How important is a 50% duty cycle?

[UTMonkey]

Hi All,

If a circuit only wishes to respond to a particular frequency and doesnt much care about the amplitude, just the fact that it is receiving highs and lows at a particular rate - is a 50% duty cycle all that important?

Infact wouldnt something firing off pulses at a frequency of 10khz with a reduced duty cycle (lets say 10%) take less power?

Thoughts?

Mark

 

[OutToLunch]

if whatever is receiving the signals is only responding to, for example, the rising edge, and doesn't care about the rest of the signal content, then yes, decrease the duty cycle and save power.

 

[Sceadwian]

If it's edge triggered you should be fine, as long as the duty cycle is high enough to trigger it. If it's truly frequency dependent you can't change the duty cycle because that changes the frequency. Even if you're only sending 10 thousand pulses per second if the duty cycle is 10 percent the actual frequency is 10 times higher, if there's an analog filter involved it will likley completly filter out the signal you're trying to pass. Frequency is a lot more complicated than just the number of rising or falling edges, per second.

 

[Roff]

if whatever is receiving the signals is only responding to, for example, the rising edge, and doesn't care about the rest of the signal content, then yes, decrease the duty cycle and save power.

You have to be careful how you define "duty cycle", since the power dissipation depends on what the load is returned to. If the signal swings from 0 to +V, the load is to zero volts, and the duty cycle is defined as Thigh/Tperiod, then a lower duty cycle will indeed save power. On the other hand, using the same signal levels and duty cycle definition, but with the load returned to +V, then a higher duty cycle will save power.

 

[indulis]

If it's truly frequency dependent you can't change the duty cycle because that changes the frequency. Even if you're only sending 10 thousand pulses per second if the duty cycle is 10 percent the actual frequency is 10 times higher...

Please explain why you think this is true. Frequency is 1/period. It is NOT duty cycle dependent by definition.

 

[Diver300]

If it's truly frequency dependent you can't change the duty cycle because that changes the frequency. Even if you're only sending 10 thousand pulses per second if the duty cycle is 10 percent the actual frequency is 10 times higher...

Please explain why you think this is true. Frequency is 1/period. It is NOT duty cycle dependent by definition.

If you measure a frequency with a frequency counter, the reading is not dependent on the duty cycle.

However.....

The frequency spectrum content of a waveform is very dependent on the duty cycle.

So a 1 :mu:s pulse every second has a frequency of 1 Hz, but you need an instrument with a bandwidth of 1 MHz to measure it.

 

[indulis]

The frequency spectrum content of a waveform is very dependent on the duty cycle.

What does spectral content have to do with the definition of "frequency"?

So a 1 s pulse every second has a frequency of 1 Hz, but you need an instrument with a bandwidth of 1 MHz to measure it.

There wasn't any disscusion about how to measure the frequency of a waveform. Beside the frequency is the frequency irreguardless of its duty cycle.

 

[Speakerguy]

Someone here needs to study fourier transforms

 

[Sceadwian]

indulis, I don't think it's true that's just the way it is.. Period only determines frequency when you're using a pure sine wave. non-sine waves are very different. I guess the easiest way to explain it would be that a Square wave with a 10% duty cycle is the same thing as a sine wave at 10 times the frequency with 9 whole cycles missing inbetween each pulse of the primary frequency. What it comes down to is rate of change. A sine wave the rate of change is constant throughout the entire signal in a sine fasion. Look at the spectral spread of a sine wave on spectrum graph and you'll see. Then look at a square wave at 50% duty cycle and you'll see a high peak at the main frequency and a gradually decreasing peak at every odd harmonic of the main frequency. Change the duty cycle and watch what happens. I'd recommend learning a little about simulators like LTspice for simulating things like this to understand what's going on.

 

[Leftyretro]

indulis, I don't think it's true that's just the way it is.. Period only determines frequency when you're using a pure sine wave. non-sine waves are very different. I guess the easiest way to explain it would be that a Square wave with a 10% duty cycle is the same thing as a sine wave at 10 times the frequency with 9 whole cycles missing inbetween each pulse of the primary frequency. What it comes down to is rate of change. A sine wave the rate of change is constant throughout the entire signal in a sine fasion. Look at the spectral spread of a sine wave on spectrum graph and you'll see. Then look at a square wave at 50% duty cycle and you'll see a high peak at the main frequency and a gradually decreasing peak at every odd harmonic of the main frequency. Change the duty cycle and watch what happens. I'd recommend learning a little about simulators like LTspice for simulating things like this to understand what's going on.

Sorry but I agree with indulis, changing a 50% duty cycle waveform to a 10% wave form does not change the frequency as measured by a frequency counter or as defined F= 1/Period. The spectrum components are a whole other subject and there will be different harmonic power content but that was not the original question.

Lefty

 

[Roff]

As others have said or implied, the disagreement between Indulis and Sceadwian is based on the fact that Indulis is talking about the fundamental frequency, while Sceadwian is talking about harmonic content.
Sceadwian, I understand your point, but

A sine wave the rate of change is constant throughout the entire signal in a sine fasion.

is technically incorrect. I'm sure you know that the rate of change of a sine wave is highest at the zero crossings, and zero at the peaks.

 

[mneary]

Energy at the "fundamental" frequency is maximized at 50% duty cycle; any other duty cycle reduces the energy at the "fundamental" frequency.

If a (resonant) detector is dependent on the energy at a particular "frequency" it will see lower values at decreased duty cycles.

If an (edge) detector is triggered on the rising edge only (NOT the falling edge) then it may function well at reduced duty cycle.

Energy at the "fundamental" frequency can become insignificant at low duty cycles, even when the repetition rate is the same.

 

[Hero999]

A 10% duty cycle won't always save you power, if the load uses both the positive and negitive parts of the signal then the power consumed will be the same as 50% duty cycle.

 

[indulis]

Sorry.. was off for the Thanksgiving Day weekend. The period of a waveform defignes it's frequency. That is by definition... duty cycle has absolutly nothing to do with it... zip, zero nada!!! 10% or 90 % duty cycle is irrelevant. The sepectral content of a waveform does not, in any way shape of form defign a waveforms frequency. Anybody that has studied this stuff at a higher level knows very well that you can recreate ANY waveform with sinusoids of the proper frequency and amplitude, but what does that have to do with defigning a particular waveforms frequency... not a thing!!!

Hey, Out To Lunch... had any contact with the "old" Datel? We're Murata Power Solutions now!!

 

[OutToLunch]

i went back to some emails from around the september timeframe - and sure enough, there was some stuff regarding Murata / C & D. i didn't really pay attention to the name change at the time - thanks for the info!