■ 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
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
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
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
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.
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
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.
The current driver runs open-loop.
I tried feedback from the current-sensing resistor but it was unstable
when driving a test inductor. This