Power dissipation through RMS or Average..

[ljy4468]

Hi All,

I have a question about power dissipation calculation.
I designed some half bridge driver and simulated that. the current peaks to about 10A only during switching period.

I calculated this current through rms and average value and found that there are huge difference b/w that.

Irms=about 1A
Iavg=about 0.15A

Of course there exist negative current, but I think quantity of these negative current is negligible.

So Could you tell me why this huge difference occurred?
And Which one should I use for my power dissipation?


And let's assume if I measure this half bridge circuit in real world with power supply.
Then what will be displayed in power supply LCD while operating this half bridge? 1A display or 0.15A display??

Thanks in advance.

 

[FvM]

Power dissipation of an individual electronic component is Pavg = 1/T*∫i(t)*v(t)dt, you need to know how voltage is related to current for this component.

In case of a four-pole, internal power disspation is the difference of input and output power. You should tell more details about the circuit for a specific answer.

 

[schmitt trigger]

Indeed; as FvM mentions:

Pavg = 1/T*∫i(t)*v(t)dt
and
Prms = 1/T* sqrt[∫{i(t)*v(t)dt}^2]

because of the squaring, all negative products that would reduce the value of an average computation now become positive.

 

[FvM]

Prms = 1/T* sqrt[∫{i(t)*v(t)dt}^2]

Please review the expression. I don't think that squared instantaneous power has a physical meaning.

Actually, there's no thing like Prms.

 

[crutschow]

For the bridge power dissipation calculations you use the RMS current value of 1A.

A current meter in series with the supply will measure the average current value of 0.15A. This average value of current times the supply voltage will equal the power dissipation in the bridge as calculated by the RMS current value.

The difference between RMS current and average current is because of the high peak current and the duty-cycle of the waveform, and because RMS power is proportional to the square of the current. Thus, for example, a 10% duty-cycle 10A square pulse will dissipate a power of 100*R during the pulse, giving an average (real) power over the waveform period of 100 *R * 10% = 10*R. This gives an RMS equivalent current value of √10 = 3.16A. The average current for this period is just 10 * 10% = 1, thus giving a factor of 3.16 difference between the average and RMS current values.

 

[mikehoopes]

The answers above are correct. I generally don't need to solve those equations manually; I let my simulator do that. In LTspiceIV, for instance, it's very easy to see instantaneous and rms power in any component. Just Alt-left-click any component after running a transient analysis to see the instantaneous power dissipation (pointer will change to a thermometer). CTRL-left-click on the waveform viewer trace name to see the average and rms power over the visible time interval. Very handy, and much easier than in a lot of other simulators, which mostly use virtual instruments in their GUI.

Mike

 

[chuckey]

The only reason for using Iav is for its magnetic effect. So if you get a 0-1 A ammeter and apply a fullwave rectified current wave form, which measured 1 amp peak, it would read .63A (mean of the sinewave). Because the mean current is a useless term, the ammeter is calibrated for the more useful RMS value which is .707, and corresponds to the heating effect of a .707 A DC current. The ratio .707/.63 is called the form factor and is 1.11 for a sinewave. You have to be clear in your mind when the RMS value flows. For things like analogue TV transmission it can get complicated as the peak transmitter power is only there during the sync pulses, the power then drops to lower value during line blanking, then drops again during the active TV line period. So to find out what the real power handling of the aerial equipment is takes a bit of calculating!
Frank