= 45 mV. However, we measured a peak-to-peak value of
173 mV in the time domain. Taking half of this as the
amplitude would be 87 mV — almost twice the 45 mV we
get from the spectrum.
This difference is because in the time domain, we
measured the total peak-to-peak value of the triangle
wave; while in the frequency domain, we are looking at
the sine wave frequency component amplitudes.
The first harmonic amplitude of this triangle wave is
45 mV, but there are a bunch of other amplitudes in the
spectrum — each dropping off with higher frequency.
These two displays of the same information — the
time domain and the frequency domain — give different
pieces of information.
When evaluating power supply noise, the peak-to-peak and the amplitude at the first harmonic components
are both important metrics.
Loading a Power Supply:
How Not to Blow Things Up
The ripple noise of 1.3% doesn’t sound like a lot, but
this is with nothing plugged into the power supply — not
how it will be used. We really want to know the voltage
and ripple when it is supplying current through a typical
load. The challenge is applying a typical load resistance
and not having it melt.
Adding a resistor across the power supply with a value
to draw the current that might be typically used in our
application will load the supply and test it in a more
realistic situation. How much resistance should be used?
This depends on the current draw in the application.
If we know the power supply voltage and the current
load, we can estimate the resistor to use to draw the same
current using Ohm’s Law. Its three forms are:
If the application calls for 12V at 500 mA, the
equivalent resistive load is about R = 12 V/0.5 A = 24
Before you immediately connect a 24 ohm resistor
across the power supply to see what the voltage noise is,
be sure to estimate the power consumption expected and
use a resistor with a body size that can handle this power
If we connect a 24 ohm resistor across this 12V
supply, the power consumption would be 12^2/24 = 6
watts. This is way more than can be dissipated by a typical
To handle this power, I mounted six one watt 50 ohm
resistors on an aluminum heat spreader as shown in
Depending on how they are connected together, I
can get resistances of 10 ohms to 300 ohms. Two resistors
in parallel have a resistance of 50/2 = 25 ohms — close
enough to 24 ohms.
When this 25 ohm resistor was connected across the
wall wart, the output voltage dropped considerably to
10.6V. Figure 6 shows the measured voltage response of
the wall wart.
The wall wart — with about 500 mA, its “rated”
current — had a voltage of 10.6V with a peak-to-peak
ripple of 0.89V. This is 8% peak-to-peak noise. Even
though it was stamped 12V at 500 mA, its output voltage
was only 10.6V with 8% peak-to-peak ripple noise. This
gives you an idea of what to expect in some cases.
July 2015 55
FIGURE 5. An array of one watt resistors on a heat spreader
capable of dissipating 10 watts.
V V V=IR I= R= R I
FIGURE 6. Measured DC and AC voltage response of the
wall wart, nominally rated for 12V and 500 mA with a 25
ohm load. The performance is far from the rated spec.