Peak and RMS current - when to use which for thermal? - Page 1
Forum section: electronics / projects

author: shadewind
author_rank: Frequent Contributor
timestamp: September 10, 2011, 10:18:12 am
content: I have a current sense resistor on the low side of a boost converter switching MOSFET for current mode control and the average current through this resistor (and thus also the MOSFET) would be D * I(Lavg) where D is the duty cycle and I(Lavg) is the average inductor current. Also, the current through the flyback diode would be (1 - D) * I(Lavg).From what I understand here, it is the average (or rather RMS) current that matters. But when does the frequency become so low that it is the peak current that matters? Some components have current limits for other reasons than thermal (inductor saturation et.c.) but thermal is what my question is about.

author: Zero999
author_rank: Super Contributor
timestamp: September 10, 2011, 10:46:58 am
content: The RMS is not the average current but the current which results in the same average power dissipation as the same DC current so 1A RMS will produce the same power into a 1R resistor as 1A DC: 1W.The point at which the peak current becomes important depends on the thermal time constant of the resistor. I've not done much research into the subject but I'd think the peak current becomes important below the lower cut-off frequency which is equal to 1/(2pi*t) where t is the thermal time constant.

author: Unknown
author_rank: Unknown
timestamp: September 10, 2011, 12:17:54 pm
content: In other words to the ones Hero used, it depends on how quickly the resistor can get rid of the heat to prevent itself from overheating to destruction.If the rate of heat dissipation to ambient is lower than your rate of power dissipation on the resistor, then obviously you cannot sustain it without damage. As you have guessed, you need to start looking at pulse characteristics. These figures closely describe what would happen if the resistor was thermally insulated from ambient, which is the case due to the thermal mass of the resistor's internal materials. If you are familiar with thermal impedance networks (there is an episode on here), then this behaviour is equivalent to adding capacitors to the nodes hence increasing the thermal time constant.Back to the boost converter, a pulse of juice through your sense resistor has finite energy. That will be equal to the plots of I through the resistor multiplied with V across (hence instantaneous power versus time) and then integrate this plot to find the dissipated energy per pulse, or more importantly the maximum energy over a given period of time. It doesnt matter if your pulse has an amplitude of 1 ExaWatt on a 1/4W resistor if it has no time duration; energy is zero. Having added some margin, check with the datasheet. Very modern DSOs will do this work for you, but you can use a workaround (excel, approximations with pen and paper etc).

author: shadewind
author_rank: Frequent Contributor
timestamp: September 10, 2011, 12:41:50 pm
content: Quote from: Hero999 on September 10, 2011, 10:46:58 amYes, I know the difference between aveage and RMS, I just messed it up this time.

author: shadewind
author_rank: Frequent Contributor
timestamp: September 10, 2011, 01:01:54 pm
content: Quote from: Alex on September 10, 2011, 12:17:54 pmYes, now, when thinking about, it makes sense. I assume the resistor has some temperature limit that it can handle. If the power dissipation is larger than the heat dissipation, this means that the temperature will rise as a function of time and if this critical temperature is reached, then it's a problem. If the RMS power dissipation is more than the heat dissipation, then it won't work no matter how short the pulses are.Is my thinking correct?

author: ejeffrey
author_rank: Super Contributor
timestamp: September 10, 2011, 01:33:07 pm
content: Quote from: shadewind on September 10, 2011, 01:01:54 pmUsually the problem with pulses comes from long pulses. A 1/4 watt resistor will not be happy dissipating 10 watts for 1 second and then being off for 59 seconds. The average power may be lower than 1/4 watt, but it will have burned out during the 1 second. On the other hand, it likely will be perfectly happy dissipating 10 watts for 1 millisecond and being off for 60 milliseconds. The data sheet will often specify a heat capacity in J/K (joule / kelvin). Calculate your pulse energy (peak power * pulse width) and divide by the heat capacity to get the temperature rise (in kelvins) during the pulse. If this exceeds the maximum allowable temperature rise you may have a problem.Sometimes you will see a rating for maximum surge current. This is a current that shouldn't be exceeded even briefly. For instance, it might be the fusing current of wirebonds or other low heat capacity element.By the way, it is almost never correct to speak of RMS power. You normally want RMS current/voltage and average power. Average current is sometimes useful (say across a diode with a constant voltage drop), but RMS power is mostly meaningless.

author: Unknown
author_rank: Unknown
timestamp: September 10, 2011, 09:38:33 pm
content: Quote from: shadewind on September 10, 2011, 01:01:54 pmYou got the first part! Have a look at Alex CAD attached. Ok, you might need some imagination.Suppose that is a resistor component. Each bucket represents a material in the resistor (2materials here).The water you pour in the first bucket is electrical energy dissipated. The water level in each bucket represents the temperature of that material. If a bucket is full, any more water will destroy that material. You can see that if your pulse of water is more water than the bucket can hold, that material will be destroyed.Water pours out of the first bucket from the hole. This drops the water level in the first bucket (material), but it does so slowly. You can see that if you pour water fast into the first bucket (high rate of energy dissipation aka power), then the hole becomes irrelevant and the only thing that matters is the size of the bucket and the amount of water you pour into it. A pulse is like dumping a volume of water in the first bucket.If the first bucket is the resistiive material in the resistor, then you can see how it is the volume of water (total energy per pulse) that matters, as it doesnt have time to escape from the hole. Of course, if you pour the water slowly, it has enough time to go to the next bucket until it dissipates to ambient as heat. When fast becomes too fast will depend on the pulse energy and resistor limiting specs.I hope this explains it intuitively.

author: IanB
author_rank: Super Contributor
timestamp: September 11, 2011, 02:24:39 am
content: Quote from: shadewind on September 10, 2011, 01:01:54 pmThis statement technically does not make any sense, but what I think you might mean is when the heat generation exceeds the heat dissipation.When current flows through a resistor electrical power is converted to heat inside the resistor. This is heat generation. Consequently the temperature of the resistor rises, and the resistor dissipates the heat by conduction and radiation. There is some temperature where heat generation and heat dissipation are in balance, and this is the steady state operating temperature of the resistor.If the resistor starts out cold and you apply a current to it, it will start to heat up. The rate which it heats up is related to the heat capacity of the resistor. Simply stated, big resistors have a bigger heat capacity than small resistors, meaning that bigger resistors will heat up more slowly from cold than small resistors.The heating process in a resistor has a time constant, or characteristic time. This thermal time constant is analogous to the RC time constant of electrical theory.Where this gets interesting is you can compare the thermal time constant to the frequency of operation of your electrical circuit. If the electrical frequency is much faster than the thermal time constant (it usually is), then you can forget about peak currents and just consider average heat dissipation produced by RMS currents or RMS voltages.For instance, suppose your circuit frequency is 60 Hz. This gives a characteristic time of 1/60 second. Now suppose your thermal time constant is 1 second. This is much slower than the electrical time constant, so you can safely consider average heat generation and not worry about peak currents. If your circuit frequency was 1 kHz the statement about ignoring peak currents would be even stronger.You may ask about how to calculate the thermal time constant of the resistor? This bit is hard, since the data is not readily available. You need the thermal resistance, measured for example in kelvins per watt (K/W). Then you need the thermal capacitance, measured in joules per kelvin (J/K). You multiply the two together to get the thermal time constant in seconds (K/W x J/K = J/W = s).As a rule of thumb, if your circuit frequency is measured in kHz you can assume within reason that you only need to deal with average power dissipations in big elements like resistors. (But if you consider the microscopic bonding wires inside an IC, then all bets are off...)

author: shadewind
author_rank: Frequent Contributor
timestamp: September 11, 2011, 11:02:40 am
content: Quote from: IanB on September 11, 2011, 02:24:39 amYes, that is what I meant but with english not being my native language, I might have gotten the terminology a bit wrong. Let me rephrase: If the heat is generated in the component faster than it is dissipated, the temperature will rise because the thermal energy increases over time. If the pulses are long enough for the tempterature to rise above some rated temperature within the time of the pulse, then there is a problem even if the average heat generated over time is less than the heat dissipated.I was not really wondering how to calculate it since I know it is complicated. In this case, the frequency is 800 kHz so I assume that is more than enough for the pulse length to not really matter.

author: Dr_Ram
author_rank: Contributor
timestamp: September 15, 2011, 04:38:47 am
content: There is a nice bunch of calculators on Welwyn's websitehttp://www.welwyn-tt.com/products/resistors/calculation-tools.aspIn the Pulse Calculator you can enter the frequency of the pulses if you choose the "Continuous Pulses" option and see exactly what it entails. I've used the Single Pulse option earlier and found it quite useful.-Ram