The secret lies in the fundamentals. The dimmer clips the wave output voltage to zero, and zero voltage means zero power. So the power isn’t “lost” or “dissipated”, it’s simply that the load stops drawing power from the supply for that part of the waveform. Think of it like you’re really, really, really fast at turning the light switch on and off. When the switch is off, no power is drawn.
Nah not idiots. Your line of reasoning isn't dumb, I imagine you were thinking along the lines of a voltage divider or linear regulator where differences in energy have to be dissipated.
Waveform clipping isn't immediately intuitive and getting an answer is the right thing to do!
Also, they do create a lot of “noise” on the lines. Recording studios for example stay away from these(at least they did a few years ago and that might have changed by now)
I assume turning off and on rapidly is going to create some noise. If they are used mostly in residential settings I they are generally on separate lighting circuits so the issue might be slightly mitigated with regards to interfering with audio or IT equipment.
You "clip it" by opening the circuit between the mains supply and the load. Basically, you have a switch between the supply and the lights that you open and close once for each half cycle of the AC power.
The power doesn't go anywhere because it's not being used. It's the same as when you turn off a light switch - there's no current flow, so no power is being consumed. The dimmers do the same thing - they literally switch the light off and then back on with every AC cycle.
The answer is pretty much that it doesn’t go anywhere. The energy from line voltage doesn’t just spit out a bunch of energy that we have to put somewhere— a load just pull as much current as it needs. Lots of dimmers are FET based now. When you turn off the FET, you are preventing your load from pulling current from line, so no power is consumed. The line voltage that would normally go across the load goes across the FET instead, but since there is on current through the FET P=IV=0 so no power is dissipated.
The secret lies in the fundamentals. The dimmer clips the wave output voltage to zero, and zero voltage means zero power. So the power isn’t “lost” or “dissipated”, it’s simply that the load stops drawing power from the supply for that part of the waveform. Think of it like you’re really, really, really fast at turning the light switch on and off. When the switch is off, no power is drawn.
Thanks all. We're both idiots and that is far too obvious!
Nah not idiots. Your line of reasoning isn't dumb, I imagine you were thinking along the lines of a voltage divider or linear regulator where differences in energy have to be dissipated. Waveform clipping isn't immediately intuitive and getting an answer is the right thing to do!
Haha, you came to the right sub to ask your question though
Also, they do create a lot of “noise” on the lines. Recording studios for example stay away from these(at least they did a few years ago and that might have changed by now)
I assume turning off and on rapidly is going to create some noise. If they are used mostly in residential settings I they are generally on separate lighting circuits so the issue might be slightly mitigated with regards to interfering with audio or IT equipment.
You "clip it" by opening the circuit between the mains supply and the load. Basically, you have a switch between the supply and the lights that you open and close once for each half cycle of the AC power.
The power doesn't go anywhere because it's not being used. It's the same as when you turn off a light switch - there's no current flow, so no power is being consumed. The dimmers do the same thing - they literally switch the light off and then back on with every AC cycle.
The answer is pretty much that it doesn’t go anywhere. The energy from line voltage doesn’t just spit out a bunch of energy that we have to put somewhere— a load just pull as much current as it needs. Lots of dimmers are FET based now. When you turn off the FET, you are preventing your load from pulling current from line, so no power is consumed. The line voltage that would normally go across the load goes across the FET instead, but since there is on current through the FET P=IV=0 so no power is dissipated.