RC Question - Page 1
Forum section: electronics / beginners

author: bostonman
author_rank: Super Contributor
timestamp: June 14, 2019, 03:18:32 pm
content: Hello,I have a question about resistors; mainly their instantaneous power dissipation.If I connect a DC voltage source to a resistor and capacitor in series to ground, after some time, the current is (theoretically) zero amps. Instantaneous though is much different because the capacitor is a short to ground, so all the voltage drop is across the resistor.To use practical numbers and part numbers. Let's assume the DC voltage source is 25V, the resistor is a wire wound 4 ohms (RWR80S part number), and capacitor is 10uF.Based on the general formula of t=RC, it would take 40us to reach a relatively flat DC level. At this point, the resistor would dissipate almost zero watts.How do I find the length of time this resistor can handle such power (based off the datasheet), and, how do I calculate the power dissipation curve between t=0 and (in this case) 40ms to know whether I'm exceeding the resistor power dissipation capabilities?

author: cur8xgo
author_rank: Regular Contributor
timestamp: June 14, 2019, 03:32:02 pm
content: Quote from: bostonman on June 14, 2019, 03:18:32 pmBTW 5 x RC is the usual amount of RC's to use to figure out when the cap is done charging, although its somewhat arbitrary.Resistors usually aren't rated in a way where you can directly determine there exact thermal behavior to pulses far beyond their DC power rating. The "unknowns" are the thermal mass of the resistor and its thermal resistance (how fast it can get rid of heat).A simple case is if the resistor is rated at 1W continuous for example, you could be pretty sure that a 2W 50% duty cycle waveform would also be acceptable and have basically the same thermal outcome. However, extrapolating that to a 10000W pulse for 100 us is not so trivial.However, to learn about this, check out the thermal impedance curves in power mosfet datasheets. Those DO specify this sort of thing from DC to very short pulses. Can give you some insight if not an answer.Also: to get the power for the time period you are interested, you could integrate over that period, or you could ballpark its average with a straight line approximation (erring on the conservative side most likely).

author: Yansi
author_rank: Super Contributor
timestamp: June 14, 2019, 04:01:17 pm
content: Quote from: cur8xgo on June 14, 2019, 03:32:02 pmWhy not using the exact energy integral?Same amount of energy as the cap will store at the fully charged voltage, will be dissipated by the resistor: Wr = 0.5*C*U2The instantaneous power on the resistor is (in a charging circuit): p(t) =e-2t/T* Usupply2/ R

author: RoGeorge
author_rank: Super Contributor
timestamp: June 14, 2019, 04:42:53 pm
content: There are two limiting parameters to take care: average power dissipation and peak power for a resistor. For your voltage and resistor values, you won't exceed the instantaneous power, which is \$P_i=\frac{V^2}{R}\$, but you should check for yourself, because I didn't looked into the resistor's datasheet while writing this. Now let's see for the average power.The simplest way will to calculate average power (without using calculus) will be to think in terms of energy.When completely charged, a capacitor will store \$W\$ Joules of energy, where \$W\$ is\$W = \frac{1}{2} C V^2\$All this \$W\$ energy will need to pass through the resistor in order to arrive to the capacitor, so the same amount of energy will pass through the resistor, too.By definition, power \$P\$ is nothing more than how much energy pass in a second, so during a charging from zero to \$V\$ the average power dissipated by the series resistor will be\$ P = \frac{W}{t} = \frac{\frac{1}{2} C V^2}{t} = \frac{C V^2}{2t} \$

author: bostonman
author_rank: Super Contributor
timestamp: June 14, 2019, 05:18:46 pm
content: QuoteMy calculus is rusty, so I'd like to figure out how to solve for the power over time. Also, what is Usupply in this case?Back to my question, if once the charging curve is derived, to assign a number for purposes of conversation, say at t=10us, the power is 100W. How do I safely say the resistor is a safe value to use?

author: Yansi
author_rank: Super Contributor
timestamp: June 14, 2019, 05:35:36 pm
content: That formula gives you the exact instantaneous power in time. Usupply is obviously the supply voltage to the RC circuit, R is the resistors resistance, T is the time constant (Tau=R*C) and the smalltis the time, from 0 to infinity.

author: bostonman
author_rank: Super Contributor
timestamp: June 14, 2019, 05:51:36 pm
content: QuoteSo I'd integrate that formula from 0 to infinity? Going off the top of my head for a moment, it would be Usupply2/R integral 0 to infinity of e^-2t/T ?Aso, is the 2 in Usupply square. i.e. Usupply^2 ?

author: magic
author_rank: Super Contributor
timestamp: June 14, 2019, 11:04:27 pm
content: Some datasheets specify allowable transient overload, for example here, for a 50 milliseconds pulse:https://www.ttelectronics.com/TTElectronics/media/ProductFiles/Resistors/Datasheets/OAR.pdfPower dissipation to the environment over such short periods is next to nothing, so the question boils down to how much heat can be dumped into the resistive material without heating it up too much. Simple wire resistors like above or wirewounds are certainly at an advantage over film resistors because most/all of their mass is the resistive material and they have plenty of it. I suppose some crude approximation can be achieved by simply multiplying the mass of the resistor times its resistive material's specific heat times the temperature delta it can withstand. Good luck finding those numbers, thoughPossibly secondary concerns may exist over mechanical consequences of sudden thermal expansion, I'm not sure. I guess nothing beats a resistor which has been tested by the manufacturer (or you) and determined to survive whatever abuse you want to throw at it.Quote from: RoGeorge on June 14, 2019, 04:42:53 pmThat's surely not a valid argument. Any energy stored in C at the end cannot have already been radiated away by the R, it's not the same energy passing through one to the other.Perhaps you are right that these two numbers happen to be equal, but that's not a proof of it.

author: bostonman
author_rank: Super Contributor
timestamp: June 16, 2019, 02:25:51 pm
content: Maybe starting with the basic will help a bit.If a resistor and capacitor are in series to ground, with 25v DC applied, the resistor is 4ohms (wirewound RWR80S) and the capacitor is 10uF, how would I calculate either instantaneous power?Or even better, how would I derive a graph over time showing the power dissipation?I imagine it would be based on the RC, but the initial power would be V^2/R; or 156W. As the capacitor charges, this power would decrease to 0W. So how would I generate a graph based on this?

author: Yansi
author_rank: Super Contributor
timestamp: June 16, 2019, 02:32:24 pm
content: Quote from: bostonman on June 16, 2019, 02:25:51 pmI think I have written you the exact formula for that, haven't I?p(t) =e-2t/T* Usupply2/ RAgain,eis the 2.71 euler constant,tis the time from zero to infinity, Usupply is your 25V suppyl voltage, R is the 4ohms resistor and the big T (Tau) is the time constant of the circuit T= R*C = 4ohm * 10uF = 40us.Stick those values in and plot it!

author: Yansi
author_rank: Super Contributor
timestamp: June 16, 2019, 02:41:01 pm
content: Just for the sake of it to verify my formula:

author: bostonman
author_rank: Super Contributor
timestamp: June 16, 2019, 03:04:57 pm
content: QuoteI apologize, it looked like you typed p(t) = e-2t/T = Usupply2 / RSo I didn't pay much attention to that formula because I thought you were giving me something linear.

author: bostonman
author_rank: Super Contributor
timestamp: June 16, 2019, 03:10:43 pm
content: The curve provides a nice visual.Out of curiosity, should that curve be integrated to give a total amount of power dissipated throughout the entire time constant?

author: DannyTheGhost
author_rank: Contributor
timestamp: June 16, 2019, 04:12:27 pm
content: If you integrate power over time you get energy, that's what you need to understand.As were stated above, to get your 'power dissipated' you simply need to get energy stored in capacitor (CU2/2) and divide it by 5*RC (how much time needed to 'fully' charge capacitor)energy/time=power

author: Yansi
author_rank: Super Contributor
timestamp: June 16, 2019, 04:40:25 pm
content: Quote from: DannyTheGhost on June 16, 2019, 04:12:27 pmI don't understand your math.You either want POWER, which in this case is changing with time (then look at my exp equation), or you want the TOTAL ENERGY DISSIPATED, which does NOT have anything in common with your time constant, not even anything like 5*Tau.Total energy dissipated is equal the energy stored in the capacitor, which is exactly W = 0.5*C*U2

author: DannyTheGhost
author_rank: Contributor
timestamp: June 16, 2019, 04:45:12 pm
content: Quote from: Yansi on June 16, 2019, 04:40:25 pmSorry, what I've calculated is average power dissipation in resistor during time when power can actually dissipate (ie there is current through resistor)

author: Yansi
author_rank: Super Contributor
timestamp: June 16, 2019, 06:52:19 pm
content: Current flows through the resistor infinitely from time t=0 to t=infinity.Average power dissipation may be relevant when supplying AC or pulsed voltage to the RC circuit. It has mostly zero meaning for designing one-shot charge or discharge circuits. Then the peak power and overall energy dissipated comes into play.For example when selecting an inrush current limiting thermistor (NTC), what you usually get from the datasheet is the pulsed energy as a relevant figure. Peak power dissipation is not that relevant, inrush NTCs are hefty device with a lot of thermal mass - heat is dissipated within almost all the volume of the device. The peak power they withstand is enormous.With switched safety discharge resistors you are usually after the allowed peak power dissipation and its duration so you can check the heat will spread fast enough within the resistor so no meltdown will occur. Contrary to the NTCs, in power resistors, heat is dissipated in the resistive material, which may be a wire (very small thermal mass on its own - temperature peaks very high locally), thin film carbon, etc. Which is why power resistors usually do not like too much power peaking.

author: bostonman
author_rank: Super Contributor
timestamp: June 17, 2019, 01:55:52 am
content: The entire story to my original question is based on creating a reliability report. The resistor I stated in my original question needs to have a maximum power dissipation value based on some length of time. It's a 3W resistor, but we need to place a value for a short (instantaneous) value.So in other words, after a long time, the resistor dissipates nearly zero watts. In our report, we'd state it's a 3W, and, after a length of time, it dissipates approximately 0W. On another line, we need a specification for instantaneous wattage, and then how much the resistor dissipates.This is why I thought taking the integral from 0s to something like 1ms would give the total power.As an example, if the resistor can handle a peak of 1000W for 1ms (although I haven't seen such a spec), then I'd imagine taking the integral from 0s to 1ms would be the correct way to provide an "accurate" number.

author: The Electrician
author_rank: Frequent Contributor
timestamp: June 17, 2019, 05:36:19 am
content: You might find some useful information here:https://www.vishay.com/resistors-fixed/high-pulse-load/

author: RoGeorge
author_rank: Super Contributor
timestamp: June 17, 2019, 07:10:52 am
content: Quote from: bostonman on June 17, 2019, 01:55:52 amSince you need to create areportfor a resistor, I will assume that that resistor is part of a critical apparatus.The only way to approach this, is to contact the resistor's manufacturer, explain them exactly how do you plan to use the resistor, what are the implications if the resistor fails, and ask them for advice. It is more complicated than simply thinking only in terms of average and peak power.Again, forget about online advice from random people, and instead ask the engineers that are making that particular type of resistor you plan to use.

author: bostonman
author_rank: Super Contributor
timestamp: July 11, 2019, 12:46:02 am
content: Unfortunately the company didn't respond.I used my work email to show that it was from a company who uses their product. My guess: they don't want to bother with such a question.I don't know why companies have email and phone numbers. Many times I send emails and call contractors, and nobody responds.