Resistor - math question

[nForce]

Trough a resistor with nominal resistance of 2.2 Ohms and nominal power of 5 W flows a square wave current with an amplitude of 2 A. The on/off ratio of a square wave is 1:5.
Is a resistor going to overheat?

So, I have solved this like this:

The power is P = I² R

The signal is on 20% of a full period. So I = 2 A * 0.2 = 0.4 A
R = 2.2 Ohm.

So we plugged this in the first equation:
P = 2.2 Ohm * 0.4² A

Is 0.352 W of heat.

So the answer is No, it's not going to overheat. Is it correct?

I have one more question about resistors, tho. Thanks for your help.

[Brumby]

For a 20% duty cycle you can't take 20% of the current and then square it.

You must use the full current value in the power calculation and then take 20% of the result.

P = (I2 * R) * Duty cycle	.     .
P = (2A2 * 2.2 Ohm) * 0.2		P = (2A2 * 2.2 Ohm) * 0.167
= 4 * 2.2 * 0.2		= 4 * 2.2 * 0.167
= 1.76 W		= 1.47 W

As long as the power rating of your resistor is over 1.76W  1.47W AND your operating frequency is fast enough, then no, it will not overheat.

Edit: Calculation reworked for correct duty cycle.

[Brumby]

The reasoning is simple - when the power is ON, the resistor is dissipating 8.8W.  What will then kill a resistor is the temperature increase that results from this.  But by having it switch off for a time, the power becomes zero and the resistor can cool down.  The duty cycle will determine how much heating up time there is and how much cooling down time there is.

The numerical result of the equivalent power over a full period is the average, which just works out as the 'on' power times the duty cycle.

These changes occur no matter what frequency the power is operating at - but if the frequency is more than 10Hz, then the temperature fluctuations will not be noticeable.  IF, however, your cycle time is 5 minutes, then you will have 8.8W running through that resistor for a whole minute before it gets a 4 minute rest.  In that minute, you can expect one rather stressed out (ie overheating) resistor.

[nForce]

Oh, ok thanks.

Another question:

For a real resistor derive an equation for impedance expressed in dB. Neglect parasitic capacitance.

So if we neglect parasitic capacitance which is paralel to the resistor, the only one left is parasitic inductance which is in series. But how can we then derive the impedance in dB?

[Brumby]

Correction:

The on/off ratio of a square wave is 1:5.

That means the duty cycle is ¹/₍₁₊₅₎ or 16.7% which will result in less heating (than the 20% duty cycle used in the original calculation).


As for your second question, I'm bowing out.  It's 3am here.

Maybe someone else will take up your homework question.


Edit: Added note re duty cycle correction

[nForce]

It's not a homework question, I am preparing for a test. 

[vk6zgo]

Impedance is not expressed in dB.
It will be expressed in Ohms,as a Resistive & a Reactive component.

By the way,your waveform in the first question is not a square wave,which by definition has a mark/space ratio of 1:1.
A square wave is a special case of a Rectangular wave
What you have is another type of Rectangular wave,

[alsetalokin4017]

Hmmm....

Wikipedia says that the average power of a periodic waveform is P=(I_rms)²R and that the RMS value of a rectangular pulse train is equal to the peak amplitude x sqrt(duty cycle).

So, the RMS value, I_RMS, of the function I(t) is the constant current that yields the same power dissipation as the time-averaged power dissipation of the current I(t).

https://en.wikipedia.org/wiki/Root_mean_square

So I_rms = 2 x sqrt(0.167) = 0.81731267 A
and P_avg = (0.81731267 A)² x 2.2 Ohms = 1.4696 W

Note that the result obtained by using these formulae (Pavg=1.4696 W) agrees with the result obtained by using Vishay's online Pulse Energy calculator here:
http://www.vishay.com/resistors/pulse-energy-calculator/


What am I missing here?

[alsetalokin4017]

But as others have said, to know whether or not the 5 W resistor will overheat, you need to know the frequency (or rather the period, 1/f) to know for how long the maximum power of 8.8 Watts will be on.  If you have, say, a period of 100 seconds (16.7 seconds at 8.8 Watts) ..... then yes, the resistor will probably overheat. If you have a period of 0.1 second, then no, it probably won't overheat. Just where the line is between Yes and No depends on other variables. Is the resistor mounted directly on a circuit board or is it on "standoffs" so that air can circulate around it? Is there fan-forced cooling or only convection? Etc. Etc.

IOW, it's a bad question.

[Brumby]

I shouldn't get into math at that time of the morning....   

[alsetalokin4017]

Just for fun.... here's as close as I could get with my cheapo DDS FG and the Rigol ds1054z:

[Brumby]

I shouldn't get into math at that time of the morning....   

No need to feel embarrassed - my method was correct after all, for the example given.  The use of the square root of the duty cycle bugged me until I sat down and did a minute of algebra....