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Note that the oscilloscope reads the peak-to- peak DCC voltage
as 14.65 volts. When this photo was taken, the output voltage
was measured at 13.57 volts DC. This is why I use a 1-volt dif-
ference between DCC track voltage and DC output voltage as a
“rule of thumb”.
Okay, now we have a DC power supply (very much like what is
in a DCC decoder). While that is necessary, it doesn’t do what
we came here for. Let’s get on with designing the circuits that
do what we really want.
Camera Power
I’m going to assume a DCC track voltage of 15 volts through-
out this example. As you saw above, the DC power supply will
put out about one volt less than the DCC input. I’m going to
use 14.3 volts in my calculations. First, let’s supply power to
the camera.
In order to design this circuit, we need to know how much volt-
age and current the camera needs. Since it was designed to
DCC Impulses Column - 3
use a 9-volt battery, the quickest way to get
this data is to connect it to one. The cam-
era owner connected a conventional battery
through an RRAmpmeter to the battery. He
read 8.6 volts at 0.12 amps.
We’ll use those numbers to figure out what
value of resistor (R1) to use to connect the
camera to our power supply. The purpose
of this resistor is to drop the voltage down to what the camera
wants when the camera is drawing its operating current.
First, we figure out how much voltage we need to drop across
the resistor. The camera seems happy at 8.6 volts and the
power supply will deliver 14.3 volts. Thus, the resistor needs to
drop 5.7 volts (14.3 volts – 8.6 volts).
4: Capacitors used in this circuit: electrolytic (top) and
ceramic (bottom). The electrolytic capacitor shown is a
much larger value than what this circuit needs.
5: DCC input waveform to power supply – 14.65 volts
“track voltage”.
4
5
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MRH-May 2013