Isolation For Current Sources

I recently had an interesting conversation with a customer about galvanic isolation. They were using a current supply for cathodic protection, but the devices were breaking in the field.

The solution from the vendor was to use an isolated DC/DC to the input of the current supply device. How does isolation solve the problem?

If you look at the circuit below, there's no reason isolation would be required. Power comes into the DC/DC converter, and then the current source pushes current through the load.

Simple current source

The schematic hides the details of the current source circuit, but the details are important to understand why this doesn't work.

Constant current source

Op-amp current source

In both current source circuits, the load is not connected to ground. It has to be connected to an internal point within a circuit. If the bottom side of the load is connect to ground, then a short circuit to ground occurs. Current through the load resistor flows directly to ground, and bypasses the control circuitry.

Short circuit

Isolation breaks the ground loop and forces current to be returned through the current source, instead of to the power source.

Isolation breaks the short circuit

Although the example circuitry shown above is fairly simple, the same principles apply to more complex circuity. I recently designed a 4 to 20 mA current output using a Texas Instruments DAC161S997.

The simplified schematic of this circuit is shown below. In the schematic, the DAC is loop powered. If the loop power is not isolated from the power input to the IC, then current will return to ground at the COMD/COMA pin, and not through LOOP-, which is wrong.

TI DAC161S997 simplified schematic