Homework Help: Measuring Power Consumption of a Light Bulb with a Dimmer Switch

1. Feb 10, 2014

baha

Hi guys,

For a science experiment I've got to measure the power consumption of my circuit of a 100W incandescent light bulb connected to the mains AC, with a dimmer switch to vary its brightness. I was planning on just using one of those plug-in energy monitors with a digital display but I've heard that when there is a dimmer in the circuit (triac) this would make the readings of these plug-in energy monitors completely wrong - errors up to 50% - is this true? If so how else can I easily measure the power consumption of such a circuit?

Thanks

2. Feb 10, 2014

The Electrician

Which plug in energy monitor are you referring to? Can you provide a link to a description, or advertisement for sale so we can see which unit it is?

3. Feb 10, 2014

baha

4. Feb 11, 2014

The Electrician

You haven't said, but I'm assuming you live in a country where the mains voltage is 230 VAC.

What equipment do you have available for making measurements at your school? Do you have voltmeters, ammeters, oscilloscopes, resistors, capacitors, small transformers, etc.?l

Do you understand how dimmers work (by narrowing the waveform of the applied voltsge)?

I suspect that the circumstance where the error of your energy monitor becomes high is where the light is greatly dimmed, and the applied voltage becomes a very narrow pulse.

What accuracy will be acceptable for your project? Will reduced accuracy be acceptable at the highly dimmed range of the dimmer?

5. Feb 11, 2014

rude man

I doubt that that's true.

Power monitors work in several possible ways.

One way samples the ac current with a very low-ohmage resistor (like 0.01 ohms). That limits the power dissipation of a 1500W load to the sampling resistor to (1500/220)^2 * 0.01 < 0.5W. The sampling ac voltage then ranges from 0 to 68 mV which is then amplified, rectified, filtered, and sent to a dc a-d converter for display.

Reducing the duty cycle of the load by chopping the ac voltage thereto will not affect the accuracy of the filtered dc voltage, hence the displayed power.

Another way is to pass the load current thru a current transformer. The load current is passed thru a 1-turn, very low-resistance primary winding with a large number of secondary windings. This sizable ac voltage is again rectified etc. Again, no problem with a chopped load current.

I use a low-cost Belkin device which has no problem handling chopped line voltages. It has < 1 mW offset and good scale factor accuracy. The only problem for such devices is reactive loads for which VA rather than W is displayed. A chopped resistive load is no problem.

6. Feb 11, 2014

.Scott

As an alternative, you could (carefully - to avoid fires):
* Thermally insulate the dimmer module with a thermometer probe and an unconnected heating resistor.
* Let the dimmer run until it's temperature increases by some about, say 10C.
* Let it all cool down.
* Connect the resistor and measure the time that it takes to get the same temperature increase with a known wattage.

7. Feb 11, 2014

The Electrician

I see a couple of problems with this. First you end up with a value proportional to the average (absolute, if full wave rectified, preferably) value of the AC current, when what is needed is the RMS value. Then how does this value for the current become a value for power? Isn't a multiplication by some voltage needed? We don't want to display the current; we want to display power.

8. Feb 11, 2014

rude man

This is just a matter of scaling. FYI for a sine wave, rms = (1/√2)(2/π) of the fw rectified voltage dc component.
The voltage for the load is known and constant. It's 220V or whatever (in the US, 115V). So again, just a matter of scaling.

Don't confuse load power with sampling resistor power. We are not interested in the sampling resistor power which would indeed require squaring the current (or voltage).

9. Feb 11, 2014

Staff: Mentor

Dimmers are often based on varying the conduction angle of thyristors or SCRs, and result in current curves that are non-sinusoidal (and often create an appreciable amount of RFI for the cheaper versions). This will complicate power calculations.

10. Feb 11, 2014

The Electrician

The load voltage and current aren't sine waves, which I thought you understood, since you said "Reducing the duty cycle of the load by chopping the ac voltage...".

Also, the OP said "...there is a dimmer in the circuit (triac)..."

I'm not confusing load power with sampling resistor power.

Obtaining load power would still require squaring the current, if we knew the load resistance.

But, since the load is an incandescent light bulb, the resistance of the bulb is not a constant; it depends on the amount of dimming. Just squaring the current won't work when the load resistance isn't known.

Which is what the OP said he has (triac rather than SCR); the standard household dimmer style.

11. Feb 11, 2014

rude man

Fellas, it doesn't matter that the load voltage does not consist of integer numbers of sinusoids. The period of the off-on cycle is on the order of seconds (my microwave is actually an unacceptable 30 sec. or so) whereas the ac frequencies are 50 or 60 Hz. Lots of whole 60 Hz load current cycles go undistorted before one cycle is distorted. And on top of that it averages out.

I don't know what else to tell electrician about his "need" to square things. Not necessary at all. Load power = Vi and you know V = 220 and i is measured.

If you really want to be critical you might do some calculations on the required op amp offset voltage precision reuired to do a 16 bit conversion, which my Belkin does. It makes me think the transformer method might be a lot better.

12. Feb 11, 2014

The Electrician

You apparently don't understand how a triac light dimmer works. The on-off period is not on the order of seconds. How would that be called a "dimmer"? The light would be on full brightness for a few seconds and then completely off for a few seconds.

A triac light dimmer is a phase controlled device. The 50/60 Hz mains waveform is turned on to the load partway through every half cycle, so the on-off period is every 100/120th of a second.

Here's a scope capture showing mains voltage (green), load current (purple) and instantaneous power (dark red) with about half power applied to a 60W incandescent bulb: