Circuit lets you isolate and measure current
Anton Mayer, Murr, Germany- June 1, 2011
You often need to measure current during circuit design and debugging. You can perform that task by breaking a path, inserting a shunt resistor, measuring its voltage, and converting the voltage to current. Unfortunately, that approach is sometimes impractical with an oscilloscope because one side of an oscilloscope probe connects to ground. Thus, you need to isolate the oscilloscope from the circuit under test.
The circuit in the figure produces a voltage proportional to current and isolates the oscilloscope from the measurement point. The circuit uses IC1, an HCPL7800 isolation amplifier, which adds input-to-output isolation of as much as 890 V. It also amplifies its input voltage by eight. The table shows the overall gain for each input range. The circuit’s bandwidth is typically 100 kHz.
The HCPL7800 has a 3% tolerance. When you are using resistors with 1% tolerance, the 3% tolerance dominates the overall uncertainty of the circuit. The circuit uses two independent voltage supplies. A 9-V battery supplies the input part of IC1. A stabilized 9-V-to-11-V wall-wart power supply powers the output side of IC1 with IC2, an LM358 successive amplifier.
When battery switch S5 closes and the voltage of the battery is sufficient for the circuit, LED1 illuminates for approximately 3 s. The duration of this illumination minimizes drain on the battery. LED2 is on when the 9-V-to-11-V power supply is operating. IC5, an L272, provides an additional ground potential halfway between the supply voltage. With this split supply, you can measure both positive and negative currents. T&MW
This article originally appeared in the September 23, 2010, edition of EDN.