# Some bulb tests (and Kill A Watt)

## July 2017

Light bulbs, LED, Compact Fluorescent (CFL), halogen incandescent and conventional incandescent bulbs, are all rated in watts. Power, measured in watts is the rate that energy is transfered to the bulb. (If you're interested, 1 watt is a rate of 1 joule per second, and 746 watts equals one horsepower) The electrical company charges you by the kilowatt hour, which is yet another way to talk about energy transfered. Do they really measure the power your equipment is using to calculate the energy you use in a given time? It may be surprising, but they don't. They measure the current draw and multiply that by the nominal voltage. But isn't that the same thing as measuring the power? After all power, by definition, is voltage multiplied by current, isn't it? Yes, and no. In a DC (direct current) circuit, power is indeed equal to the current times the voltage. But things get a little more complicated with AC (alternating current), which is what your house outlets provide. With a resistive load, such as a conventional light bulb or a heating element (think toaster, electric stove, etc.), the power is the current times the voltage. However, if the load is reactive (has significant inductance or capacity), or the load has a non-linear relationship between the current and voltage (the shape of the current waveform doesn't mirror the shape of the voltage waveform), the power is no longer equal to the current times the voltage. In such a circuit, there will be more apparent power (current times voltage), measured in volt amps (abbreviated VA) than there is actual power. Since you are actually paying for apparent power, rather than the actual wattage used, your bill is higher than you'd expect. There is a unitless value that's the ratio of the actual power used divided by the VA used known as the power factor (PF). Essentially a measure of financial efficiency. This will be a number between 0 and 1. With a PF of 1 (resistive load, conventional light bulb, toaster, etc.), you're getting all the power you pay for, with a completely reactive load (a capacitor or inductor across the line), you're paying money for electricity, and getting nothing in return. In the real world, a power factor of 0 would be unheard of, but getting down around 0.5 isn't uncommon, meaning that you're paying for twice the power you actually use.

What about LED and compact fluorescent (CFL) lights? Do they have a power factor of 1 like a regular bulb? Or is it lower, meaning that you pay more than the wattage would lead you to believe? Well, I knew that the PF was not 1 on these devices, but how bad is it really? I took 2 LED bulbs, 2 CFL bulbs, and for a control, a conventional bulb. I then measured the power factor of these bulbs.

There is a handy device, available at most larger hardware stores, called a "Kill A Watt." This measures the wattage, the voltage, the apparent power, and power factor of whatever is plugged into it. But can an inexpensive device like this (usually about \$19) be trusted? I used a Kill A Watt and an oscilloscope, which I trust more than an inexpensive hardware store device, to collect data about these bulbs. I looked at the results for each bulb and compared the oscilloscope measurements to the Kill A Watt.

Here's the information about the bulbs, I just chose 5 random makes of bulb, there are way too many brands to take a comprehensive look at them all.

Bulb 1, Great Value LED rated 9W (60W replacement, claimed 800 lumens) (the most dazzling bulb tested)

Bulb 2, GE LED rated 11W (60W replacement, claimed 800 lumens)

Bulb 3, Great Value CFL rated 13W (60W replacement, claimed 870 lumens)

Bulb 4, Great Value 43W halogen (60W equiv, claimed 620 lumens)(the dimmest bulb tested)

Bulb 5, Sylvania CFL rated 13W (60W equiv, claimed 850 lumens)

First, I'll cut to the chase and give the results and some observations, then, for those who are curious, I'll describe how I made the measurements with the oscilloscope. There isn't much to tell about making the measurements with the Kill A Watt, you simply plug it in, plug the device to be tested into it, then push the buttons.

Here are the results in tabular form, each bulb has two entries, one from the oscilloscope (the upper one), one from the Kill A Watt (the lower one), with the differences between the measurements in yellow.

 Device Power VA PF 8.9W 14.1VA 0.63 1 9.2W 14.4VA 0.63 Difference 3.40% 2.10% 0.00% 14.0W 18.1VA 0.78 2 13.9W 17.9VA 0.76 Difference -0.71% -1.10% -2.60% 14.5W 24.6VA 0.59 3 13.3W 26.0VA 0.56 Difference -8.30% 5.70% -5.10% 45.2W 45.2VA 1 4 45.5W 45.5VA 1 Difference 0.66% 0.66% 0.00% 13.4W 21VA 0.64 5 13.1W 23.5VA 0.61 Difference -2.20% 11.90% -4.70%

Bulb 3, the Great Value bulb, a very common brand, scored the lowest on power factor. With this bulb you'd be paying nearly 80% more for the electricity used than the specs on the light bulb claim. Truly sad. Even the best energy saver bulb tested here uses almost 30% more electricity than claimed. That's just the price of using these bulbs, but 30% over is a lot better than 80% over.

The only bulb that had a PF of 1 was the incandescent bulb. However, it used the most power by far and had the lowest light output, which was very obvious to the eye. (I don't have any way to measure the light output of the bulbs, that would be very interesting)

As you can see, most of the readings between the Kill A Watt and the oscilloscope are fairly close. Since there are several possible sources of error in the oscilloscope readings, and surely the same in the Kill Watt, these readings are fairly reasonable. I'm surprised that a few of the differences in readings are far greater than others, and more so that some of them are off in different directions. I tested over and over again and kept getting very similar results.

Using the claimed light output, what is the range of lumens per VA?

 Device Lumens VA ('scope) Lumens/VA 1 800 14.1 56.7 2 800 18.1 44.2 3 870 24.6 35.4 4 620 45.2 13.7 5 850 21 40.5

There is quite a difference in efficiency. Most interesting is that the bulb with the greatest subjective intensity (bulb 1) used the least apparent power (VA) Also, despite having the best power factor, the halogen incandescent bulb is by far the least efficient and I'd only recommend using the halogen incandescent bulbs if you have a lot of circuits using older style light dimmers. Which I have a lot of...

I can't test every bulb available here, while this is interesting and sort of gives you "the lay of the land", it's not too useful to make buying decisions. First of all, you'd never find the exact same bulb I'm using, also, bulbs of the same brand could well have different circuitry inside, yielding different results. It does show that all of the CFL and LED bulbs cost less to operate than the halogen. If you're curious as to which of the bulbs available to you is most efficient, your best bet is a Kill A Watt and the purchase of a lot of different bulbs to test. Not too practical, but there you have it.

How did I use an oscilloscope to measure the wattage, apparent power, and power factor of the bulbs? That's a lot deeper than the results themselves, here it is if you want to read it (and I hope you do, it's interesting). First, we start with the circuit used:

The isolation transformer isolates the "hot" and "neutral" pins of the secondary from ground, they float. This is for safety, both for the user and the test equipment.It's still fairly dangerous to play around with line voltage on the bench, and I wouldn't recommend it to an inexperienced person, but the transformer helps somewhat.

The circuit is connected to the transformer with a "suicide cord", a line plug on one end, and alligator clips on the other. Again, dangerous and not for the inexperienced. We used to use these all the time when I worked in TV shops 30+ years ago. The fuse on both sides of the suicide cord might seem overkill, but with the circuit floating, it's not obvious which side to put the fuse in, so why not cover both possibilities?

The voltage across the bulb is measured between the reference point and the top of the bulb. The actual voltage across the bulb is less by the voltage across the sense resistor. The voltage across the sense resistor is subtracted from the total voltage in the software, so it's not an issue. It's small enough that it wouldn't make a serious difference if ignored.

The current through the bulb is measured by reading the voltage across the sense resistor. With a one ohm resistor, one amp of current through the resistor yields one volt across it.

The actual value of the resistor is important, so I measured it. You can't hook an ordinary ohmmeter to a 1 ohm resistor and expect to get a useful reading. 10% or more of the reading will be the resistance of the test leads and clips. So how to measure it? You use the "Kelvin 4 wire method." You simply pass a current through the resistor and use a voltmeter connected directly to the leads of the resistor and measure the voltage. This way any resistance in the wires supplying the current to the resistor will still cause a voltage drop, but it won't be included in the reading on the voltmeter. Ohm's law is used to find the resistance where R = E / I, with R being the resistance, E the voltage, and I the current.

I used a current regulated bench supply and two digital multimeters. I set the current as close to 1 amp as I could and read the voltage. I divided the voltage by the current and came up with 1.01 ohm. (1.003 volts and 0.991 amp) You could have just as easily set the voltage as the current, the results would be the same. The meter used for current is rated at 0.7% accuracy and the meter used for the voltage is rated at 0.5%. Obviously, we only know the value of the resistor to a little worse than 1%, still better than the 5% rating of the resistor. Since the oscilloscope has a 3% of full-scale reading.the oscilloscope is the main source of error.

Then I set up the oscilloscope and measured the voltage and current of each bulb, voltage on one trace, current on the other. I used a 10x probe for the voltage and a 1x probe for the current. I then saved the data for each trace, as displayed the oscilloscope, to a CSV (comma separated value) file. Then I edited the CSV files to represent one cycle of the waveform. You can use a spread sheet or a capable text editor for this.

I wrote a program that analyzed the CSV files to calculate the power, RMS voltage and current, the apparent power and power factor. These are the values reported above. The program listing is at the bottom of this article and may be run in either Linux or Windows, using Perl and, in Windows, the Linux Subsystem (alternatively, a Perl native to windows exists, I haven't used it in many years).

The first thing I noticed was that the current waveform was very non-linear. I didn't expect a sinusoidal current waveform from a switching supply, that's why I tested the PF of these bulbs in the first place, but three of the five bulbs tested didn't even come close. The GE LED bulb drew current when the voltage was non-zero but still very non-sinusoidal. Of course, the halogen incandescent was a resistive load, and the current did follow the voltage. The other bulbs drew current in smaller portions of the waveform. In a weird way.

The applied voltage is not a pure sine wave either, I wonder if all the non-linear load from all the switching supplies in the house distorted the waveform? Or is the isolation transformer producing a non-sinusoidal output?

Bulb 1 waveform:

Bulb 2 waveform:

Bulb 3 waveform:

Bulb 4 waveform:

Bulb 5 waveform:

Here's a listing or the program used to calculate the needed values, with comments to explain how it works:

```#!/usr/bin/perl

# makes the warnings and errors at run time far more useful
use strict;
use warnings;

# declare the variables, when used with strict, this catches a lot of typos
my (\$file_name);
my (\$current_line);
my (\$time);
my (\$current);
my (\$voltage);
my (\$voltage_over_10);
my (\$sense_r_voltage);
my (\$sense_r);
my (\$count);
my (\$p_sum);
my (\$in_file);
my (\$av_power);
my (\$i_sq);
my (\$i_sq_sum);
my (\$e_sq);
my (\$e_sq_sum);
my (\$i_rms);
my (\$e_rms);
my (\$va);

# The measured value of the sense resistor
\$sense_r = 1.01;

# get the csv file name
print "Enter file name: ";
\$file_name = <STDIN>;
chomp (\$file_name);     # remove the carriage return from the name of the file
unless (-e \$file_name)
{
die ("***ERROR*** '\$file_name' does not exist.\n");
}

# open csv file
unless (open (\$in_file, "<", \$file_name))
{
die ("***ERROR*** Could not open '\$file_name' for read, \$!\n");
}

# zero out values, just to be safe
\$p_sum = 0;
\$i_sq_sum = 0;
\$e_sq_sum = 0;
\$count = 0;
while (\$current_line = <\$in_file>)
{
chomp (\$current_line);

# \$time is in mS
# \$voltage_over_10 is voltage at high end of the bulb /10 due to 10x probe
# \$sense_r_voltage is the voltage across the sense resistor
# split line into sections at commas
(\$time, \$voltage_over_10, \$sense_r_voltage) = split (/,/, \$current_line);

# current is E/R
\$current = \$sense_r_voltage / \$sense_r;

# voltage aross the bulb is the voltage on the high side of the bulb
# minus the voltage across the sense resistor
\$voltage = \$voltage_over_10 * 10 - \$sense_r_voltage;

# used to calculate the rms voltage and current, which is the sum of
# the individual values averaged, with the square root taken after
# the summation is complete
\$i_sq = \$current ** 2;
\$e_sq = \$voltage ** 2;
\$i_sq_sum += \$i_sq;
\$e_sq_sum += \$e_sq;

# increment the instantaneous power, for averaging later
\$p_sum += \$current * \$voltage;

\$count++;
}

# close file
close (\$in_file);

# caculate average power from summation
\$av_power = \$p_sum / \$count;

# calculate rms current from summation
\$i_rms = sqrt (\$i_sq_sum / \$count);

# calculate rms voltage from summation
\$e_rms = sqrt (\$e_sq_sum / \$count);

# calculate volt amps
\$va = \$e_rms * \$i_rms;

# print results
printf ("\nAverage power: %.1f W\n", \$av_power);
printf ("rms current: %.3f A\n", \$i_rms);
printf ("rms voltage: %.1f V\n", \$e_rms);
printf ("VA: %.1f VA\n", \$va);
printf ("Power Factor: %.2f\n", \$av_power / \$va);
```