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## Hot answers tagged esr

27

An equivalent circuit for what you describe is: simulate this circuit – Schematic created using CircuitLab Note that the resistors aren't in parallel, so we can't use the usual parallel resistors equation: $$R_\text{effective} = {1\over 1/R_1 + 1/R_2}$$ What we do have, however, is two impedances in parallel, so we can use the very similar ...

12

The peak ripple current for C1 is approximately I(out)/D where D= duty cycle. If the Duty cycle is say 50% at your 30 V output then the ripple for C1 is 3.3/0.5 = 6.6 A. As the duty cycle is reduced this gets worse. If the duty cycle was 10% = 0.1 then the current peak is 33 A. If you then use your ESR value the power dissipated is about 0.4 W, much ...

12

For your purposes, you should read that statement as asserting two things. 1) Ceramic capacitors have low ESR. 2) You can use a ceramic capacitor with this regulator because it is stable with low ESR caps. Due to a lack of historical perspective, you didn't quite understand what this statement meant. It is not your fault. Once upon a time, most regulators ...

11

Both of your examples commonly care about ESL and ESR of a capacitor. Along with many other examples like LDO output stability etc. I have done designs like this using Tantalum instead of ceramics for the exact reason. Stuff from China will most likely use Electrolytic ones to save money. There are still applications where the higher ESL and ESR of an ...

10

One graph says it is displaying |Z|, whereas the other graph says it's displaying "R". Obviously the 2 graphs are different, Impedance ($Z$) has both resistive and reactive components: $$Z = R+jX$$ So if the ESR is small compared to the reactance of the capacitor (as it should be, especially for low frequencies) then $\left|Z\right|$ has almost ...

8

It's just one of many ways to specify the same thing. Other ways include: dissipation factor (DF) quality factor (Q) $$\tan \delta = \text{DF} = {1\over Q} = {\text{ESR} \over X_C}$$ It's simply a matter of which is more convenient for the task at hand. The choice of which measure is used might be influenced by the expected application of the capacitor,...

8

Without the equations (like you're 5), an ESR zero 'speeds up' the response, that is it reduces the phase shift, around the feedback loop, and so allows you to make an unstable loop stable. Consider a sudden step of current into the output capacitor. With a pure capacitance, no ESR, the output voltage will start to ramp, as the current gets integrated. ...

8

but what's the meaning of the second one? Is it ESR? It's a combination of ESR and dielectric losses lumped together and computed as a series resistance to the capacitor. So, at 100 Hz the capacitive reactance is around 160 kOhms and the series loss resistor is around 700 ohms i.e. a Q of around 230. At 100 kHz Q has dropped to around 100 and keeps falling ...

7

ESR isn't necessarily physically a resistor in series with a capacitor. It's whatever resistor value gives the best approximation to the behavior of the capacitor when it's modeled with a series combination of ideal C, R, and L elements. Since the resistor is the only one of those elements that causes power to be absorbed by the device, the ESR generally ...

7

It's quite useful for repair purposes. The voltages are low enough that semiconductors don't come into play. When there are parallel e-caps, you should be comparing a total ESR of the parallel combination, but you're really looking for gross differences not even the last 2:1. The typical situation is an older piece of electronics where the electrolyte dries ...

7

Yes, you can simply put a resistor in series with a ceramic capacitor. The lower the better from the point of view of bypassing, so I would aim at 0.5 to 1 ohm. If you have lots of space, the electrolytic is fine (in fact you can parallel the two), and they are cheap. There are low ESR electrolytics and ones that are not-so-low, read the datasheet. If no ...

7

The ESR of the capacitor forms a zero at frequency fz = $\frac{1}{2\pi\cdot\text{ESR} \cdot C}$, which tends to reduce the stability excessively if the ESR is too low. The ESR also should not be too high, of course, so usually there is a recommended range. Older parts that were designed before high-capacitance ceramic parts were ubiquitous may omit ...

7

It would probably be better for you to choose a crystal oscillator rather than a crystal. You just buy it and it works, and is guaranteed to start and conform to stated specifications. If you are using a crystal then you need to consider drive power, load capacitance, accuracy as well as other things. In the datasheet they recommend you to refer to the ...

6

The PIC's datasheet says less than 1 Ohm ESR and it specifically mentions tantalum as being OK. The regulator provides power to the core from the other VDD pins. A low-ESR (less than 1 Ohm) capacitor (such as tantalum or ceramic) must be connected to the VCAP pin (Figure 27-1). This helps to maintain the stability of the regulator. The recommended ...

6

Generally, low ESR can be used to replace general purpose capacitors, but there are situations where the low ESR capacitor could cause oscillation due to the use of a finicky regulator. For example, the LM1117 is a very common semi-LDO regulator (mostly because it's cheap and you can get them very easily). It has requirements as follows (from the ...

6

The following link sheds some light on explaining why the ESR changes with frequency: (Source: http://www.murata.com/products/emiconfun/capacitor/2013/02/14/en-20130214-p1) Low-frequency region: |Z| in regions with a low frequency decreases inversely with frequency, similar to the ideal capacitor. ESR shows a value equivalent to dielectric loss from ...

6

Your assumption is incorrect. 4.7 uF ceramic capacitors are widely available. Mouser, the same site where you seem to have found your power supply, has on the order of 100 listings for through hole 4.7 uF ceramic capacitors, and over 1000 for SMT. So the best thing to do would be to simply purchase 4.7 uF ceramic capacitors, as recommended by the datasheet ...

6

As it seems, the dampening effect of a series resistance (remember what ESR stands for!) is required to make the control loop, that a DC/DC switching converter controller actually is, stable. Without that, charge might just be stored in the capacitor without significant slew, and then the control might (probably) overshoot quite a bit, leading to a suddenly ...

6

W just means $\Omega$, so Ohms. The small letter omega: $\omega$ looks a bit like a "w". It is possible that a "W" was typed at the Y-axis of the graph but that this wasn't converted into the Greek character $\Omega$ when the pdf file was made.

5

Your first assumption is incorrect. To consider that a bipole is in parallel with another one, each pair of terminals should share the same voltage. That is not true to both ESR, because the voltage of the terminal connected to the capacitor depends on the capacitor characteristics. So they are not in parallel, you cannot apply the stated law. Of course, if ...

5

While ESR is not always given it is often easy to find on a better manufacturers website. Many have online or downloadable tools that will show you all of the information you could ever want about a cap. With the rise of the need to simulate PDN for everyday boards, people have been demanding exactly this kind of information. For example take a look at ...

5

Specification says, p.12 "A minimum of 1 µF capacitance with an ESR of 1 Ω or less is recommended to ensure the stability of the ADP151" and "Any good quality ceramic capacitor can be used with the ADP151, as long as it meets the minimum capacitance and maximum ESR requirements" You will be very-very hard pressed to find a 1uF small ceramic cap ...

5

First, I doubt that the ESR measurement at low frequency is accurate. Ceramic caps have much lower ESR than Tantalum, which can affect phase margin and transient response since your ESR zero moves way higher in frequency. Second, ceramics caps can show very large reductions in capacitance with DC bias, so depending on the dielectric you have you may have ...

5

It depends on the loop design, some power supplies (Particularly older LDOs) required a certain minimum ESR to provide a zero so the loop compensation would work (Often they specified tants for the output cap to provide this). More modern parts tend to be designed for ceramic output caps and are stable down to a few milli ohm ESR levels (usually you can ...

5

Your inductor is 6 uH ? The datasheet tells us that the max. frequency is 100 kHz and that the peak output current in the switch is 1.5 A. How does the buck work? During the "on" time, energy is stored in the inductor, and during the subsequent "off" time, that energy is released into the load (powering it while the input is effectively disconnected). ...

5

If you are stepping down from 5.0 V to 3.3 V, the converter is operating with a duty cycle of 66%. At 100 kHz, your on-time is 6.6 us. With 6 uH inductor and 500 mA load current, you can expect a peak inductor current of 2.12 A, which is much higher than your load current. I assume here that you are operating in continuous conduction mode. I think for such ...

4

If you know the diameter of the core and the inductor is close-wound on it, then the length per turn is $$Lt = \pi \times (D +d),\$$ where: $Lt$ is the length of the turn, $\pi$ is about 3.14, $D$ is the diameter of the core, and $d$ is the diameter of the wire. The length of the wire used to wind the core will then be:  Lw = n \...

4

You are correct in assuming that going outside the stability regions is a 'bad idea'. It's difficult to predict exactly what will happen if you operate outside the stability region of a fixed-feedback regulator like the one you're describing. The regulator may start oscillating immediately, or appear fine but start oscillating if there's a sudden input or ...

4

You are confusing ESR, that stands for Equivalent Series Resistance, and the leakage. The first is modeled as a series resistor, and take account of leads resistance, leads-internal plates resistance and so on, and is ideally zero. The second is modeled as a resistor in parallel with the capacitor and takes account of small leakage currents in the dielectric,...

4

You can't rely on readings when the capacitor-under-test is in circuit because other components (such as other similarly sized capacitors) may be in parallel. I'm talking about power rail circuits because they are a common usage for large electrolytics where you might be particularly relying on ESR being low. In places other than power rails, electrolytics ...

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