Nuts & Volts - June 2007

■ FIGURE 4. A differential amplifier measures the current. An integrator supplies the vertical signal.

reference inductor. As the current ramps down from plus full scale to minus full scale, the inductor voltage becomes -1V, giving the desired 2V peak-to-peak signal. As the test signal ramps up for 1 ms and down for 1 ms, the repetition rate is 500 Hz.

Making a Ramp

linearly to match the increasing current. This is achieved by feeding the inductor voltage to an integrator. This cancels the differentiation of the input current by the inductor. The result is a curve whose slope represents the permeability.

In an integrator, any input offset causes the output to grow without limit. As there should be no net DC voltage across the inductor, the mean Y voltage must be zero. This is emulated by placing a 1 megohm resistor (R24) across the integrating capacitor. This doesn’t materially alter the displayed curve but any net DC Y output should be ignored.

in its horizontal sections reveals a resistive imbalance that should be trimmed out. In practice, I found I could adjust VR2 to give a plausible B-H display in the normal test mode. Loops whose tips cross themselves show misadjustment.

Design Details

This tester arose from my interest in inductors for small DC-DC converters. Thus, the test range is 10 to 1,000 µH and the maximum test current is 500 mA. The latter limit let me power the tester from a conventional wall-transformer and to use rather skimpy heatsinking.

The time taken to sweep the current from -500 mA to +500 mA depends on the voltage one wants across the maximum inductance. I chose the peak-to-peak voltage for a 1 mH inductor to be 2V.

A voltage (V) across an inductor (L) causes the current through it to increase at V/L amps per second. One volt across 1 mH causes a 1 A/ms current change. The opposite also applies; a 1 A/ms rate of current change induces 1V across a 1 mH inductor.

The maximum test current ramps from -0.5A to + 0.5A in 1 ms. This induces a +1V signal across the

I avoided a lot of board clutter by using an ICL8038 signal generator chip (U3) to create the ramp waveform. This chip contains the necessary switched current sources, comparators, and buffer amplifiers to generate a ramp having equal positive and negative peaks. It also has sine and square wave outputs. The latter supplies the test signal for setting VR2.

The ramp length is controlled by two equal resistors and a capacitor. The capacitor (C5) is 0.047 µF and should be plastic foil, not ceramic. The resistors (R1 and R2) should be 1% types and, ideally, should be matched. The six-way switch (SW1) selects one of four output peak amplitudes: 50 mA, 100 mA, 200 mA, and 500 mA. It also selects the square wave and zero.

Driving Current

Real Inductors

Any real winding has some resistance so the voltage across an inductor has a term proportional to the drive current. Since the X voltage is a linear function of the current, we can subtract a little of it from the coil voltage to correct for the winding resistance. VR2 lets you make this adjustment for windings up to 2. 5 ohms. Switch SW1 has a calibration setting which applies a current square wave to the inductor under test.

With the oscilloscope in its normal mode, the resulting Y output should also be a square wave. Any tilt

An ordinary op-amp can’t supply a bipolar current output. Here, I used an old trick. An op-amp drives a load resistor and its power pins supply positive and negative drive currents to a push-pull output stage. The op-amp used should have an output stage with matching NPN and PNP transistors. The TL081 would be ideal, but I only had TL082 (dual) and TL084 (quad) amplifiers in stock. I used both halves of a TL082 and gave each half its own 200 ohm load resistor.