# Tag Info

1

If we suppose you applied a step function at the input over an RLC system, you can estimate parameters with the levenberg marquardt algorithm. That way you get realistic value of R, L, C and see then variation of R and C to dampen the oscillation. As Sredni Vashtar, the equation of a rlc system is of the following form: V(t) = A - A exp(-t/tau) sin(wt + ...

1

The manufacturer has many different models, and they know how the internal circuits and transformers operate inside the module. So based on how they operate internally, they can suggest the best configuration for the EMI capacitor. The difference may be which transformer leg is driven with square wave, or how the transformer is wound up. The variants 2 and 3 ...

1

It really doesn't much matter because the input and output terminals respectively have rather large capacitors across them (internally) in comparison to the added cap, so they are at virtually the same AC potential. Adding two seems like a pointless thing to do.

1

The use of 5V power supplies goes back to the days when TTL logic chips were used everywhere. TTL was introduced in the 1960s, long before anybody had ever heard of supercapacitors. These days, 3.3V logic is also common. Unfortunately, that doesn't work well with supercapacitors, either.

1

Supercaps come at 2.7V because that's what chemistry says, and 5.5V is just two of them in series. Lithium ion batteries come at 3.2-4.2V, so don't they have 4.2V motherboards? Obviously they can't be 2.7V and 4.2V at the same time. Answer: Why should they? Why would computer motherboards care what voltage lithium ion batteries have, or supercapacitors? They ...

2

I could really use some insight. A voltage rating of 5.5 volts for a capacitor is the maximum voltage that should be applied i.e. it is a top limit and, any excursion above that top limit is not recommended. Given that a lot of systems run at 5 volts and the 5 volt rail might have an output accuracy between 4.75 volts and 5.25 volts (across load and ...

3

Well, 5.5V (or 6V or 6.3V) are a common specifications for parts designed to work at 5V with a bit of tolerance. 5.5V for ICs is about as close as you want to call it (10% more than nominal). For example, the 74ACH00. Often the 5V is biased a bit higher to account for conductor losses at full load, so you might find a regulated 5V wall wart that outputs 5.1V ...

0

ppm means parts per million. 1 ppm is therefore equivalent to 0.0001%. 10% is equivalent to 100,000 ppm. Since this is such a large number, tolerances of this magnitude are not normally quoted in ppm but in percent.

2

Google is an amazing thing https://www.tdk-electronics.tdk.com/download/530780/7a11ce85afa1498543bf984153733752/pdf-markingandorderingcodesystem.pdf Anyhow, if you don't wanna read that The one you can read is "39 * 10^3 pf" or 39,000pf or 39nF, 250V NOTE: THe OTHER one has "630V" partially visible. i.e. THAT'S A HIGHER VOLTAGE CAP. ...

1

For a servo position control loop then using a tacho-generator can reduce the speed at which the motor approaches the optimum demanded position. This means that there is less overshoot because the mass of the "thing" being moved is less likely to significantly overshoot the demand point because it doesn't have as much momentum/energy. Because a ...

1

The equation you give is the theoretical reactance of an ideal capacitor. It comes from the mathematical descriptions of capacitance. I don't see how you get the idea that it is a "rule of thumb." $$Z_C = \frac{1}{2 \times \pi \times f \times C}$$ Clearly shows that reactance becomes smaller as the frequency increases - and that is true for all ...

1

An MLCC capacitor has smaller ESR than electrolytic or tantalum. Due to physical construction and size, the ESL is also lower than on electrolytics. The MLCC curve still approximately follows that curve, but due to smaller parasitics, it performs better, has higher resonance frequency, and narrower resonance peak, before the inductance makes the impedace to ...

2

MLCC capacitors generally have a much lower effective series resistance and therefore will have a sharper series resonance compared to Electrolytics/Tantalums: - Picture from this wiki page. "this equation is not ( in math terms ) "rigorously correct", it is actually a crude "rule of thumb", which has some built-in assumptions that ...

1

Tantalum capacitors are specified to make life exciting and to help remind you of the bad old days :-). A Tantalum capacitor has good capacitance per volume and low ESR, at the expense of a propensity to (any or all of) smoke, shriek, burst into flame and explode when subjected to small voltage excursions above rated value when connected to an other than low ...

-2

So i was also questioning this and did some research and i came up with something that i'm not sure is correct Electron flow is the reason the LED lights up then fades but not from electrons that originated from the battery, but rather from the metal plates located inside the capacitor keep note the series circuit is not actually closed, the capacitor has ...

3

The discharge equation is: $V_c = V_0\cdot e^{\frac{-t}{R\cdot C}}$ If this is re-arranged to isolate t: $t = -R\cdot C \cdot ln(\frac{V_c}{V_0})$ if R is a true short and there is no series resistance then the capacitor will instantly discharge (with infinite current). In a real-world situation with a finite resistance it will discharge "quickly&...

3

It's pretty much legacy information. For instance, the AD7192 on page 36 says the same thing and that dates back to 2009. The AD7124 is 2015 vintage and it appears to have just copied over some of the same words: - If I could be bothered to go back further I'm sure I'd find that pretty much all similar devices that are older say the same thing. You can use ...

2

MLCCs have mostly better ESR and ESL compared to tantalum capacitors, but tantalum capacitors are a bit stronger against voltage and temperature fluctuations. For example, an MLCC's capacitance may change 40% at 50% of rated voltage. Plus, some converters need a bit high ESR and a bit high ESL for stabilization (e.g. to prevent unnecessary oscillations).

7

The LDO built into the chip needs an ESR that is high enough to ensure stability under all conditions. You can use a 1uF ceramic capacitor with a series resistor of 5 or 10 ohms in place of the tantalum capacitor. And keep the 100nF capacitor in parallel.

1

I simulated the same circuit and current seems fine and I used LTspice XVII. Maybe the issue is your version of LTspice. I think you might try to delete your version and upload the latest again.I hope this fixes your problem.

3

Suppose I want to calculate the ESR @ 100 kHz for the 220 µF capacitor : You can't just use the $\mathrm{tan\delta}$ value, which is mostly given for 120Hz in the datasheets, for calculating ESR at 100kHz. Because, as it's described in the calculation method, the equivalent series inductance, L, is neglected at frequencies up to 1kHz. ESR values given in ...

2

We actually do not know at which frequency the dissipation factor is given, but i supposed that it is given for 100kHz. We do; it's 120 Hz as stated on the front page of the data sheet: -

1

The capacity of the capacitors can be back-calculated. Let's do some math. The capacity of the battery is 2900mAh. Let's say that the phone will run for 6 hours on this. Roughly the phone is then consuming 480mA. Also, a typical Li-ion battery goes from 4.2 ish volts down to 3.4 with a hard cutoff at 3.0v. Armed with these facts suppose we want the phone to ...

2

The MCU in Arduino has a successive approximation ADC. It works by briefly taking a voltage sample via a multiplexer into a small storage capacitor to handle multiple input channels with one ADC. With a high source impedance, the sample/hold capacitor may not have time to fully charge, and thus the sample of the voltage does not resemble the actual voltage. ...

0

The meter will have either a real capacitor or a software-simulation of one internally. If it didn't, you'd never get a stable number to be displayed and it'd be impossible to read. You can be sure this has nothing to do with any 'resistance mismatch'. A little story.... When I was in college, I was taking a CMOS class - We laid out integrated circuits and ...

0

Normally such values are used to suppress RF noise due to leakage in the power transformer. It is not so much about any shock hazard as it is to remove static from the speakers that may exist on the AC power line. The power transformer has capacitive leakage (100s of pF) and these are an attempt to neutralize that problem.

1

What makes this circuit work in that situation? Base current to the driver transistor is limited by the 15k resistor, which in turn limits peak charging current due to current gain of the transistors. Typical HFE of the BC847A is ~180 at 30 mA, and the BD140 is ~40 at 1.5 A. With a control current of ~170 uA this would supply a charging current of 0.00017 x ...

5

The circuit you mentioned could be described as a pwm current source. It uses its relatively large 2 ohms current sensing resistor, as well as the resistor R3 at the base of Q1 to limit its output current. A quick simulation shows that there is significant power dissipation in the other Q1 (BD140) during the on phase of the pwm cycle. So to answer your ...

-1

After doing some reading online, it seems that you can't really build a step-down converter with only a filtering capacitor and without an inductor. Because the losses occuring during charging of the capacitor would lead to low efficiency. This type of DC-DC converter is known as a "charge pump". The capacitors in a charge pump are not for ...

1

Search for motor-start capacitors. They are non-polarized electrolytic capacitors that are specifically designed for that duty. However 2uf seems small for a start capacitor. It also seems unlikely that a fan would have a start capacitor. Start capacitors are usually used for motors that require a high starting torque. Fans have a very low starting torque ...

1

You want a film capacitor rated for 350VAC or higher. I see 12 different types in stock at Digikey at prices from a few dollars USD up. For example, Kemet C274ACF4200LF0J, not the cheapest but a convenient package perhaps.

2

As per the manual, x is a special variable and for C it's the voltage across the device (so $i=C\frac{\mathrm{d}v}{\mathrm{d}t}$, thus $v$ is derived), and for L it's the current through it (so $v=L\frac{\mathrm{d}i}{\mathrm{d}t}$, thus $i$ is derived). But x is not a mandatory variable, that is, just because it's there it doesn't mean you have to ...

1

For the ceramic capacitor (X7R), you need to use either a much larger value than recommended (i.e. 10uF ceramic) or a voltage rating about 4x higher than your output voltage. I doubt you'll find an NPO cap at 2.2uF that you could afford to buy; especially since you seem to be sensitive to the price of a tantalum cap. https://www.maximintegrated.com/en/...

2

No. Ceramic caps other than C0G/NP0 have reduced capacitance with DC bias. Tantalums do not have DC bias but explode at the slightest overvoltage. I don't see why you can't use an polymer electrolytic. Tantalums have higher volumetric capacity than ceramics, but better frequency characteristics than electrolytics...but as mentioned, the are very sensitive to ...

0

yes I agree that limiting the large current spikes with a 150 Ohm resistor will in the long term save the switche's contacts.

1

R34 is to limit discharge current noise the switch. As the switch contacts bounce a few times they produce a tiny amount of local RF noise which the 10 nF capacitor suppresses. R34 takes this further and keeps the noise from the IC RESET pin. You do not even want the trace to the switch to have noise on it as it can migrate to adjacent traces. If the cap is ...

2

For many reasons. First of all, the external 10k pull-up is not connected by default, and the 10nF capacitor is not populated either. So forget that those exist. Without knowing any details what the reset connection looks like in the module, it is quite typical for microcontrollers to have a capacitor on their reset pin, so inside the module the MCU could ...

4

You could use @Spehro Pefhany's answer, but there is another way to do it: since you only need a fixed amplitude sine, don't use a behavioural source, instead, use a simple voltage source with SINE(0 {disp} {f}). This is because behavioural sources are a bit more tricky and need tinkering. If the displayed waveform is still not satisfactory, then disable the ...

5

That voltage is rather low. Simulate->Control Panel and set Absolute Voltage tolerance to something smaller like 1E-7 or 1E-8.

0

I finally got my device to turn on without using a battery source. I have charged the capacitor to a certain voltage, then had a switch to connect to the load when above a certain threshold. To do so, I needed to break down my calculations to Energy (Joules/sec). My device can turn on for at least 2 minutes before shutting down and having the capacitor to ...

3

Why are you immersing your assembly in oil? The only reason I can think of is that you intend to submerge it. So, I'll work on that assumption. First, any component that has an airspace is a problem. All those airspaces need to be filled with the fluid to balance pressure. Knowing this, and knowing how most electrolytics are made (they're layer stacks rolled ...

1

Dielectric constant of oil will be much higher than air, 2-3x more probably. If your circuit has high frequency tuned circuits or fast edges the additional capacitance and cross-talk might have some negative effects. That may affect some circuits. I would be particularly worried about the electrolytic capacitor seals. In fact Chemi-Con says: As @Hearth ...

1

This is actually a simple result of Gauss Law. Consider a cylindrical closed surface whose cross section is shown below (in dotted lines): Due to the symmetry of the plates the electric field is pointing in direction perpendicular to the surface of the plates as shown in the figure. So how much flux crosses the cylindrical surface? The answer is zero ...

2

I'm going to use a different capacitor data sheet to more easily explain this. Consider a 100 uF 35 volt capacitor such as this one from Farnell: - The important things to note are in red above. It has a quoted ESR of 0.16 ohms and the dissipation factor is calculated as: - $$DF = \dfrac{\text{ESR}}{X_C} = \text{ESR}\cdot 2\cdot\pi\cdot f\cdot C$$ Corrected ...

1

I think you may have created an anomaly in the way you have drawn your picture. Capacitor plates that are parallel and of the same size will have equal and opposite charge. But for there to be unequal charge on two capacitor plates there needs to be a difference in the plate areas and, this creates "fringing" to a third party (usually ground) like ...

2

It looks like a simple design if you can learn to choose a driver transistor for these parts. The design for 300mA might appear simple with these values using the model I made for this IR LED nominal values. But getting there takes some trial and error with good NPN switches and effort in simulating datasheets. Necessary details You won't be able to work ...

0

analogsystemsrf has already hinted at my cheeky answer: if you need to work with non-sinuoidal signals, rather than a nice sinewave specified by a single frequency, the response of your circuit will be a spectrum, rather than a single "phasor" :-) A spectrum of phasors across some frequency range of your interest... This is because your input ...

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