where VL is the DC supply voltage ( 3.3V in
At this point, all the inductor current is
flowing through the switch. After holding
the switch closed for some period, TCLOSED,
the switch is then opened. At this point,
the inductor current has risen to the value:
Now things get interesting. When the
switch is opened (Figure 3), the inductor
current is no longer able to take the switch
path, but the inductor current must keep
If you’re not used to working with inductors, they
are the dual of a capacitor. Whereas a capacitor is
perfectly willing to instantaneously change its current
but tries to maintain whatever voltage is already across
it, the inductor is perfectly willing to instantaneously
change the voltage across it, but wants to maintain the
current flowing through it. In this case, the inductor will
increase its voltage until it finds a new path for its
current — which happens to be through diode D.
In fact, this is how the ignition on your car works. If
we replace the capacitor in Figure 3 with a spark plug,
we have an automobile ignition system. In that case, an
inductor — referred to as the coil — is charged by
momentarily closing a switch that shorts the coil across
the car’s battery. When the switch is opened, the built-up current has to go somewhere.
The inductor voltage at the opened switch rises until
it’s high enough to “jump the gap” of the spark plug.
This provides the inductor current a path back to the car’s
“ground” and — in the process — creates a spark within the
engine cylinder, igniting the air-gas mixture in the cylinder.
Okay, back to our boost circuit. Remember that we
assumed an initial capacitor voltage of 0V. So, when the
switch opens, the inductor voltage at the
inductor/diode/switch junction need rise only a few tenths
of a volt for the diode to start conducting and the capacitor
to start charging.
At this point, the capacitor is charging as the inductor
discharges (that is, the energy stored in the inductor during
the closed switch is transferred to the capacitor during the
This closed-switch/open-switch cycle repeats, with the
result that the output voltage, +V, rises higher and higher.
When it finally reaches the desired high voltage, we’ll need
to adjust the switch “on” and “off” times to maintain the
output at that voltage. Wait a minute — just how do we
control the on and off time?
Fortunately, there are integrated circuits that are
designed to do exactly that. If you do a search on the term
“boost” at an IC distributor’s website, you should find lots
of choices. In this case, we are dealing with a need to
produce less than 100 mA at a 30V output; there are lots
of boost regulator ICs capable of handling that.
Because the required current is relatively low, physically
small and inexpensive inductors and capacitors should
suffice. The boost control IC datasheet will give more
particulars about just what these components should be.
Refer to Figure 4.
Driving the Ultrasonic
Transmitter from the
Okay, so we can generate the high voltage with relative
ease. So, how do we take the 3.3V or 5V signals from our
microcontroller or other device that generates the
transducer’s frequency signal and use that to turn the high
voltage on and off? What we are looking for is a circuit like
the one in the Figure 5 block diagram.
Basically, we just want two fast switches that can
SERVO 03.2018 37
Figure 3. Boost
circuit releasing its
energy to the
Figure 4. Complete boost circuit diagram.
IL = TCLOSED