Heard on the radio this morning
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Apparently, a cell phone charger still uses a substantial amount of electricity even it is not charging the phone. (this was received by the radio host by email, from a person in the UK). Now I only did 1st year engineering electronics, but that smells a lot like cow dung to me. The logical question is, where is the energy going if the phone is not charging? I can understand energy loss while the phone is being charged, but how is energy drained from the charger if it not charging?
xacc.ide - now with IronScheme support
IronScheme - 1.0 alpha 1 out nowThere's always some resistance in any circuit, resulting in power converted to heat. Transformers are a very efficient way of converting power between different voltages, but they're not 100%. Switched-mode power supplies are more efficient on load than linear regulators but still have losses when not loaded. To stop the losses, you have to actually switch in a very high resistance such as an air gap ;) One of the problems in electronics is the difference between the idealised circuits that our maths tells us will work, and the flawed components we actually have to build the circuits out of. Real wires have finite small resistances - the reason that power lines are high-voltage is that the loss in them is proportional to the square of the current (I-squared times R), and increasing the voltage allows the same amount of power to be carried by a smaller current. The higher our working frequencies become, the larger these problems tend to be. People have had problems with circuits which were fine in analysis and when synthesised, but failed in practice due to coupling of traces and RF pickup. Indeed many small Wi-Fi devices simply use a trace on the circuit board as the aerial.
DoEvents: Generating unexpected recursion since 1991
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Why not relocate the isolator switch near your front door, then you can throw the switch to the 'off' position when you leave the house, simple! (Think of all the pennies you'll save).
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Provided I don't have any food in the freezer or fridge and don't want to record anything off tv while I'm out, why not? :) Seems a tad extreme though... :P
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go back and read your books leppie. A phone charger dependant on design converts AC votage 120/240 volts into DC votlage anywhere between 3 and 12 volts DC. The conversion is primary by a transformer which steps the high voltage down to a smaller voltage by transfer of an eletro-magnatic field. This field transfer incures loss through heat and inefficient transformers. So now we have an AC voltage of 5 volts say; this is then clipped so the negative voltage is lost(more heat loss) and the voltage is smoothed a bit using capicators so it near a constant voltage to charge your phone. All eletronics product heat and the cheaper they are the usually more wasteful they are. AC: Alternating Current a sine wave and changes on a based frequency 50-60 hetz (times per second)A 120 volt AC goes between -120 and +120. DC: Direct Voltage is a constant value ie like a battery.
Grady Booch: I told Google to their face...what you need is some serious adult supervision. (2007 Turing lecture) http://www.frankkerrigan.com/[^]
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Apparently, a cell phone charger still uses a substantial amount of electricity even it is not charging the phone. (this was received by the radio host by email, from a person in the UK). Now I only did 1st year engineering electronics, but that smells a lot like cow dung to me. The logical question is, where is the energy going if the phone is not charging? I can understand energy loss while the phone is being charged, but how is energy drained from the charger if it not charging?
xacc.ide - now with IronScheme support
IronScheme - 1.0 alpha 1 out nowChargers will have a no load current , both the transformer and switch mode type will draw some current when the load is disconnected , quite how much will vary tremendously . Without having a circuit diagram infront of me the only real test is : If it gets hot or warm then its possibly significant , if it doesn't it isnt significant .
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Funnily enough I read this on the britishgas.co.uk website this morning! It says that you will save £4 over the course of a year if you unplug your phone charger when not in use. Hardly substantial, but worthwhile nontheless.
That works out about 4W ( say 10p per KW/h) . That would be quite warm to the touch .
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I understand that. What I dont understand is the claim that a lot (or any non-nominal amount) of energy is lost on an open circuit.
xacc.ide - now with IronScheme support
IronScheme - 1.0 alpha 1 out nowI guess it depends how you quantify alot. Per transformer the electrical power used may be very small but when you take into account the fact that many people have a charger at work and one at home and the number of people who own mobile phones, the savings could be vast if every person just took the fraction of a second it would take to flick that switch.
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Why not relocate the isolator switch near your front door, then you can throw the switch to the 'off' position when you leave the house, simple! (Think of all the pennies you'll save).
WPF - Imagineers Wanted Follow your nose using DoubleAnimationUsingPath
I am going to spend several thousand $$$ to maximize efficiency in my home office, making sure that all electical outlets are controlled from a master console that will detect if I am using any electical apparatus. I figure to break even on my electric bill savings right about the time the sun goes supernova. :-D All that being said, most eletrical appliances do have some draw even when turned "off" The "Energy Star" certification here in the US has led to "low standby energy use" and the new rating in 2007 requires 80% power supply efficiency. BobP
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Apparently, a cell phone charger still uses a substantial amount of electricity even it is not charging the phone. (this was received by the radio host by email, from a person in the UK). Now I only did 1st year engineering electronics, but that smells a lot like cow dung to me. The logical question is, where is the energy going if the phone is not charging? I can understand energy loss while the phone is being charged, but how is energy drained from the charger if it not charging?
xacc.ide - now with IronScheme support
IronScheme - 1.0 alpha 1 out now -
go back and read your books leppie. A phone charger dependant on design converts AC votage 120/240 volts into DC votlage anywhere between 3 and 12 volts DC. The conversion is primary by a transformer which steps the high voltage down to a smaller voltage by transfer of an eletro-magnatic field. This field transfer incures loss through heat and inefficient transformers. So now we have an AC voltage of 5 volts say; this is then clipped so the negative voltage is lost(more heat loss) and the voltage is smoothed a bit using capicators so it near a constant voltage to charge your phone. All eletronics product heat and the cheaper they are the usually more wasteful they are. AC: Alternating Current a sine wave and changes on a based frequency 50-60 hetz (times per second)A 120 volt AC goes between -120 and +120. DC: Direct Voltage is a constant value ie like a battery.
Grady Booch: I told Google to their face...what you need is some serious adult supervision. (2007 Turing lecture) http://www.frankkerrigan.com/[^]
Frank Kerrigan wrote:
this is then clipped so the negative voltage is lost(more heat loss)
It's only actually clipped in old very cheap power supplies. By using 4 diodes instead of one (full bridge rectifier) you can flip the negative peak of the input into a second positive peak. This almost doubles efficiency (the almost comes from extra voltage drop from passing through 2 diodes instead of one) and allows for the use of a smaller capacitor to smooth the output.
Otherwise [Microsoft is] toast in the long term no matter how much money they've got. They would be already if the Linux community didn't have it's head so firmly up it's own command line buffer that it looks like taking 15 years to find the desktop. -- Matthew Faithfull
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go back and read your books leppie. A phone charger dependant on design converts AC votage 120/240 volts into DC votlage anywhere between 3 and 12 volts DC. The conversion is primary by a transformer which steps the high voltage down to a smaller voltage by transfer of an eletro-magnatic field. This field transfer incures loss through heat and inefficient transformers. So now we have an AC voltage of 5 volts say; this is then clipped so the negative voltage is lost(more heat loss) and the voltage is smoothed a bit using capicators so it near a constant voltage to charge your phone. All eletronics product heat and the cheaper they are the usually more wasteful they are. AC: Alternating Current a sine wave and changes on a based frequency 50-60 hetz (times per second)A 120 volt AC goes between -120 and +120. DC: Direct Voltage is a constant value ie like a battery.
Grady Booch: I told Google to their face...what you need is some serious adult supervision. (2007 Turing lecture) http://www.frankkerrigan.com/[^]
Frank Kerrigan wrote:
A 120 volt AC goes between -120 and +120.
Actually 120V AC means the root mean square is 120 Volt, therefore the peak voltage is sqrt(2) times larger. :)
Luc Pattyn [Forum Guidelines] [My Articles]
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I understand that. What I dont understand is the claim that a lot (or any non-nominal amount) of energy is lost on an open circuit.
xacc.ide - now with IronScheme support
IronScheme - 1.0 alpha 1 out nowHi, Think of how a permanent magnet attracks nails from a distance: there is no such thing as an open circuit in magnetism; flux lines are always closed, they don't need a "conductor", they search the path of least resistance, just like everything else in nature. Inserting the mobile unit just changes the path and the "resistance". :)
Luc Pattyn [Forum Guidelines] [My Articles]
This month's tips: - before you ask a question here, search CodeProject, then Google; - the quality and detail of your question reflects on the effectiveness of the help you are likely to get; - use PRE tags to preserve formatting when showing multi-line code snippets.
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I understand that. What I dont understand is the claim that a lot (or any non-nominal amount) of energy is lost on an open circuit.
xacc.ide - now with IronScheme support
IronScheme - 1.0 alpha 1 out nowleppie wrote:
What I dont understand is the claim that a lot (or any non-nominal amount) of energy is lost on an open circuit.
Again, In a voltage transformer there are 2 (two) circuits: the input and the output. There is no physical connection betwen them. Only a magnetic field. When a charger is not connected to the phone, but connected to the power supply, the input circuit is connected and looses power for heat.
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I understand that. What I dont understand is the claim that a lot (or any non-nominal amount) of energy is lost on an open circuit.
xacc.ide - now with IronScheme support
IronScheme - 1.0 alpha 1 out nowYou need energy to maintain the magnetic field, the wee led, the diode which clips the voltage, the capictors have to discharge and charge on the upper wave. In short its lost transfering AC into DC. You could hook up an AMP meter to a cord extension and plug your charger into it. You might find it will only consume arround a few miliamps which is only a few pennies a month; but every little helps.
DEVELOPER DAY SCOTLAND 10th MAY 2008 http://www.developerdayscotland.com/[^]
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Apparently, a cell phone charger still uses a substantial amount of electricity even it is not charging the phone. (this was received by the radio host by email, from a person in the UK). Now I only did 1st year engineering electronics, but that smells a lot like cow dung to me. The logical question is, where is the energy going if the phone is not charging? I can understand energy loss while the phone is being charged, but how is energy drained from the charger if it not charging?
xacc.ide - now with IronScheme support
IronScheme - 1.0 alpha 1 out nowI would imagine that the DC converter in the charger initiates a loss through heat death of the Universe.
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