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RMS or average current to determine resistoring heating?

M

mook Johnson

Jan 1, 1970
0
RMS has always been described to me as the AC current level to create the
same amount or heat in a resistive load as a DC current. This as always
been with sinusoidal currents.

In this case I have narrow 50uS pulses of current at 5 amps with repetition
rates of 500uS.

If I calculate the RMS current vs. the average current there is a sizable
difference. the average current is 0.5A the RMS current(from spice
simulation) is 1.6A.

Which if the two should be used when determining the heat generated by a
power resistor.

Assume the resistor chosen can handle the instantaneous power during the
current spikes.

I'm more concerned about the heatload this will add to the heatsink.

There is a 10X difference in power between the two.


thanks
 
M

Mike Silva

Jan 1, 1970
0
RMS has always been described to me as the AC current level to create the
same amount or heat in a resistive load as a DC current.  This as always
been with sinusoidal currents.

In this case I have narrow 50uS pulses of current at 5 amps with repetition
rates of 500uS.

If I calculate the RMS current vs. the average current there is a sizable
difference. the average current is 0.5A  the RMS current(from spice
simulation) is 1.6A.

Which if the two should be used when determining the heat generated by a
power resistor.

Assume the resistor chosen can handle the instantaneous power during the
current spikes.

I'm more concerned about the heatload this will add to the heatsink.

There is a 10X difference in power between the two.

thanks

RMS is correct. In your case (and assuming a 1 Ohm load for
simplicity), you have 25W generated for 50us, and 0 for the remaining
450us, giving an average over 500us of 2.5W. The current
corresponding to 2.5W is 1.58A (your 1.6A figure).
 
M

mook Johnson

Jan 1, 1970
0
RMS is correct. In your case (and assuming a 1 Ohm load for
simplicity), you have 25W generated for 50us, and 0 for the remaining
450us, giving an average over 500us of 2.5W. The current
corresponding to 2.5W is 1.58A (your 1.6A figure).

That was simple enough.

thanks.
 
R

Robert Adsett

Jan 1, 1970
0
mook Johnson said:
RMS has always been described to me as the AC current level to create the
same amount or heat in a resistive load as a DC current. This as always
been with sinusoidal currents.

In this case I have narrow 50uS pulses of current at 5 amps with repetition
rates of 500uS.

If I calculate the RMS current vs. the average current there is a sizable
difference. the average current is 0.5A the RMS current(from spice
simulation) is 1.6A.

Which if the two should be used when determining the heat generated by a
power resistor.

Think of it this way.

Heat generated is proportional to power (I*I*R).
The average is then the average of the I*I*R terms.
Therefore you are looking at a RMS term.

Obviously cooling figures in here so frequency matters but over a
relatively broad range of frequencies RMS will give you a good place to
start and will probably be sufficient.

Also take a look at your resistor's power derating curves. They will
sometimes provide some additional information that will help.

If the resistor will dissipate the heat between pulses then your
calculation may rely more on the pulse rating of the resistor than the
power rating and of course there is some current for which the resistor
is underated no matter how short the pulse but only a few parts have
that value fully specified that I have seen.

Robert
 
P

Phil Allison

Jan 1, 1970
0
"mook Johnson"
RMS has always been described to me as the AC current level to create the
same amount or heat in a resistive load as a DC current. This as always
been with sinusoidal currents.

** No matter what the wave shape - the rms value is the one for heat
dissipation in a resistor.
In this case I have narrow 50uS pulses of current at 5 amps with
repetition rates of 500uS.

If I calculate the RMS current vs. the average current there is a sizable
difference. the average current is 0.5A the RMS current(from spice
simulation) is 1.6A.

Which if the two should be used when determining the heat generated by a
power resistor.


** The rms one.
Assume the resistor chosen can handle the instantaneous power during the
current spikes.

I'm more concerned about the heatload this will add to the heatsink.

There is a 10X difference in power between the two.

** Correct.

BTW:

The most dramatic example of this I have seen in a commercial product is
with " Glow Plug " drivers for model size methanol engines ( 2 or 4 stroke).
Model size glow plugs are made with platinum wire and draw up to 5 amps at
1.2 volts - so the best way to power them is from a single Ni-Cd or NIMH
cell.

However, some modellers like to use their 12 volt lead-acid starter motor
battery to drive the glow plug too.

But as 5 amps at 12 volts is 60 watts, a 1.2 volt linear regulator is not
practical - a switching regulator would be ideal but there is a much
cheaper ( and nastier) solution that the Asian makers of cheap glow driver
use. Simple PWM.

Yep - the 12 volt battery is switched ( with a FET ) straight onto the 1.2
volt glow plug at a high (but audible ) frequency with a very low duty
cycle - about 1%.

This means the current pulses are in the order of 50 amps !!!!

The average current draw from the 12 volt battery is then 50/100 = 0.5
mp - while the glow plug has 5 amps rms flowing through it.




........ Phil
 
J

John Fields

Jan 1, 1970
0
RMS has always been described to me as the AC current level to create the
same amount or heat in a resistive load as a DC current. This as always
been with sinusoidal currents.

In this case I have narrow 50uS pulses of current at 5 amps with repetition
rates of 500uS.

If I calculate the RMS current vs. the average current there is a sizable
difference. the average current is 0.5A the RMS current(from spice
simulation) is 1.6A.

Which if the two should be used when determining the heat generated by a
power resistor.

---
If your pulse is rectangular, then the current won't be RMS, it'll
be peak, and you should use average.
---
Assume the resistor chosen can handle the instantaneous power during the
current spikes.

I'm more concerned about the heatload this will add to the heatsink.

There is a 10X difference in power between the two.

---
Just for grins, let's say your pulses have an amplitude of 5 volts.

That means that if they're pushing 5 amps through the resistor the
resistor must have a resistance of one ohm, and the power being
dissipated in the resistor, when the pulse is on, must be:


P = IE = 5V * 5A = 25 watts.


Now, let's further assume that your pulses are on for 50µs and off
for 50µs.

That means that the resistor will be dissipating 25 watts for 50µs
and 0 watts for 50µs and, since the pulses have rectangular edges
and current has the dimension of time, the resistor will be
dissipating an average of 12.5 watts.

Assuming a constant load impedance, then, your example of a 10% duty
cycle will result in 1/10th of the peak power being dissipated by
the load and transferred to the heat sink.
 
P

Phil Allison

Jan 1, 1970
0
John Fields is LOSING IT BIG TIME



** Wow - now THAT is really crazy bullshit.

Rectangular pulses have no RMS value ??

The average value of a rectangular pulse stream gives the heating effect
voltage ?



The song for John:
----------------------

They're coming to take me away, HA HA

They're coming to take me away, HO HO HEE HEE HA HA

To the funny farm

Where life is beautiful all the time

And I'll be happy to see

Those nice, young men

In their clean, white coats

And they're coming to take me away, Ha-haaa!




....... Phil
 
G

Greg Neill

Jan 1, 1970
0
Any waveform has an RMS value, including rectangular
pulses. Of course, for rectangular pulses that are
always greater than or equal to zero the average
will work out to be the same as the RMS value. For
waveforms that traverse zero, the average can be
any value between the peaks including zero (consider a
square wave with equal positive and negative peaks).

[snip]
---
Just for grins, let's say your pulses have an amplitude of 5 volts.

That means that if they're pushing 5 amps through the resistor the
resistor must have a resistance of one ohm, and the power being
dissipated in the resistor, when the pulse is on, must be:


P = IE = 5V * 5A = 25 watts.


Now, let's further assume that your pulses are on for 505s and off
for 505s.

That means that the resistor will be dissipating 25 watts for 505s
and 0 watts for 505s and, since the pulses have rectangular edges
and current has the dimension of time, the resistor will be
dissipating an average of 12.5 watts.

Assuming a constant load impedance, then, your example of a 10% duty
cycle will result in 1/10th of the peak power being dissipated by
the load and transferred to the heat sink.

The RMS value for your 505s/505s waveform is about 3.536 A.
With a one ohm load the power dissipated is then
(3.536 A)^2 * 1 Ohm = 12.5 W, same as using the average.
 
P

Phil Allison

Jan 1, 1970
0
"Greg Neill"
"John Fields"
If your pulse is rectangular, then the current won't be RMS, it'll
be peak, and you should use average.
---

Any waveform has an RMS value, including rectangular
pulses. Of course, for rectangular pulses that are
always greater than or equal to zero the average
will work out to be the same as the RMS value.


** WHAAAATTTT ???????????????

For single polarity, rectangular pulses, the average and rms values are
NOT the same !!!!!!


I av = I pk times duty cycle.

I rms = Ipk times sq.rt duty cycle.


Get real, dickhead.




....... Phil
 
J

John Fields

Jan 1, 1970
0
Any waveform has an RMS value, including rectangular
pulses. Of course, for rectangular pulses that are
always greater than or equal to zero the average
will work out to be the same as the RMS value. For
waveforms that traverse zero, the average can be
any value between the peaks including zero (consider a
square wave with equal positive and negative peaks).

[snip]
---
Just for grins, let's say your pulses have an amplitude of 5 volts.

That means that if they're pushing 5 amps through the resistor the
resistor must have a resistance of one ohm, and the power being
dissipated in the resistor, when the pulse is on, must be:


P = IE = 5V * 5A = 25 watts.


Now, let's further assume that your pulses are on for 505s and off
for 505s.

That means that the resistor will be dissipating 25 watts for 505s
and 0 watts for 505s and, since the pulses have rectangular edges
and current has the dimension of time, the resistor will be
dissipating an average of 12.5 watts.

Assuming a constant load impedance, then, your example of a 10% duty
cycle will result in 1/10th of the peak power being dissipated by
the load and transferred to the heat sink.

The RMS value for your 505s/505s waveform is about 3.536 A.
With a one ohm load the power dissipated is then
(3.536 A)^2 * 1 Ohm = 12.5 W, same as using the average.
 
M

Mook Johnson

Jan 1, 1970
0
BTW:
The most dramatic example of this I have seen in a commercial product is
with " Glow Plug " drivers for model size methanol engines ( 2 or 4
stroke). Model size glow plugs are made with platinum wire and draw up to
5 amps at 1.2 volts - so the best way to power them is from a single
Ni-Cd or NIMH cell.

However, some modellers like to use their 12 volt lead-acid starter motor
battery to drive the glow plug too.

But as 5 amps at 12 volts is 60 watts, a 1.2 volt linear regulator is not
practical - a switching regulator would be ideal but there is a much
cheaper ( and nastier) solution that the Asian makers of cheap glow driver
use. Simple PWM.

Yep - the 12 volt battery is switched ( with a FET ) straight onto the
1.2 volt glow plug at a high (but audible ) frequency with a very low duty
cycle - about 1%.

This means the current pulses are in the order of 50 amps !!!!

The average current draw from the 12 volt battery is then 50/100 = 0.5
p - while the glow plug has 5 amps rms flowing through it.




....... Phil


Thanks

I too run RC boats (IMPBA) and have one of that hobbico power panels that
make the glow plugs queel. :) I never considered the heating affects of the
50Amp pulses through the batteries internal impedance. Good thing we only
need them for a few seconds during startup.
 
P

Phil Allison

Jan 1, 1970
0
"Mook Johnson"
I too run RC boats (IMPBA) ...


** How did you know I was ( once) a model boat racer ?

It's the AMPBA here.

I ran a couple of monos and a tunnel hull (aka catamaran), 3.5cc and 7.5cc.

and have one of that hobbico power panels that make the glow plugs queel.
:) I never considered the heating affects of the 50Amp pulses through the
batteries internal impedance. Good thing we only need them for a few
seconds during startup.


** Those "power panels" must be the WORST possible way to light up a glow
plug.

Some of the evil problems include:

1. The 12 volt battery's voltage sags and hence plug wire temp DROPS when
you use the electric starter, often so much so that the engine will not
start. Particularly true if the engine is a little flooded.

2. Accidental contact between the starter motor frame and the engine sends
12 volts straight to the plug = instant plug death.

3. When the engine starts and you throttle it up briefly, the plug wire
becomes white hot and may fail. This is because the power delivered to the
plug by the panel increases with resistance, which is suddenly raised when
the engine runs.

None of this happens with a Ni-Cd cell - even a 1400mAH sub C size is
fine.

Waaaaaay better than any stupid MOSFET power panel.



....... Phil
 
G

Greg Neill

Jan 1, 1970
0
So I was wrong?

Your method works fine for a postive-going
waveform (as in this particular case), but would
not be correct for a general waveform. If the
OP's pulses were rectangular but not anchored
on zero volts, the average and RMS currents would
be different, and the RMS value would have to be
used to find the power dissipated in a resistive
load.
 
M

Mook Johnson

Jan 1, 1970
0
Phil Allison said:
"Mook Johnson"



** How did you know I was ( once) a model boat racer ?

It's the AMPBA here.

I ran a couple of monos and a tunnel hull (aka catamaran), 3.5cc and
7.5cc.




** Those "power panels" must be the WORST possible way to light up a glow
plug.

Some of the evil problems include:

1. The 12 volt battery's voltage sags and hence plug wire temp DROPS when
you use the electric starter, often so much so that the engine will not
start. Particularly true if the engine is a little flooded.

2. Accidental contact between the starter motor frame and the engine
sends 12 volts straight to the plug = instant plug death.

3. When the engine starts and you throttle it up briefly, the plug wire
becomes white hot and may fail. This is because the power delivered to the
plug by the panel increases with resistance, which is suddenly raised when
the engine runs.

None of this happens with a Ni-Cd cell - even a 1400mAH sub C size is
fine.

Waaaaaay better than any stupid MOSFET power panel.



...... Phil

7.5 and 15CC outriggers here.

Yup had the same problems you discribe. Was using 4AH gell cells and the 90
boat would start relaibly. Switched to low impedance 7.5Ah gell cells (low
internal resistiance) and that went away but the field pack got heavy.

We run 50% - 60% nitro and the plugs are COLD and we run rich. So we need
the 4 - 5A get them glowing. I had a setup that used a Alkaline D cell for
lighter. Worked well but then I "graduated" to a power panel. :(

Might be time to flunk back a grade. That D cell seemed to last FOREVER.
(my boats start pretty easy)
 
B

BobW

Jan 1, 1970
0
Phil Allison said:
John Fields is LOSING IT BIG TIME
[snip]

...... Phil

Phil,

Everyone's allowed one mistake in their life. Take your mother, for example.

Bob
 
F

Fred Bloggs

Jan 1, 1970
0
RMS has always been described to me as the AC current level to create the
same amount or heat in a resistive load as a DC current. This as always
been with sinusoidal currents.

In this case I have narrow 50uS pulses of current at 5 amps with repetition
rates of 500uS.

If I calculate the RMS current vs. the average current there is a sizable
difference. the average current is 0.5A the RMS current(from spice
simulation) is 1.6A.

Which if the two should be used when determining the heat generated by a
power resistor.

Assume the resistor chosen can handle the instantaneous power during the
current spikes.

I'm more concerned about the heatload this will add to the heatsink.

There is a 10X difference in power between the two.

You shouldn't be concerned with either one. Power dissipation is the
rate at which the resistor is converting electrical to thermal energy in
joules per unit of time. When the periodicity is small compared to the
thermal time constant of the heat sink, then you can you can use the
average value of joule dissipation per cycle, because much like a low
pass filter, the heat sink temperature will not respond to the high
frequency component of the thermal dissipation profile, it will only
respond to its DC value, and another way of saying 'DC value' is
'average'. You might notice that computing average power dissipation per
cycle is the exact same thing as computing RMS current and multiplying
by R. Forget about the formulas and consider what it is you really want
to know.
 
J

John Fields

Jan 1, 1970
0
John Fields is LOSING IT BIG TIME




** Wow - now THAT is really crazy bullshit.

Rectangular pulses have no RMS value ??
 
P

Phil Allison

Jan 1, 1970
0
" John Fields is LOSING IT BIG TIME "


** Completely insane.


They're coming to take him away, HA HA

They're coming to take him away, HO HO HEE HEE HA HA

To the funny farm

Where life is beautiful all the time

And he'll be happy to see

Those nice, young men

In their clean, white coats

And they're coming to take him away, Ha-haaa!



....... Phil
 
P

Phil Allison

Jan 1, 1970
0
"Greg Neill"
"John Fields"
So I was wrong?

Your method works fine for a postive-going
waveform (as in this particular case), but would
not be correct for a general waveform. If the
OP's pulses were rectangular but not anchored
on zero volts, the average and RMS currents would
be different, and the RMS value would have to be
used to find the power dissipated in a resistive
load.

** WHAAAATTTT ???????????????

For single polarity, rectangular pulses, the average and rms values are
NOT the same !!!!!!


I av = I pk times duty cycle.

I rms = Ipk times sq.rt duty cycle.


Get real, dickhead.




....... Phil
 
M

Michael A. Terrell

Jan 1, 1970
0
BobW said:
Phil,

Everyone's allowed one mistake in their life. Take your mother, for example.


She made more than one:

1. She let his dad get her pregnant.
2. She didn't have an abortion.
3. She didn't flush Phil down he toilet they day he clawed his way out
of her body.
4. She didn't have him committed, for life.


--
aioe.org is home to cowards and terrorists

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