# Tag Info

22

No component is ideal, and inductors tend to be the least ideal of the passive components--and relays and solenoids are even less ideal than your average inductor, because they aren't designed to be close to ideal. All materials (other than superconductors in their superconducting phase) have some positive resistivity. For copper, this is a very low ...

13

All inductors have series resistance. Well, unless they are made with super-conducting wire, I guess. But coils designed for DC operation have sufficient DC resistance to limit the DC current to a reasonable level for continuous operation. This is convenient so that the user of the coil does not have to figure out a way to limit the current. It is like ...

9

It seems like you're wondering why there are only three fundamental linear passive elements, so I'll explain that: You have two physical quantities at play at the circuit level: voltage and current. Resistance defines a linear relationship between voltage and current. Inductance defines a linear relationship between voltage and the rate of change of current. ...

3

When you have two resistive elements in series, like a coil and resistor in a DC circuit, then each one consumes power proportional to its resistance. That is why you don't put a resistor in series with a relay or motor. You want to deliver power to the relay or motor instead of wasting it in a resistor. The magnetic force produced by a relay or motor coil ...

2

2.4 GHz is a fairly low frequency when it comes to trying to make useful L or C-equivalents with microstrip structures. But instead of trying to make direct equivalents, you could consider some other options that will achieve the required functionality. Instead of capacitors C19 and C20, you could make a directional coupler structure that blocks DC while ...

2

The Q is given by the parallel resistance of the coil. so simply add a much larger high quality foil cap in parallel to the coil to bring down it's resonance frequency to a measurable range. And determine the impedance at resonance.

2

The main tradeoffs are as follows: Higher inductor ripple: Pros: Smaller inductor size, better transient response to load step and release. Cons: Higher core and AC losses therefore lower efficiency, and higher output voltage ripple for similar output capacitance. (Or more output capacitance needed for the same output voltage ripple). The converter may not ...

2

What is the mathematics on calculating the phase shift of voltage for a capacitor/ current on an inductor? Basically you have to learn circuit analysis. The math isn't difficult, but explaining and, more importantly, justifying it is beyond the scope of this forum. There are a lot of resources available on the web. The general progression of topics in a ...

2

the i vs t graph above only reveals the current that has already been slowed down by the back emf but not the current before the drop. Stop thinking that there has to be some initial current to cause the back-emf. There was no current 'before the drop'. It started at zero exactly as the equation says it must, because that is the definition of inductance (if ...

1

First, I will assume the angular frequency is $10^5\text{ rad/s}$ as you state and that the $105$ in the problem statement is wrong. This is a steady-state circuit so you can just use phasor math. You know the following, $$2\pi f = 10^5 \text{ rad/s}$$ $$X_L=2\pi f L=j4.6 \Omega$$ Letting our reference phasor be $sin(105t)$ then the current down into ...

1

You have used an impedance triangle (and pythagoras) to calculate the total impedance. That technique would be used for a series RL circuit but not for a parallel combination. To calculate the total (parallel) impedance use product/sum or 1/RT = 1/R + 1/XL Using this method I get 3.385 angle +48degrees for the total impedance.

1

The inductor will oppose the change in steady state, hence, at the instant the switch is closed, the current through the inductor will be the same as the steady state value. Therefore $\small I= \:...$

1

Would it be possible to have a solenoid connected to a DC source be used to "slow down" the current of a separate DC circuit with a wire going through the center of the solenoid with the current travelling in the opposite direction to the magnetic field. There will be capacitative coupling between the two wires, but no magnetic coupling (assuming ...

1

Ideal inductors would indeed act the way you describe: if you stall an ideal DC motor, or apply DC to a solenoid, they would produce infinite force and consume infinite current. If you're talking about real-world components, then "short" is only defined in a context: in some schematics, 1 kOhm is a short, while in other 1 mOhm is a significant ...

1

The correct DC voltage across a solenoid or relay is important to ensure it actuates and doesn't overheat. A resistor in series with a low voltage solenoid may be useful to allow it to be operated from a higher voltage. For example the KEMET EC2-5NU is a 5V relay with a coil resistance of 178 ohms. To run it off 12V a series resistor of about 250 ohms would ...

1

Just make sure the +/- convention you are using is consistent with your current direction convention. With KVL and KCL you are free to decide the orientation of your voltage and current measurements for each component. With the current direction pointing from + to - the formula is $V = L\frac{dI}{dt}$. If the current orientation is pointing from - to + you ...

1

Best way to measure Q of very high Q coils is to use the venerable Q Meter. Two typicl examples come to mind, the Boonton 260 which is a late 1950's box or the HP4242A. Both use high impedance voltmeters. You simply place the inductor across the terminals and tune the internal variable cap until the meter peaks, then direct read from the meter the Q. The ...

1

To resonate a printed coil or wired coil with 2pF at 433 MHz , you can't use a ground plane, which ends up adding to the 2pF significantly. You have to decide if you want it shielded, coplanar, spiral formed on a screw (metric or imperial) or square printed on FR4. So you have a lot of options but 1.5 turns sounds right for an air core with ~0.6" ...

1

For example, if I made two air core inductors and placed them axially aligned in close proximity, what kind of ranges of coupling coefficient could I hope to achieve? And importantly, how accurate could I get that coupling coefficient to be? Way back in my early engineering days, we needed to implement a 50 MHz and a 500 MHz bandpass filter, with very good ...

1

LC filters are a pain in this regard: you need to account for all the possible loads, in and out, and once calculated they stay that way. Therefore a change in either the input or the output needs recalculating the values. Then you have an even order Chebyshev. As you probably know, these filters have the peculiarity of equiripples in the passband (or ...

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