# why the dissipated power of my resistor are not equal?

Discussion in 'Analog & Mixed-Signal Design' started by imijoon, Jun 29, 2016.

1. ### imijoon Thread Starter New Member

Jul 1, 2013
18
1
Hi my all friends,

In the circuit bellow I have two similar resistor (Wirewound resistors) one for charging and other in discharge path. when the switch U1 is ON (at the time of 0.5s) the capacitor starts charging through R1. In this time a huge instantaneous power is dissipated in R1 and after about 1sec when the capacitor is fully charged the dissipation power exponentially goes to zero.
Now my capacitor is charged to 380VDC.
At the time of 4sec U1 turns OFF and U3 switches ON. At this time the capacitor discharges to R2. I expect that two resistor (R1 and R2) dissipate the same power but why R1 can withstand and R2 fails?

2. ### DGElder Member

Apr 3, 2016
347
87
The power rating for a resistor is based on the max temperature before failure. Instantaneous power can't heat a resistor since the amount of energy delivered to the resistor approaches zero as the delta t approaches zero. It is the amount of energy delivered to the resistor over some span of time minus the amount of energy(heat) it can dissipate in that time that causes heating. The power rating on a resistor is the average power it can dissipate without failure.

Power through the resistors is proportional to the square of the voltage they see. And the rms voltage is what matters in the average power dissipation, not an instaneous voltage.. The rms voltage of the supply is 380/sqrt(2), the rms voltage of C1 after charging is initially nearly 380V. So initial power dissipation in R2 is greater than R1 by about a factor of 4.

Last edited: Jun 29, 2016
3. ### imijoon Thread Starter New Member

Jul 1, 2013
18
1
Thanks dear DEGlder,

But the voltage that my resistors see are not completely ac or DC. R1 see an ac voltage that exponentially is damped and R2 see a DC voltage that exponentially is damped. How can I measure the exact dissipated power of two resistors? I want an exact amount of dissipated power to compare mathematically.

4. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,971
1,135
Let LTspice do the 'measurement' and maths for you.

5. ### DGElder Member

Apr 3, 2016
347
87
Power is the rate of energy delivery or dissipation. The power dissipation in the resistors changes as the capacitor voltage changes, so it is a function of time. You would construct the function describing power vs time and then integrate it over some period of time to get the total energy dissipated; and then divide by that time to get the average power.

6. ### DGElder Member

Apr 3, 2016
347
87
The energy dissipated by R2 is easy since it is just the total energy in the capacitor = 0.5*C*V^2. Divide by one second and you have average power dissipation over one second, or you could calculate over 5 RC time constants for example which would give a higher average power dissipation over a shorter time. It all depends on what you want to know and why you need that info. Also how frequently will this charge discharge cycle be executed as this will impact the heating of the resistor.

Are you trying to derive a formal equation for power dissipation in the two resistors or just trying to come to a practical solution for a safe power rating for the resistors? If the former you need to know calculus (at least for R1), if the later an approximation will do.
.

Last edited: Jun 29, 2016
7. ### imijoon Thread Starter New Member

Jul 1, 2013
18
1
I want to prove that the stress in R1 is less then R2. I want an equation to prove it.

8. ### crutschow Expert

Mar 14, 2008
13,488
3,372
Below is my LTspice simulation of the circuit.
It shows an energy dissipation of 101.6J for R1 and 132.5J for R2.
An explanation for the difference is left to the reader.
Hint: It's not the dissipation in the diodes, which is a little over 1J total for the 4 diodes.

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Last edited: Jun 29, 2016
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9. ### m zaid Member

Jan 9, 2016
46
5
Anything burning is caused by the amps.
btw those are large numbers

10. ### imijoon Thread Starter New Member

Jul 1, 2013
18
1
big thanks,

if there is only 31.5J difference between two resistor why R1 can withstand and R2 fails? my resistor is SWA06 Firstohm.

11. ### crutschow Expert

Mar 14, 2008
13,488
3,372
Well, 31.5J (31.5 watt-seconds) is a difference of about 30%, which is significant if you are operating near the resistor limits.

Last edited: Jun 29, 2016
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12. ### DGElder Member

Apr 3, 2016
347
87
"if there is only 31.5J difference between two resistor why R1 can withstand and R2 fails? my resistor is SWA06 Firstohm."

What do you mean it fails? By what criteria?
What is the power rating? Is this pulse a one off or does it repeat every 5 seconds?

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13. ### BR-549 Well-Known Member

Sep 22, 2013
2,178
419
I get nothing but zero results for that resistor.
Have you a data sheet?

14. ### imijoon Thread Starter New Member

Jul 1, 2013
18
1
Fail means that my resistor burns (damaged) and as it’s wirewound it opens. It's a 6W wirewound resistor. The pulse is repeated every 20sec. here is datasheet.

15. ### imijoon Thread Starter New Member

Jul 1, 2013
18
1
But I don't know my resistor power limits.

16. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
511
"It's a 6W wirewound resistor."

17. ### DGElder Member

Apr 3, 2016
347
87
R2 is averaging 27W over 5 seconds which would be 6.7W average with 20 seconds between pulses. So a 6W resistor isn't good enough. You need about a 10W for R2 and the 6W resistor for R1 is marginal so I would go larger for that one as well.

Last edited: Jun 30, 2016
18. ### crutschow Expert

Mar 14, 2008
13,488
3,372
That average is only 6.7W over the 20s interval but the peak power into R2 is nearly 3kW with an energy of 132.5 Joules so the resistor likely doesn't have the thermal capacity to handle such high surge energy and is burning out from that.
You probably should go to a 10W or larger resistor as DGElder suggested.

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19. ### DGElder Member

Apr 3, 2016
347
87
I agree with crutschow, 10W is the minimum I would try. Without an applicable power de-rating curve from the manufacturer you will have to estimate based on experiment. Start with 10W and measure the peak instantaneous temperature of the resistor under worst case conditions. You want to be well below the max temperature rating. Because of the pulsed nature of your circuit the internal temperature of the resistor will be higher than what you measure on the surface - more so than would be the case with a steady DC current. So be conservative. How conservative depends on many factors you have not shared, so it's up to you.

Last edited: Jun 30, 2016
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20. ### BR-549 Well-Known Member

Sep 22, 2013
2,178
419
I would also look for a resistor with a better voltage rating.