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Name: Rich
Status: other
Grade: other
Location: CO
Country: USA
Date: Spring 2013

An overhead kitchen light is controlled by a wall dimmer switch. Works as advertised. When light fixture is taken out and a digital multimeter is used to measure voltage across common and hot, I get 120 volts regardless of position of dimmer switch (circuit is always hot. Can you explain this?

Hi Chuck,

Hi Rich,

The reason your digital voltmeter (DVM) reads full voltage is simply because its impedance is extremely high (over 10 megohms), so it draws almost no load current at all (only a couple of microamps). This is such a small load, that the normal circuit capacitance in the dimmer circuit is enough to capacitively couple the very few microamps from the mains, directly to the DVM, completely bypassing your dimmer circuitry.

The only way to get an accurate reading is to put a load on the dimmer such as a small light bulb, resistor or similar, and measure the voltage with the load connected.

Nonetheless, remember that the dimmer outputs a pulsed AC waveform; it does NOT output a steady (reduced-voltage) AC output. As a result of this, a DVM is of little use in measuring the dimmer's output. An oscilloscope would be a far better investigative tool.

Regards, Bob Wilson

Hi Rich Light dimmers work by changing the amount of current that runs through the bulb, and this is typically done with a semiconductor triac. This device works by preventing current from flowing until it is triggered. Once triggered, the device will continue to conduct current until the current stops. The current stops flowing when the current sinusoidal wave form crosses zero. The process repeats 120 times a second due to the bidirectional nature of 60 Hz AC current. When the bulb is not at full brightness, the triac is triggered later in the cycle - so instead of being triggered at zero degrees for full brightness, it may be triggered at, say 90 degrees, which will produce a current wave form that is zero for one-quarter cycle, and full on for the other quarter. The process repeats when the current changes direction, accounting for the full 60 Hz wave.

If there is no load presented to the triac, as it would be if the bulb were removed, there is no current, and the voltage present at the triac would be the line voltage of an open circuit. As soon as a bulb is put back in the circuit, the circuit is completed with a path for the current, and unless the triac is conducting, the current through the bulb would be zero; the voltage at the triac would therefore be zero. Hope this helps.

Bob Froehlich

It is predominately caused by leakage through the Triac (the semiconductor device in the dimmer that switches power on/off a certain fraction of the time each 60 Hz cycle). Normally when there is a relatively heavy load in the fixture, it is irrelevant. In electrical engineering terms, the leakage resistance of the dimmer (triac) forms the top resistance of a voltage divider, and the fixture load is the lower resistance. If the only load at the fixture is a (maybe) 100 Megohm input resistance of the meter, and the off-resistance of the Triac is (maybe) a megohm or less, the output of the voltage divider is still nearly the input, i.e., still ~ 120 VAC. If you applied even a small load at the fixture, say a 7 Watt lamp, you should see the voltage across the fixture drop down to something sensible.

Paul Bridges

Hi Rich,

Thanks for the question. For a very detailed discussion of dimmers, I would recommend reading the Wikipedia entry on "dimmer". Without the mathematical models/equations, I suspect that the dimmer only generates a voltage drop when there is a resistive load (i.e., kitchen light) attached. I would measure the voltage difference when the kitchen light is attached.

I hope this helps. Please let me know if you have more questions. Thanks Jeff Grell

Hi, Rich

Thanks for your question. I am an electronic engineer who designs power circuits, some of which are similar to a "thyristor type" lamp dimmer.

Your question omits a couple of important information points:

1) First I do not know what type of light fixture you have? The most common would be incandescent (or quartz halogen which is similar for our purposes.) Some LED type lamps may also be dimmable in the same socket, but the circuit could be completely different if you have some sort of fluorescent fixture. I will assume that your light fixture is designed for incandescent or halogen lamps.

2) A second important point is the type of digital multimeter which you have. Incandescent and quartz halogen lamps are resistive so respond to the Root Mean Square (or "RMS") voltage that is applied. If your meter is a "true RMS reading" then it should be able to provide an accurate indication of the voltage that is applied to your lamp. Other types should be able to show some variation but the reading is likely to be inaccurate at voltages which are less than full. This is because many meters "assume" a sinewave voltage. The output of most "thyristor" type lamp dimmers (when they are dimming) is not a sinewave. Please see the waveform of the output of a thyristor dimmer at

At this point I must warn you of the danger of probing or measuring this voltage as it is hazardous and capable of electrocuting you. Please do not continue with further measurements unless you know how to do them safely.

My reasons #1 and #2 above are not the most likely explanation of your (erroneous) measurement. The most likely reason is as follows:

Many circuits such as this require a load in order to work correctly. In normal operation, the load is the light bulb. The "thyristor" (if it is that) switch cycles on and off with each cycle of the 50 Hz or 60 Hz AC line voltage. "Timing is everything" in this operation, and without a load connected to the dimmer, the timing is likely to be wrong.

Due to several possible reasons there is likely to be a leakage current path in parallel with the dimmer circuit. This leakage path could be caused by a "snubber" or by leakage current which flows either through the thyristor or through the internal timing circuit when the thyristor is supposed to be "off". If you have no load or negligible load applied to the dimmer circuit, this leakage path will act as though the thyristor is "on" when it should be "off".

Another slight possibility is that something is retriggering the thyristor to turn it on when it should be off. This happens more easily than you might expect.

The best way to see the lamp voltage vary with dimmer setting is to measure the voltage applied to a working light bulb when it is being driven by the lamp dimmer. The light bulb provides the load which the dimmer circuit probably requires to operate properly.

Again please do not attempt any of this unless you know how to do it safely.

Best regards, Bob Zwicker

The light fixture is in parallel with the "+" and "-". The resistor in the dimmer switch and the light bulb have the same voltage drop. What changes is how much of the current passes through the light compared to the amount of current that passes through the light. When one goes up the other goes down and vice versa. You can find a detailed explanation if you search "parallel circuits".

Vince Calder

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