# Tag Info

0

B for sure in this case, with a separate doc for the placement (or even a diagram copy-pasted into the schematic.) For a big SoC it’s really the only way to represent the mixed values designers sometimes use to get good low-to-high frequency coverage. The A drawing obscures the intent and is harder to read. The cap lines don’t even land on chip pin pairs. ...

1

This is what I know: the charging of a capacitor from a dc source happens instantaneously. Since there is no series resistor to limit the current, then what actually prevents the current to become infinite and burn the capacitor at charging time? Alas, if a capacitor will really "burn" at high currents, it will burn regardless of which way the current ...

0

Electrically they are the same, as long as we speak schematic only. When it comes to layout, that's a different story, as each of the connections (wires) will be traces and each of them are linked to certain parasitic components (e.g. inductances) depending on how you design the circuit board. In the schematic it's typically all about readability and my best ...

6

Both are the same electrically, but B is more compact and most people will find it easier to understand. If you want to emphasize that each cap is associated with a specific ground/power pair, you can use A to indicate that, especially if a different person is doing the layout.

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What limits the charging current effectively preventing the burning of a capacitor That would be my question to you, because I don't know your power supply source or the wiring from the source to the capacitor. This impedance matters. If your wire is 6" of 2/0 AWG wiring from a truck battery, that will have a very different impedance (effective resistance)...

0

No, charging of the capacitor does not happen instantaneously...not in real life, and not in a properly constructed circuit analysis. Connecting a discharged capacitor directly across a voltage source violates the definition of "parallel" and KVL. Likewise for discharging. There must be resistance. In essence, you are correct that if there is no resistance ...

2

The DC source has some internal resistance, aka it can't physically source infinite current (even if you directly short the terminals). Additionally Caps have ESR (Equivalent Series Resistance) that also limit current. Do note that sometime it is prudent to put current limiting resistors at the gate of MOSFETS for example, to reduce current spikes.

2

The 104 is the value code. Similar to the resistor code. 1, 0, 4 zeros, picofarads = 1 0 0000 = 100000 pF = 0.1 uF

3

It's ceramic, 0.1uF, 50V, with a Z5U dielectric.

3

I suggest you start using water analogies for describing the working principles of electric components. In my experience, water analogies are best for understanding electricity. Here's a very good analogy for DC electric circuit with capacitor. Source: https://www.slideshare.net/avikdhupar/robotics-workshop-ppt Souce: https://ppt-online.org/22777

0

For your information, if the resistor is not there then the NAND gate will never gets a logic high in PIN 1 and always remains low and the resistor also increase the rise time and fall time so the circuit will not even act as a low pass filter, so we cannot decode the intention of the designer.

1

The easiest way to measure the charge, in coulombs, would be to calculate it by measuring the capacitance of the object (or by calculating the capacitance, given the object's size, proximity to ground and the air's dielectric constant), and measuring the voltage. Use an electrostatic voltmeter to measure the potential on the sphere, and from that, calculate ...

0

We can best imagine the processes in this circuit with the help of the well-known hydraulic analogy of communicating vessels. For this purpose, I have superimposed stylized "vessels" (in green) over the circuit. Voltage bars (in red) are there also and you can see the idea behind them - they correspond to the water columns inside vessels. How to investigate ...

4

If you're happy with a capacitance variation of perhaps only 2:1 or 3:1, with a large temperature coefficient, and a capacitance in the order of 1uF or so, then you could use small package modern ceramic capacitors as varicaps. Dielectrics like Y5U have a ferocious voltage coefficient. Unfortunately, a designation like that does not specify the dielectric ...

8

Voltage controlled capacitors (varicaps or varactors) are diodes. A reverse biased diode acts like a (small) capacitor. The capacitance varies depending on the applied reverse bias voltage. Varicaps are designed to have a larger capacitance than normal diodes. Still, you are limited in the amount of capacitance you can squeeze into the junction of a ...

0

At T<0 Energy stored in C1 is 500u Joules Joules = ½ QV where V=10V Q=(C * 10V) so Q=10uF * 10V At T1 and T2 simple use the decay function where Vss is steady state (of 5 volts ...which is half of the orig 10 evenly divided between the caps) and Vinit is 10 volts I suspect each cap (ideally) will retain 5 volts each, as each could/should ...

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 ...

2

The doubler bootstraps its bias voltages across the capacitors until it reaches a steady-state where the output is double the half-wave amplitude of the AC input. You can visualize the doubler's behavior using a simulation, right here using the CircuitLab tool. Here's an example: simulate this circuit – Schematic created using CircuitLab And here's ...

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How to investigate the circuit operation during the startup The processes in this circuit after the startup develop rapidly over several periods of mains voltage. In order for our slow-thinking mind to understand what happens during this time, we must somehow slow down the circuit operation. For this purpose, I have replaced the AC source with a DC source - ...

4

Searching internet on the terms: "murata capacitor impedance resistance" gives me as first hit: https://www.murata.com/en-eu/products/emiconfun/capacitor/2013/02/14/en-20130214-p1 Today's column describes frequency characteristics of the amount of impedance |Z| and equivalent series resistance (ESR) in capacitors. Understanding frequency ...

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 ...

1

If you are really worried about power supply noise from the "consumers" to affect others, a common-mode choke circuit of this type can be very effective: Source: https://www.murata.com/~/media/webrenewal/products/emc/emifil/knowhow/26to30.ashx If high-frequency switching noise from digital ICs is your concern, you may want to look into ferrite beads as, ...

3

You need to desolder it from PCB before you can measure it, as capacitors cannot be measured in-circuit. And discharge it too, but a multimeter can be used to measure if it is already discharged.

0

Actually it all depends on Kirchoff's Voltage Law. It indicates all voltages sum through a closed loop must equal to zero. So we start dividing voltages to branches. I suggest you to take a look at both Kirchoff's Current Law and Kirchoff's Voltage Law.

0

Filters are usually voltage dividers. A voltage divider has series and shunt components. Your capacitor is the shunt. Now insert an overt series impedance: 1K ohm resistor, or 1 milliHenry inductor. And ..... you must ground the capacitor at the MCU gnd pin at the IC package, or on the ground plane near the MCU.

1

The "typical" 0.1µF capacitor you refer to is used to 'kill' very short pulses (in the oreder of nanoseconds) of low current (in the order of tens of milliamperes), as they occur when a logic IC switches its output state. The LM2596 is a step-down converter. All it does is generating one current pulse after the other, but the operating frequency is around ...

1

According to datasheet section 9.1.1 about input capacitor, it prevents large voltage transients at the input, and provides instantaneous current for the switching.

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You could use an NTC resistor, the same kind used to limit inrush current on power supplies (hint: they're used to limit the charging current to the input caps.) An NTC will start at a high resistance, then heat up and lower its resistance as it discharges the cap. This is kind of what you want to avoid sparks.

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I think the capacitor inside the power adaptor has some contributions on delay of switching. Try modifying the adaptor either by removing or replacing the capacitor with lower uf.

0

Mohit has the right idea. The series capacitor is used to block DC and pass the higher frequency AC signal. The term "transfer function" is used to describe the ratio of Output to Input in a circuit. In terms of voltage, this would be Vo/Vi and is dependent on the configuration and values of components within a circuit. Here the circuit may be thought of as ...

1

Theoretically, zero current will flow when the switch is open. This is because once the capacitors fully charges up, they will not conduct any further. Or you can say, Capacitor blocks DC. Practically, a negligible amount of leakage current flows through capacitors, which may be insignificantly low for most applications. Capacitor don't need a Resistor to ...

1

But it doesn't really make sense since $C_I$ is charged by input current, not the opamp, so the opamp doesn't "see" the integrating capacitor. Incorrect, the input current flows into the resistor, through the node Vx, through the capacitor and into the op-amp output. Therefore the op-amp load is $C_I$.

0

I was thinking here that if I connect the capacitor to a DC battery, it is going to charge, right? We know that: - $$I = C\dfrac{dv}{dt}$$ And, at the instant the battery is applied to the capacitor, the rate of change of voltage has to be infinity therefore, the current into the capacitor is also infinity. But, in practice there are series resistances ...

1

If the switch is open, will the caps still draw charge/current? Yes, it will be very small current. See below for an example. Will this only happen when the load draws current from/through them? (Is the load required for the caps to charge?) When you close the switch the load gets connected, depending on the amount of current drawn by the load, the ...

0

simulate this circuit – Schematic created using CircuitLab Figure 1. OP's circuit redrawn. If we redraw the circuit to lay it out in a conventional fashion (reading left to right with positive supply at top it becomes a little more obvious what is going on. If the switch is open, will the caps still draw charge/current? It should be clear ...

1

By them selves the capacitors will have slight self discharge and with the battery connected there will also be a small leakage current.

0

Yes, since real capacitors do have a finite resistor parallel to the capacitance. The value of this resistor depends on the type and quality of the capacitor. Electrolytic capacitors as shown in the circuit usually do have a low resistance compared to film capacitors thus drawing more leakage current.

0

In theory, i.e. in the case of ideal capacitor and ideal voltage source, without resistances or inductivities or radiation of HF or other losses involved, the capacitor is strictly parallel to the source, which is a contradiction, since the initial value/voltage of the differential equation would be defined to be V_source and V_capacitor at the same time t0. ...

0

If V(t0)=0 when you connect a DC voltage (V) to the capacitor, there it will see a voltage difference of V-V(t0)= V, and in practice, the capacitor will charge to DC voltage V in a minimal amount of time. There will be some current for a small amount of time, and once the capacitor charged to voltage V than current will be zero.

13

A binary data signal isn't a single frequency. If it's truly random, it will have frequency content from near DC all the way up to about half the data rate (i.e. a 1 Gbps signal has content from DC to ~500 MHz). Using a smaller capacitor value would block some of the low-frequency content of the signal, resulting in excessive wander of the binary signal. ...

0

RC forms a high pass filter whose cut off frequency should be in the range of minimum frequency you want pass. For example if the minimum frequency for your signal is 200MHz. For this the 3dB cut off frequency of high pass filter should be little below 200MHz to pass it without any attenuation.

20

It's a HIGH-pass filter, not a low-pass filter. That cutoff frequency you calculated is the LOWEST that can get through, not the highest. Frequencies lower than that are blocked, and frequencies higher than that are passed.

1

In a perfect world, the current would be zero yes because once the capacitor "fills up" its charge, it'll behave like an open circuit and thus the current would be zero and your voltage V(t) would be equal to V(t0). Think of that as a taller water tank emptying itself out into another water tank. There will be equilibrium between the two water tanks. (...

1

You can replace the series capacitors with their equivalent total capacitance and then determine the voltage across and current through this equivalent capacitance as a function of time, just as you would if the capacitors where in fact the single equivalent capacitance. Since the capacitors are in series you know that the current through them must be ...

1

L3 and C33 Further to the above answers note that your schematic shows an inductance of 10uH which is almost certainly wrong. As others have commented, L3 should be a ferrite bead. These are described by their impedance at a certain frequency and not their inductance. Typically they have near zero resistance to DC current but high impedance at the ...

0

Assuming ideal caps all around: If $V_0$ is the total voltage across the caps, and $C_t$ is your total capacitance, the voltage between the caps could be anything. It might be 100V across $C_1$ and -99V across $C_2$ for a $V_0$ of 1V. The overall system will work the same. The homework may be asking you to assume the capacitors were also ...

0

As shown it's pretty dubious. You'll get a big pulse the first time the diode conducts and not much afterwards after some pulses in that direction, and then reverse pulses due to the diode capacitance, which may be larger than the forward ones. If you just put an RC (eg. 1M/1uF) in front of the meter on DC volts you should read the voltage of 5mV average ...

1

Ok, I found the real manufacturer of this cap, or at least the logo matches 100%. It is AID Electronics Corporation out of Taiwan. As you can see, they use it in the same fashion and design. I was looking for information on a different film capacitor, and the logo caught my eye. They are NOT a paper film cap.

0

C30 is a decoupling capacitor. It reduced the variation caused in supply voltage. L3 and C33 are also used to reduce the variations in check by forming a low-pass capacitor.

0

To me, your LTspice (blue) model seems not to be far away from the real data (circles). If you closely look at the real data, you see a sine component with approx. 30ns period whereas your model seems to have about 5-10ns period. So you might want to increase the Cs or Ls in your model in order to get a lower resonance frequency and to get a better match.

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