Hot answers tagged

8

I want to point out one thing that the other answers are ignoring, or at least glossing over. In this measurement, we stop the flow of current in our inductor, and this causes current to flow in the earth below it. That means we're effectively using the earth itself as the secondary coil of a flyback transformer. In a flyback transformer, the primary ...


8

Can you collapse an inductors magnetic field instantaneously? NO ….you can neither create nor collapse a magnetic field instantaneously since there is always some L and R time constant involved. You CANNOT reverse the voltage on an inductor either (at least not in any practical circuit) and get lower discharge times. Perhaps a simple example may help: ...


5

In a permanent magnet or shunt-wound DC motor torque is directly proportional to current, and the torque constant (Kt) is the inverse of the velocity constant (Kv). If your motor does 12000rpm at 12V unloaded, then (assuming no-load current is 'small') Kv = 1000rpm/V or 105rad/s/V. The inverse of that is 1/105 = 0.0095N-m/A or 97g-cm/A. At 1.2A it would ...


4

There's no simple, accurate way to do it from a graphical plot. The two alternatives are (a) simplest, set the RBW to 3 kHz, which will encompass essentially all the power in the signal, now read the marker or (b) hard work, get a numerical dump of the display, and use anything from Excel to your favourite computer language to integrate the power across ...


3

No, of course not. The torque that a motor can generate is related to the strength of the internal magnetic fields and their geometry — especially the radius of the rotor. The magnetic field is proportional to the current, and the max current also puts a limit on the max torque, regardless of RPM. If you want to trade off RPM for torque without ...


2

How would one actually collapse a large field quickly in practice? I am looking for the nuances of collapsing an inductive load that is large which potentially 1kV+ of kickback. As you acknowledged in your first update to your question, it is unrealistic to expect the instantaneous collapse of the magnetic field generated by the "exciting" inductor. However,...


2

The problem is probably interference between the power supply and the telephone line. Telephone lines and power lines should be isolated from one another, but cheap devices (the DTMF decoder, or the power supply, or both) may skimp on isolation - if it were a circuit with a speaker, you would hear a loud humming noise. This goes along with the USB ...


2

I would expect at 1M ohm input impedance and with 0dbm, to get an ~.244 V amplitude, yet this is not the case. At 0 dBm (1 milliWatt) into 50 ohms I would expect an amplitude of: 224 mVrms = 316 mVp = 632 mVpp. However, you removed that 50 ohm load! So now you get double that voltage: 448 mVrms = 632 mVp = 1.267 Vpp This schematic explains what happens: ...


1

It is all about electrical power and energy. An arc need a higher voltage so that the jolt of electricity can jump over(or through) air. This is why more voltage is needed than current. And a heater needs more current than voltage. When current goes through a wire it can produce heat. The heat power is p=I^2.R More the current is there, more the conductor ...


1

The good and proven recipe is in the laptops. They use buck regulators everywhere on the motherboard. So I would recommend it to you as well. Just use buck regulator with parameters you like.


1

This answer does not attempt to address your real question but is only to clarify your understanding. This is my problem, I bought this circuit to receive caller ID over protocol DTMF. As JRE has pointed out, you are confusing DTMF (dual-tone, multi-frequency) with Caller ID. Figure 1. DTMF frequency matrix. Pressing any button transmits the row and ...


1

If you assume that the opamp is a rail-to-rail device you can then say from inspection: The configuration has unity gain. At zero load current the output limits at approximately +/- 4.3V (one V(be) drop from each supply). The configuration will clip at approximately +/-70mA and the output voltage will reduce linearly (the 10 Ohm resistor) above that current....


1

Here is a schematic from others that use TDEM's to do research, which is an H-bridge with TVS diodes to short out the high voltages that may come across the inductor when it is disconnected: Source: https://www.researchgate.net/post/How_to_evaluate_the_performance_of_a_time_domain_electromagnetic_TDEM_or_TEM_system This would be a starting point for a ...


1

Power output capacity: 75 VA into a 3 ohm resistive load So your power amplifier can deliver the highest power into a 3 ohms load. If you would connect all speakers in series you would get 30 x 8 ohms = 240 ohms, then the maximum power to the speakers is limited by the voltage that the amplifier can make. This is about 15 V, into 240 ohms that means P = U^...


1

I don't think this will cause any issues. Although technically this is a loop, ground loops are usually a problem when some device has multiple "ground" connections which may differ slightly in potential. This won't occur here since there are no devices in the loop.


1

You could probably do this, the problem is if you did, a way would be needed to make sure you don't send more than 2A through one TPS2215A. The other problem is matching between devices, if one device has say 10% more resistance, then the other device would conduct more current and dissipate more heat. But TI also has a TPS2120 that is rated for 3A which ...


1

Another way to collapse the field - connect it to a capacitor. Wait until the capacitor voltage rises to a peak and then disconnect it. The voltage can be calculated from the energy of the inductor eg LI^2 = CV^2. The fall time is determined to be half of one sine wave whose frequency is 1 / (2*Pi* (LC)^0.5) You also get to save most of the energy for the ...


1

They ain't collapsing no field instantaneously. At best it's limited by the speed of light; in reality they're generating some high voltage discharge, or the energy of the field is being dissipated into the soil. In open air, you'd discharge the coil into a spark gap. "Instantaneous" is literally wrong, but figuratively, if you get the coil strength down ...


1

where would the overproduced electricity go, if the consumer has estimated his consumption poorly? It doesn't go anywhere - if the consumer under-consumes compared to the original prediction then, that under-consumption is physically less-current being taken and less-power being supplied by the generator. Electricity isn't like a delivery van setting ...


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