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RC Question
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bostonman:
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?
cur8xgo:

--- Quote from: bostonman on June 14, 2019, 03:18:32 pm ---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?

--- End quote ---

BTW 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).
Yansi:

--- Quote from: cur8xgo on June 14, 2019, 03:32:02 pm ---BTW 5 x RC is the usual amount

--- End quote ---

Why 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*U2

The instantaneous power on the resistor is  (in a charging circuit): p(t) = e-2t/T * Usupply2 / R
RoGeorge:
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}  \$

bostonman:

--- Quote ---p(t) = e-2t/T * Usupply2 / R
--- End quote ---

My 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?
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