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Eh up I'm not an electronic engineer, so I'm no expert on this. Im a metallurgist using a laser for materials processing. We use duty cycle to describe the laser pulsing characteristics. Its usually expressed as a percentage like this

duty cycle = t/T*100%;

at least in the case of lasers

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The time it takes for the welder to cool itself down so that it doesnt overheat.

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Duty cycle in some equipment such as laser printers is a specification that no one seems to be able to define or explain. Peter Resch

Duty cycle in laser printers is the theoretical average number of pages the machine should be used for in a given time period, generally one month.

Maximum Duty Cycle in a LaserPrinter would be the highest number of pages the printer should be used for in a given period of time, again generally one month.

These numbers are more a marketing tool than anything practical.

To determine the actual practical duty cycle of a laser printer you need to know how many pages a toner cartridge is good for and how often a maintenance kit needs to be installed. I can cite one example where a customer printed over 1,000 pages a day on a particular laser printer, the machine required service every 6 months, but had over 1,000,000 pages on it the last time I serviced it. The OEM specifies the duty cycle of that printer at 65,000 pages a month, so the customer was using the machine at half its potential, but, if the machine had been used at its full rated duty cycle, a maintenance kit would have to be installed every 3 months. (list price on that kit is $295, the printer sold for around $1,000)

HPRepair

I'm not entirely sure what you're trying to say here... anyway let's work some things out here.

An office laser printer is a SoHo type device that isn't expected to run 24/7, as some industrial printing presses might. The duty cycle is probably referring to how many pages you could reasonably get out of a heavily subscribed one that's printing at the maximum page rate, continually, through a working day, with the requisite short breaks for restocking paper, replacing toner cartridges, "installing" maintenance kits (IE giving it a cleanout and emptying or replacing waste toner bins, replacing worn fuser or imager parts), and indeed waiting for fresh data to come down the pipe from the print server. It's a metric designed to give you some idea of whether the machine is suitable for your needs, relative to how much printing you expect to do, or how many you need to buy if you intend to have a roaming-accesss network of them (ie users send their work to a central server, then type in a code or swipe a card, either on a printer or an attached terminal, to release their queued document(s) to that particular unit).

65,000 pages a month is equivalent to 780,000 per year, or about 15,600 a week (if we assume a total of two working weeks lost to national holidays and such throughout the year). At five days per working week, 3120 per day, which is a fair amount (more than 6 standard reams). If the machine is in use for maybe 9 hours a day (someone gets an early start, others wait around for a while at the end for their jobs to complete) then that's an average of 347 per hour, or 5.8 pages per minute - just over 10 seconds per page. For a standard grade machine, with all the other considerations to take into account, that's coming pretty close to the actual physical maximum. If we consider that they may be officially rating the machine for a 50% duty cycle but it's over-engineered to just-about withstand 100%, or something close to it, without breaking down, then that means it may be an actually fairly quick 12ppm (5 sec/page) model (the general human, maintenance, stocking and data transfer/rendering factors accounting for the aggregate downtime in-betwen periods of full-bore output).

Given just how much printing that implies, running it at only 1/3rd of it's total potential still sounds like quite a lot of print to me. I work somewhere with a heavily used open-access printer net, and even the most congested, heavy duty, industrial spec models in the central locations rarely need a full paper restock (consuming about 5 to 7 reams depending how empty they are and how well you distribute the paper between trays) more than every 2 to 3 days; ie, also about 1000 pages per day, max (the smaller ones which only have a 2 1/2 ream capacity can get annoying by firing off low paper alerts on a daily basis, though). The very most heavily subscribed one (and it is one particular machine!) being identifiable with your eyes shut because it exudes that unmistakable stench of ozone from the imager assembly almost constantly, whereas the others only occasionally emit a slight whiff, and most never get pounded hard enough for it to build up to detectable levels. That one, we tend to leave spare boxes of paper next to, and the trays unlocked, as the risk of theft is so much lower due to there being one or two people stood next to it very regularly, and representing a pretty small percentage of the total use anyway.

Anyway, how old is that example model? You say it's up to a million pages, but then try to say this represents it being used at a third of its potential. Erm? Anyway, going on the 1000 pages a day thing (indeed, about a third of 3120), that would mean it would have been in use at that DC for 1000 working days. At roughly 250 of those in a year, it'd be four years old by then, and upto probably its 8th or 9th service? Seems fair, really. People often don't appreciate just how hard working, and often long-lived, office laser printers can be, with a relatively very low failure rate (even though it seems high because of how heavily used they are). They're truly marvellous examples of pragmatic engineering. A school I worked at only had, as far as I could tell, two cheap Kyocera mono laser printers (one in the science prep-room and used for all the class worksheets, another in the main office for official letters etc) and a handful of crappy colour inkjets in the classrooms, which rarely ever worked right even when you could convince the drivers that you had authorisation from the (never-seen...) network admin to print anything (!). One of my jobs was to print and distribute all the requested worksheets for various lessons for the week (at half size, double sided - yay duplex units!, and either folded or chopped up on a hand guillotine as necessary) to the teachers in their "home" classrooms. Often this was something that would have to be stratified by the day each one was needed, and teachers leant on to get their requests in as early as possible the week before (especially for monday) to avoid a hideous rush on monday morning. It would often be late tuesday or early wednesday by the time all of them were ready through to that friday, and the printer could be given a break (and all of us sharing that room, from the smell, and from having to interrupt other jobs to go up, grab everything from a particular job and put it in a separate tray for processing, and un-pause the next one in the queue, having set all of them to "pause" as they went into the spooler... and even to start laboriously putting more things into the queue as the computer was both very slow and had very little memory, so couldn't deal with too many jobs at once). But it just kept on plugging. Must have cost all of about £150 including V.A.T. but rarely skipped a beat. The page counter was well on the way to 2 million (or possibly 3 mill?) by the time I left there. I mean, do you even realise just how many pages are needed for a lesson where there's 30 pupils and they have three different 8-page experiment booklets to fill out... each... with an accompanying pair of 4-page information packs? Even half size duplex, that's a 240 page run, or a half hour what with the duplex delay... continuously, through the day, 3 or 4 days a week. Hardcore. No inkjet would put up with it. 1/3rd duty for one of those things is still a huge job of work. 51.7.16.140 (talk) 13:20, 28 August 2016 (UTC)[reply]

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In the equation:

   duty cycle = t/T x100

where

   D is the Duty Cycle (% when the signal is active or max peak) 
   t is the duration that the function is non-zero (in one cycle or period only)
   Τ is the period of the function.

Ej: In ones period of 6sec The LED brights for 2sec.

   Duty Cycle= 2s/6s x 100 = .333x100 = 33.3%
   Duty Cycle= 33.3% (the LED brights 1/3 of the time in each period)

"D" does not occur in the equation, and tau does not occur in the "where" explanation, nor in the figure.

Further " D is the duration that the function is non-zero" in nonsensical.

Can we get an improvement? Thank you!

Hans Leander

This should now be better, please feel free to edit the article to improve it further. - CyrilB 19:43, 23 June 2006 (UTC)[reply]

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Duty cycle is appicable to only square wave form or other forms also?

This equation and its description doesn't make much sense to me either. Although, I do not know exactly what the duty cycle of a periodic function really means. I have a program which can create a triangle wave, and it has a function to manipulate the duty cycle of the triangle wave. It basically changes the slope of the wave:

duty cycle = 0

|\|\|\|\|\|\|\|\|\|\

duty cycle = 0.5

/\/\/\/\/\/\/\/\/\/\

duty cycle = 1

/|/|/|/|/|/|/|/|/|/|

(pardon the cheesy ascii graphics)

The description of duty cycle in this article is obviously incomplete and needs to be altered by an expert.

69.140.29.251 23:00, 9 April 2007 (UTC)[reply]

I think the confusion here probably arises from a poor description in your software. Really what it's doing is changing the relative steepness of each side of the triangle, and in a somewhat poor manner at that as you can't alter them independently of each other (ie controlling the period of both the rising AND falling side without affecting the other), presumably so you can set a certain base frequency and have that maintained as you twiddle the shape. In effect, you've got a three-point wave bounded by the nadirs of the troughs on each side (which are immovable, having been defined by the base period), and what you alter is the relative temporal position of the peak apogee between them. In this case, "duty cycle" is being misused as a name for... well, I don't know what it would really be called, but presumably something like Attack:Decay Ratio (ADR?) or similar. When that = 0, the attack is instant (zero period), and when it's infinite, decay is instant - and the opposing value in each case is equal to the base period itself. And indeed, where it's 1, they're even. Therefore it's actually simply describing how much of the total wave is represented by the rising edge, between 0 (0%) and 1 (100%), and the triangle type is obviously 0.5 (50%). So you could sorrrrrt of say it's the *Attack* duty cycle, but it still seems wrong.

However, it can still be explained in terms that make some kind of sense, even if they haven't really given it a properly clear name.

The true meaning of it is supposed to be what proportion of the available maximum time in a cycle is occupied by doing a particular thing, normally supplying a "high" signal, doing useful work, or anything else that involves the transfer of energy and thus the potential for heating, frictional wear and tear, buildup of resonances, depletion of capacitors or battery banks, etc (or even accumulation of waste substances in muscles that are working at a higher rate than can be removed by blood circulation). And it can be sometimes aggregated in terms of percentage of maximum signal / effort / etc multiplied by how long is spent at that output. For example you might have an industrial engine that can supply its absolute maximum output for maybe 75% of the time only, and has to spend time at low load or idling in between to allow built-up heat to dissipate, particularly from parts like crank bearings that might be under high load but aren't able to shed heat quite as easily as others which are actively rather than more passively cooled. But if it's de-rated / governed to give only 75% of that maximum (or maybe slightly less to give a little extra safety margin), and installed as effectively a one-quarter less powerful prime mover, it can sit at that output continually, possibly getting quite hot in the process but not actually overheating as everything achieves a thermal equilibrium (and still gets to cool down occasionally when its power isn't needed, or it reaches a scheduled maintenance point). Which is why some things like heavy truck, locomotive or marine engines seem to have rather poor outputs for their size - they are expected to give maximum effort for extended periods, perhaps several long minutes - hours - days on end respectively. If you had a smaller automobile engine (or a chain of them, making equivalent peak power) and tried to run it the same way, it would overheat or suffer very premature terminal wear as a result, as it's just not made for that kind of duty cycle. Rather than 75% (or 100% when de-rated, or whatever), they might be expected to withstand the equivalent of no more than maybe 50% with quite high cooling demand (eg motorway cruising with lots of air rushing in), or less where they can't be cooled as well... for example the conditions under which the big engines operate, moving slowly and pulling a heavy load up a big hill. Which is why towing heavy trailers through the mountains is a particular overheat-risk for a typical family car, as you may need to apply maximum power for a full minute or more just in order to get up to a speed where you can engage third gear and ease off the throttle a little because you're no longer in danger of being overtaken by a cyclist. By which point your coolant has gained about thirty degrees...

In other words, it's what fraction of maximum power is being transmitted through a system, whether at a steady level, or as it's more usually used, in a pulsating form (as few loads are absolutely stable, and those which ARE stable usually have a 100%-duty supply of suitable size matched to them). Hence how high frequency PWM is often used to adjust the brightness of lights and the speed of motors these days, as it not only means the same thing but it's also generally both more efficient (no need to insert extra resistances) and can be controlled FAR more accurately. Wider pulse = higher duty cycle (more of the entire on/off cycle is "on") = more power flows, in a linear fashion.

It just so happens that this is important for audio because our own hearing system picks up the somewhat virtual extra, shifting harmonics produced by this variance, when the base cycle frequency is held the same, as a different "timbre". Outside of audio, it's pretty much just a technical thing for indicating how much something is used. 51.7.16.140 (talk) 13:20, 28 August 2016 (UTC)[reply]

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The article on duty cycles appears substantially accurate. In a general sense, “duty cycle” is simply the proportion of any time period for which an apparatus is active. For example, to generate the desired speed for an electric pencil sharpener, the supply of voltage to the sharpener’s electric motor might be limited to specific intervals of time. The electric motor might only receive voltage 60% of the time. The “duty ratio” for this electric motor is said to be 60%. By increasing the percentage of time a motor receives voltage, greater speeds can be achieved.

One method for achieving a desired duty ratio utilizes periodic signals, such as the “sawtooth” signals that your program generates. Although labeled as “duty ratio” by your program, it is more accurate to say that these sawtooth signals are used in the derivation of a duty ratio. These signals, when used with a MOSFET, or other similar component, are logically compared against a “control voltage.” If the sawtooth signal is less than the control voltage, then the electric motor receives voltage. If the sawtooth signal is greater than the control voltage, then voltage to the motor is cut off. In this manner, the electric motor will only be active for a period of the time - - hence, its duty ratio.

28/06/2006

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As an electronic engineer I would largely agree with the comment immediately above. Whilst you may refer to the duty cycle of a sawtooth waveform or control signal of any system as the period for which it is "active" you would effectively be talking about it in a binary sense, either "active" or "inactive". For a non binary/logic signal the threshold/control level needs to be defined; In the sawtooth example given it seems that you are comparing to a threshold level of zero, but if a different threshold is given then the resulting duty cycle would be different also. Duty cycle is generally used to describe logic signals and a non binary/logic signal should be converted to this form by comparison with a fixed reference before the duty cycle can be defined.

While I see discussion of incompleteness I see no discussion of current factual dispute. I have therefore removed the dispute tag from the article in the course of grammatical clean-up. If you have some actual factual dispute with this article, then please reapply the tag and leave a note on the discussion page (with a subject, omg!) Phlake 22:27, 14 August 2007 (UTC)[reply]

Duty cycle could be applied to almost anything with on and off states, and is not generally applied to any other waveforms (as has been pointed out earlier in this discussion, where any duty cycle would have to be derived from setting a threshold). Not until I started reading this page and the one on switched mode power supplies have I ever heard the phrase "duty ratio."

I think this is pretty much complete, and I'm not sure we need the "needs attention from an expert" tag any more.

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I agree that that article seems relatively complete. Remember that this is an open source encyclopedia and if someone wants expert information on the subject, they will likely have to consult a text in order to have proper references. As far as a common knowledge goes, the article as it stands is plently. (128.63.18.10 (talk) 17:24, 17 January 2008 (UTC))[reply]

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Re: duty cycle and sinusoidal waveforms

sadly, I DO hear "duty cycle" used occasionally when describing sinusoidal brain oscillations (such as alpha and theta rhythms)... but I think it's an innaccurate use of the term, as, in the case of theta rhythm,there are no on or off states... simply peaks and troughs. There also are data suggesting that, during memory formation in the limbic system, neural activity on peaks and troughs of theta simply perform different "duties" - so it's never really "off" and has no "off" states. It's an archaic term that isn't a good description of brain oscillations (unless they explicitly have something that functionally may be considered an "off" state, e.g. some neurons have "up" and "down" states, so if the down state makes a neuron unresponsive, that may be defined effectively as an "off" state, and the term duty cycle could apply). But I avoid it... it irritates me. I still see it used correctly with physiology studies(see Burke RE (2008) Local duty cycle and muscle spindle density. Brain Res Bull. 2008 Mar 28;75(5):501. Epub 2007 Oct 29), where it applies, but with brain oscillations, it makes me think of the guys who look like they're from Bell Labs (circa 1963), complete with short sleeve white shirts and horned rimmed glasses... —Preceding unsigned comment added by 24.243.50.75 (talk) 23:48, 13 April 2008 (UTC)[reply]

Might it be relating to the varying *intensity* of those waves, or even varying frequency? IE time spent dwelling at a particular wavelength, or how long they spend at high amplitude vs lower? 51.7.16.140 (talk) 13:22, 28 August 2016 (UTC)[reply]

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It makes no sense to have work ratio redirect here. 16:11, 31 May 2009 (UTC) —Preceding unsigned comment added by Younge1986 (talk • contribs)

De gustibus non est disputandum but this reference may look more professional. If you care ....--Bougainville (talk) 00:35, 30 June 2008 (UTC)[reply]

The duty ratio seams to be an important parameter when studying artificial/robotic locomotion... but no mention in the article ? XApple (talk) 15:47, 5 January 2009 (UTC)[reply]

For no apparent reason (which math hounds prolly understand) the example period changed from minutes to seconds. At first glance it appears to claim that an arc welder with 20% duty cycle must run two seconds on, then 8 seconds off.
quote:

In equipment such as a welding power supply, the maximum duty cycle is defined as the percentage of time in a 10 minute period that it can be operated continuously before overheating.[6] ............ snip........Based on this analysis the signal is 1 V for 0<t<2 and 0 V for 2<t<10, therefore the period is T= 10 seconds, and the duty cycle is then ON/Period of the signal i.e., 2/10 = 0.2, which gives a duty cycle of 20%

Wasn't period already industry-standardized at 10 minutes? So why the swap? (I'm not questioning the arithmetic, just the communication device.) Why any mathy stuff at all here?

I'm removing this, perhaps it can be made more realistic?

For example, let's say a DC signal of 1 volt starts at time t=0 seconds, and stays there for t=2 seconds, at which point the DC signal goes to 0 volts and stays there until t=10 seconds. At time t = 10 the signal goes back to 1 volt, and the process repeats over and over again. Based on this analysis the signal is 1 V for 0<t<2 and 0 V for 2<t<10, therefore the period is T= 10 seconds, and the duty cycle is then ON/Period of the signal i.e., 2/10 = 0.2, which gives a duty cycle of 20%

--68.127.84.95 (talk) 23:54, 25 August 2011 (UTC)Doug Bashford[reply]

Although the term, "duty cycle," has its origins in engineering, it is also used by many physiologists to analyze biological data such as muscle contraction and bursting by neurons. Therefore, I would like to propose broadening the scope of this article to include living systems, a section on biological data, and to omit "in engineering" at the beginning of the first sentence. Comments welcome. danielkueh (talk) 16:58, 16 December 2011 (UTC)[reply]

Seems fair, though really it applies to any mechanical (inc. acoustic, thermal...) or electric (inc nervous, photonic...) system that can be modelled, particularly in terms of energy flow and accumulating stress factors (waste heat, combustion/respiration products, etc). It's a fairly universal idea. OTOH what you describe could be called BioEngineering :) 51.7.16.140 (talk) 13:26, 28 August 2016 (UTC)[reply]

The term "square wave" is used in a couple different places in the article to refer to what I'm sure would be more correctly termed a Rectangular Wave or Pulse Wave (depending on whether the "off" state is taken as -1 or 0, vs "on" being +1). A square wave being a rectangle or pulse with a duty cycle of 50%. Any variation from that, and, well, the wave isn't "square" any more (with a typical scaling that has a square scope and the amplitude and wavelength both normalised to fill the screen, and the wave being Pulse, ie off is 0 rather than -1, when on is 1), but two unequal rectangles. It also has a particular tone which, as demonstrated by the varying pulsewidth animation, is discernable from other duty cycles simply by ear. 51.7.16.140 (talk) 13:31, 28 August 2016 (UTC)[reply]