# Tag Info

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Most failures are ohmic, meaning materials heat up and degrade, and melt. This can be caused by overvoltage or overcurrent depending on the component. In my experience, part failure occurs most when the voltage is too high and it leads to breakdown and heating. Most of the time it's heat that kills a part, which is why when powering circuits on for the ...

1

Yes, you can have kilovolts on very small value capacitors. This is a Wimshurst machine. It generates around 30kV. It has no explicit capacitors for storing charge. It has only the inherent capacitance between the two iron bars with the large balls on the ends. That capacitance is some few picofarads. It is still enough. It charges up, and ...

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You measured the loudspeakers' DC resistance at 12 ohms each, which means they're "16 ohms" speakers. Note the speakers' impedance varies with frequency, it includes an inductive part and reactive parts due to mechanical resonances, cabinet, etc. So, it is normal you don't get 16 ohms on your multimeter for a "16 ohms" speaker. Here's a "8 ohms" speaker ...

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Found someone with the exact same issue here EEVBlog Caps suggested RS 871-2748 CPC MAL205858472E3 I would still be interested if anyone has comments about how to determine the specs for a replacement

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When designing a capacitor, the maximum voltage is determined by the insulator between the metal plates. Roughly speaking, the thicker the insulator, the higher the voltage it can stand, though some insulators are better than others. But the thicker the insulator, the lower the capacitance. So high capacity capacitors (especially supercapacitors) tend to ...

0

The capacitance does not decide how much voltage it can store, what decides the voltage is how well insulated the capacitor itself is to handle the voltage. The capacitance will decide how it charges/discharges and the electric field it can store. I will give you an example, ceramic capacitors can be found on that voltage range, you gotta specify if it is ...

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You can use high voltage ceramic capacitors , they can handle 1-30 kV . But I don't know what you going to do with them so I cant tell 18nF is enough or not. Also you don't want a bigger capacity for that much voltage even these are quite dangerous .

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How can you damage DMM if you haven't discharged a capacitor? Most multimeters test capacitance in one of two ways: Injecting a small AC signal into the capacitor and measuring what happens, and/or Paralelling an inductor, giving it an electronic "kick", and measuring the oscillating frequency. Calculation from the reference inductor value = capacitance ...

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The multimeter says CAT III 600V for the capacitance measurements, however, usually capacitance measurements on a meter injects charge and then measures the response (sometimes a frequency and then measured response). Either way these are sensitive measurements and probably use different circuitry than the voltage measurement and are not equipped to handle ...

1

According to Wikipedia, one of the uses of a ferromagnetic generator is indeed charging a capacitor bank: EDFMGs are especially well suited as seed power sources for explosively pumped flux compression generators and can be used for charging capacitor banks. I don't think you are expected to come up with a practical, finished circuit. Given the voltage ...

1

Note: This is not how to use pulse generator to charge the capacitor, rather a circuit to charge it from mains. L1 is the choke that limits the current, then it goes troug step up transformer and diode rectifier. simulate this circuit – Schematic created using CircuitLab With voltage doubler, two caps simulate this circuit

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a Capacitor behaves short circuit in ideal when t=0 (initially) it means that if physical properties of such capacitor was suitable then the current flowing through cap would be infinite, but because of ESR it is not possible for real caps and for such ESR value initial current will be Vcap/ESR which is a very high amount of Amps For high amount capacities, ...

1

The gradient (in this context) is the small but uniform change in oxide thickness across the surface of a wafer. If you have two capacitors formed from the same oxide layer you should lay them out so that their long axes are parallel. This will cause the gradient to affect both capacitors in about the same way, so the ratio of capacitance values is more ...

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Summary: The VCAP powers the Core. Hence, the core frequency matters other than system frequency Connecting two 2.2uF capacitors in parallel also helps in reducing ESR because the ESR will be half of the individual parts. A win win situation. you can either mount one 4.7 uF or two 2.2 uF sharing same copper pads. 4.7uF capacitors options are mentioned below ...

1

You can get Tant. to meet the same C=4.7uF and ESR =100 mOhm in a 1206 case while 805 case is twice that ESR. Price increases dramatically near and below 1us = T = ESR*C in any electrolytic while T<10us is low ESR and T>100us is General Purpose. https://www.digikey.ca/products/en/capacitors/tantalum-polymer-capacitors/70/page/2?k=4.7uf%20tant&...

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The voltage rating is tied up to the lifetime of the capacitor, so doing a short test will not give definitive answers immediately, which is unfortunate for you. However I would play with the DC-bias characteristic. When applying a DC voltage across the capacitor to capacitance value will drop significantly. Especially DC-bias close to the voltage rating. ...

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Using the general method, rather than doing the differential equation analysis, you have to work out the initial and final values of the variable of interest. You also have to work out the time constant, which is easy for the above circuit, but often needs Thevenin or another circuit analysis method, as the components comprising $\small RC$ or $\frac{1}{... 0 Here's the circuit in your link:- Initially SCR2 is triggered to charge the capacitor through the load. Once the capacitor has charged up to the supply voltage SCR2 will turn off when current drops below its holding current. If SCR1 is then triggered to power the load, the capacitor will discharge through the diode and inductor (which is now connected to ... 0 Summary: it is really not possible to guarantee the characteristics of the capacitor merely by testing only a few random samples from the unknown batch. If these parts are going into a design in to the field, I would never advise to do that. Capacitor voltage ratings are not highly controlled parameters. So it is difficult to gauge the voltage rating of ... 3 edit /correction. ... A step voltage cause a step current max Ic (t)=Vcc/ESR. (@ t=0) when contact is made from discharged cap to an ideal voltage source then .... Current decays at a slope by ~60% of initial surge max Ic at t=0 when t = ESR*C =$\tau$So consider an expensive big cap with Vcc =12V, and ultra low ESR was 12 milliohm that in theory ... 1 In my years in surge suppression it is oxidation of the metal film that leads to failure of metal film capacitors. The precursor to this is lack of an airtight seal. This combined with an applied mains voltage can cause rapid oxidation of the metal film. Measurement of failed capacitors show a value of about 1% of rated value. The seal must be air tight ... 2 In my experience, overheating tends to cause bulk failure of the metalized film capacitors. As pointed out by Marko above, voltage spikes (thunderstorms, heavy duty appliances switching on and off, switching on and off the equipment itself) produce spikes that can damage those caps over time b y reducing the effective capacitance. They are called soft-fail ... 12 In theory when an empty capacitor is connected to a ideal voltage source, infinite current flows until capacitor is charged to the supplied voltage. Basically same thing when disconnecting a powered inductor, in theory the voltage goes to infinity. Many devices like computer power supplies, LED lamps and phone chargers have inrush current limiting because of ... 0 When you connect$C_1$and$C_2$in parallel they will share the same voltage$V'$. A portion of the initial charge from$C_1$($Q_{1_0}$) will flow to$C_2$, so that in the end we'll have:  Q_1'+Q_2' = Q_{1_0} \text{ (conservation of charge)}   \frac{Q_1'}{C_1}=\frac{Q_2'}{C_2} = V' \text{ (same voltage for two components in ... 2 At the current state of our universe, charge is conserved. (This wasn't necessarily always the case. See this article on dark matter, for example, discussing the possibility that charged particles created shortly after the Big Bang lost their electric charge during the inflationary period.) This means that the sum of two relative charges held by the two ... 1 The scenario you describe is nonsensical and cannot be analyzed using normal circuit analysis techniques. Suppose you have two ideal capacitors with two different voltages across them. The voltage across a capacitor cannot change instantaneously because an infinite current would be required. So if you connect the two capacitors together with ideal wires ... 1 (1) I'd say the notation$I={Q \over t}$is confusing. As the Wikipedia article states it: An electric current is the rate of flow of electric charge past a point or region. Right there we should key on the phrase rate of flow. And it's pretty intuitive too -- when we think about current, it is simply moving charge from one place to another, so all ... 1 The notion of "resistance to which it discharges itself" is not very meaningful. It's better to think about the path of the current. When Vin goes to 1V, the left end of the MOSFET becomes the source and the$V_{GS}$is a high voltage so the MOSFET becomes conducting. Current flows from the capacitor, through the MOSFET, and into the voltage source. In ... 0 Yes. You are correct, if they all have the same cros sectional area and same doping concentration in the semiconductor. For a MOS capacitor, the capacitance value in inversion region is given by the the series combination of insulator capacitance ($C_{ox}$) and the semiconductor capacitance in inversion ($C_{s}$). Since all the three capacitors have ... 6 (Thanks for all people commented, having my answer improved greatly). These are called bypass capacitors (see fragment from What-is-a-bypass-capacitor): A bypass capacitor is a capacitor that shorts AC signals to ground, so that any AC noise that may be present on a DC signal is removed, producing a much cleaner and pure DC signal. A bypass ... 2 You are somewhat misguided about the purpose of decoupling. It is not mainly the noise that is problematic. ICs, especially digital ones due to the harmonics of the current pulses that they draw from the power supply, need a low impedance power source. This is because the MOSFETs inside the ICs draw virtually all of their current when they are switching and ... 4 The stuff you see around the capacitors is an epoxy potting compound to keep them from making noise while the PSU is operating (it’s on the filter toroid too, for the same reason.) When big electrolytics fail, they vent out the top. There is an ‘X’ scored in the top to allow them to vent if that happens. This would be super-obvious (and smelly) if that ... 5 That tan colored buildup is a 'hot glue' and for hotter temps than regular hot glue. You can also see the glue holding down the inductor. Usually capacitors will exhibit corrosion or a bulge when defective, it usually comes out from the top or bottom on electrolytic capacitors. The capacitors look fine from the pics shown above. Here is what a bad cap ... 1 There is the brute force method or exact science method The brute force usually relies on experience of which caps have a maximally low ESR over a 1 or 2 decades of freq or high SRF, so values are spread this far apart over the spectrum of interest using an impedance ($Z$) map vs$f\\$. for a more exact design, you need to know how to test a prototype for ...

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The two capacitors complement each others’ strengths and weaknesses. Large electrolytic capacitors work better at lower frequency than they do at high. This is due to their internal resistance, referred to as Equivalent Series Resistance, or ESR. They also have relatively high series inductance, or ESL. Both of these limit the electrolytic capacitors’ ...

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Selecting decoupling capacitors isn't an "exact science", the exact values do not matter. Many engineers just use capacitors which they already use elsewhere in their design (to keep the BOM shorter). Small value capacitors need less compromises to fit an amount of capacitance in a small space so in a small value capacitor we could for example use thicker ...

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Okay, I find this from quora: answer from Rafal Muszynski Capacitors are connected on input and output side to secure stable (oscillation-free) operation and improve step response (behavior during sudden changes of the load or input voltage). Most of the times the required capacity is in order of 1uF-100uF and such values can only be practicaly ...

1

I don't know what power factor requirements there are for such small appliances, but the cap is almost certainly there for EMI suppression, not for power factor. There's a good chance that it's 220nF because the company that designed it in has a standard input circuit, and that's it. If it were there to correct power factor, it would be there to ...

1

Class-X and Class-Y capacitors are safety-certified capacitors generally designed and used in AC line filtering in many electronic device applications. These safety capacitors are also known by other names, including EMI/RFI suppression capacitors and AC line filter safety capacitors. (EMI stands for electromagnetic interference and RFI stands for radio-...

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C6 is unusual and would usually be detrimental. R1 is on the high side and especially so if C6 is installed. Gate time constant = Trc_= R1 x Cg should be small wrt switching times but not very small. Related: Move D5 to immediately physically adjacent to Q1/Q1A - as close to the GS leads as possible. It clips negative gate ringing peaks and takes energy ...

2

This is sometimes added if a specific instance is susceptible to dv/dt turn-on (layout, lower current drive, installation). A collector-gate current, via the parasitic capacitance: Ccg flows at every switching event. If the gatedrive does not have enough drive capability to absorb this charge then the IGBT can inadvertently turn on. By adding a capacitor ...

0

their ESR is relatively constant (perhaps high not low) over a range of frequencies involving audio. this makes them easy to pair with inductors for a range of freqs stable rather than high frequencies.

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If you use a transimpedance charge amplifier, the input capacitance of the amplifier is to a virtual ground so it only has an effect (to first order) on noise. simulate this circuit – Schematic created using CircuitLab R1 provides a DC path for the bias current and should be very high value Ignoring R1, output voltage change for input charge q is -...

0

If the energy inside the cap is decreasing linearly, how can the voltage be also decreasing linearly according to that equation? The initial conditions aren't specified which are important for any capacitor calculations. But the current source will try to force 100mA through the capacitor, the only way to do this is to continually increase (or decrease ...

1

You will see a linear decrease in the capacitor voltage, but that is because you are drawing constant current, not constant power. The power you're drawing from the capacitor decreases with decreasing voltage. Energy is the integral of power over time, so it makes little sense to refer to "constant energy" unless no power is being drawn at all.

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If the energy inside the cap is decreasing linearly, how can the voltage be also decreasing linearly according to that equation? In your circuit, the current source is absorbing or supplying energy, so the resistor is not the only place the capacitor energy can be transferred to. When the capacitor is charged above 0.1 V, the current source will be ...

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But if the dielectric is thick enough to handle higher voltages, how can a lower voltage, like 30v, attract a charge on the other side? Why not? The process is completely linear. There's no "threshold effect". I can use a capacitor rated at 50V to couple a signal that might be only a few µV in amplitude. Also, it isn't the electric field that causes ...

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Some LDOs will overheat, as they try to charge up a large capacitor, and then SHUTDOWN for a period of time so the silicon die can cool down.

2

I need for peak current 2A SIM800C module. Your 7805 seems to be rated for less than 2A. So, pick a voltage regulator designed for the current you need. I want to use two lithium 18560 You're battery powered, so power efficiency should be of importance to you: The 7805, as a linear regulator, is undesirable because it converts the complete voltage ...

3

You can add the capacitor if you don't exceed the max power dissipation of the L7805. The L7805 is current limited, which means the low impedance of the capacitor won't kill it initially, if the temperature doesn't get too high in the part. This will depend on the input voltage. Because typically a big capacitor will cause problems for the load. The L7805 ...

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