a 10-Ω sense resistor (R10). A low resistance value minimizes the voltage dropped across it and allows the loop to operate with low loop supply voltages. Q1 is connected as an emitter follower in a feedback circuit that controls the current flowing in the loop. The feedback circuit forces both A3 amplifier inputs to the same potential. Consequently, the voltage across the sense resistor (V6) is equal to the voltage across the feedback resistor (R8). Because no current flows into the amplifier inputs, current I3 flowing through R8 must equal the sum of the currents flowing through R6 and R7. Iloop is therefore controlled by I1 and I2.

I2 sets up a 4-mA loop-current offset, while Iloop is controlled by the DACPORT to add a further 0 to 16 mA. The loop current is:

I\[subscript\]loop (mA) = 4 + 16(D) where D is a fractional representation of the digital word applied. It may be programmed between 0 and 4095/4096.

The circuit contains an offset and full-scale adjustment facility. This offset potentiometer (P1) should be trimmed until a 4-mA current is measured flowing in the loop. This is done with all 0's loaded to the DAC. With all 1's loaded to the DAC, the gain-adjusting potentiometer (P2) should be trimmed until a current of 20 mA - 1 LSB (19.996 mA) is measured flowing in the loop. Note that 1 LSB of current is equivalent to 16/4096 mA = 3.9 µA.


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