First of all a note of caution: you should embark on such a project only if you are very familiar with electricity and the dangers of handling line voltages. In certain unfortunate conditions, a person can be electrocuted by the line voltage of 120 Vac and die. I assume no responsabilities!
PLEASE, PLEASE...... BE CAREFUL!
I am observing only a fraction of the advertised predicted life of 10,000 hours; consequently I decided to investigate the possibility to prolong the life of these complex lamps by reducing their power consumption, which, of course, is also an added bonus to the household expenses. It is a well known fact that less voltage and, therefore, current, equates normally to longer life of any electrical component. The reason is a reduction of power losses typically dissipated in heat.
To test my hypothesis, I used a Variac (pic#0) to test a 100 W equivalent CFL, which consumes nominally 23 W. A Variac is a variable transformer, which allows me to vary the voltage to the bulb. I use a CFL that I bought at COSTCO. The brand is: FEIT “Conserv-Energy”. The technical specifications are:
Power: 23 W (equivalent to a 100 W incandescent lamp)
Voltage: 120 Vac
Current: 350 mA
To my surprise, the bulb lights up brightly at 60 V! By increasing the voltage the light emitted increases also, but not by a significant amount.
Emboldened by this discovery, I thought how to reduce the line voltage without wasting additional power. There are various techniques, each one with advantages and disadvantages:
1. Autotransformer: The most expensive proposition
2. Capacitor in series: cheap, causes a phase advance in the line current
3. One diode in series: very cheap, but causes DC current in the line
4. Resistor: expensive because it wastes power.
I decided to investigate solution 2. and solution 3. I have to add that my line voltage is 125V and not 120V, therefore in order to compensate for the difference I will multiply my power measurements by 0.92 (120/125^2).
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Step 1: Capacitor in Series
Capacitor in series .
A capacitor is a reactive component that does not consume real power, but only apparent power; its impedance is: Xc = 1/ (2*π*f*C), where f is the line frequency, in USA 60Hz and C is the capacitance, typically expressed in microfarad (μF). The circuit is depicted in pic#1. The waveforms needed to calculate the power is depicted in pic#2. With a 8μF capacitor in series the power consuption at the bulb has been reduced from 23 W to 14.2 W, a net reduction of 38.3%. assuming an average cost of $0.14 per KWh, for a bulb life of 10,000 hours the cost would be: 0.023KW*10,000Hour*$0.14 = $32.20, but with the capacitor, the cost is reduced by 32.6%, that is: $12.30. At this point I need to add a comment regarding the capacitor effect on the line. A capacitor creates a phase advance in the current. The utilities would like to supply power to users without phase shift between voltage and current. Said that, the typical loads of a household are: electric motors, washers, kitchen appliances, etc. which create a phase-lag, being inductive. By using this technique of adding a capacitor, we actually “help” the utilities, by compensating the phase-lag of a typical household with some phase advance! In pic# 3, I show a bank of 4 CFCs all connected in parallel and then in series with two capacitors in parallel (I had them in my electronic junk) for a total of 32μF.
On the web the cost of a starting capacitor 8 μF, 250Vac is around $2-3. The price is not much different for higher capacitance values if multiples of 8 μF is used for more than one bulb.
Step 2: Capacitor Implementation
1. Turn off circuit breaker, so that there is no voltage where are you working, better to be safe than sorry !
2. Open ceiling lights housing
3. Identify live (black) and neutral (white) feed wires.
4. Cut live wire (black)
5. Insert FAST-ON female connectors on each end of cut wire.
6. Insert female FAST-ON connector into the male receptacle on the capacitor and isolate with electric tape.
7. Reassemble ceiling light housing
8. Turn on circuit breaker
9. Job done
Step 3: Diode in Series
Diode in series
A diode is a component that allows current to flow in one direction only. If we connect a CFL in series with a diode, the power utilized will be sourced only by the positive or the negative portion of the voltage present at the line. Again, the CFL lights up brightly even after the insertion of the diode. Pic#4 depicts the circuit and pic#5 the waveforms. Now the power used by the CFL is 17.30 W, that is a reduction of 24.9%. The energy cost saving for the life of the bulb would be: $8.02. I need to comment on the DC current extracted from the line. The utilities do not like that, but a concerned user could rig up CFLs in the household in such a way that on average half uses the negative and half the positive portion of the line supplied power, and in this way balance the load. The cost of a diode is around $0.25. What diode to use? Any diode that can handle > 2 A (peak current), and >250 V (peak voltage) This for a single 23W (100W equivalent) bulb.
Step 4: Diode Implementation on a Standing Light
1. Remove power plug
2. Expose feed wires by opening the outer insulation sleeve of the power cable
3. Identify the live wire
4. Cut it and strip back insulation to expose internal conductor
5. Solder diode (polarity not important, but read explanation in order to balance the load) beween the two exposed copper wires and isolate with shrink tube
6. Cover diode with high quality isolation electric tape
7. Job done
Step 5: Conclusions
For reason of completeness I added also pic#6 that shows the waveform of the CFL without any addition. The nominal power is: 23 W, I measure 23.9 W, which multiplied by 0.92 equals 22 W; that is in good agreement with the specification. Of course, to the aforementioned savings, I should add the cost saving due to the prolonged life of the CFLs, but I do not have any hard data, therefore I leave it to you, reader, to estimate it.
In the case of a bank of lights, I suggest the "capacitor method" for practical reasons. Because the capacitance tolerance is quite large in the motor-starting capacitors, in the range of +/-20%, to be sure one has at least 8μF, a capacitor with a nominal capacitance of 10μF is needed per bulb. For example with a bank of four bulbs one needs a capacitor with a nominal capacitance of 40μF.
In the case of a single bulb, I suggest the diode method for practical reasons, although the power saving is less.
In conclusion, according to reliable statistics: http://www.eia.doe.gov/cneaf/electricity/esr/table5.html
the average usage of electricity in 2008 in USA was 920KWhr/month, of which 12% was due to lighting. This represents a cost of 920*0.12*12*0.14 = $185/year ($0.14 = avg. cost KWhr). By replacing incandescent bulbs with CFL, one would save 77% and implementing my suggestions with capacitors: 86% and diodes: 83%. …and remember less power we use and less carbon dioxide goes into the atmosphere.