# Tag Info

1

As you say the inductor cannot allow current flowing through it, to change instantaneously. When you turn off your switch it is equivalent to raising the series resistance for the switch and hence the inductor in series. If the current is not to change instantaneously when you change the resistance then the voltage will rise across the inductor to achieve ...

4

The catch diode insures a proper return path for the inductors current. Without the diode there is risk of damage to the MOSFET switch and a greatly reduced output. There is also a possibility of incorrect polarity at the output, possibly causing damage or drawing excessive current from the source. The diode solves many problems on both the ON and OFF cycle ...

0

Neutral stays close to the same potential as 'earth' because they are shorted at your local power source. The AC of live is plus or minus compared to neutral. Regarding shock, consider that one end of 'you' is at same potential as earth (e.g. your bare feet on the ground). Then only if the other end is at a big enough potential difference away, THEN you ...

0

Get a 12v5amp wall wart. Use 2 or more dc-dc buck converters to get the 5v and 3.3v supplies. Plenty off-the-shelf modules available these days at low cost. Use lot's of capacitors.

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As regards electricity, what matters is the potential difference (this is the reason why you need to touch two points to get shocked). Usually, the neutral and ground are at the same potential and a simultaneous contact is harmless. On the opposite, touching the live and one of ground or neutral is to be avoided. There can be faulty situations where there ...

13

We force it to be that way Mains power is wired as an isolated system, with an asterisk. The asterisk came about for some very good reasons. The "safeness" of neutral is a side-effect, and an optional one. If mains power were an isolated system (And I've run it that way, and it works), and you are grounded presumably... then it wouldn't matter if you ...

19

When I probe my city mains with one probe in the live and one in earth (which should be 0 volt) it shows a voltage of around 250 V. That's correct - if a little high. But when I probe the neutral and the earth it shows no voltage. That's good too. That line has been "neutralised" by a connection to earth at your supply transformer and, depending on ...

10

The neutral does not act like the live because the neutral is tied to ground at a single point somwhere. You are imagining the neutral and live both move above and below each other about ground. Relative to each other, the neutral and live do move above and below each other. This is if you think of them in complete isolation relative to each other and only ...

1

Well the cleanest option is to have it use a single 12VDC coming in from a wall AC/DC converter. Then you would have voltage regulators to get the specific voltages you need within the unit. Easiest way to do that is to have a custom PCB made, enabling the regulators to be done on board. If that isn't an option, you should be able to get pre made voltage ...

0

There are 100 V 5 W Zener diodes on market. You can construct a voltage limiter circuit. All you need to do is building the limiter circuit with less voltage level. For example consider a 5 kV DC with +250 V peak to peak ripples, a voltage limiter with 4.9 kV will kill all the ripples. This solution is likely cost you more because larger diodes that on ...

0

I know it's cliche, but you could use a solar panel, there's a lot of variables on the water when it comes to wind. Specifically how are you going to manage your windmill extracting energy from the motion of your boat when you're trying to get somewhere (relative wind from the boat's driving through the air), and what are you going to do when there's a storm?...

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Your load must be smart enough to go to sleep so when the Under Voltage is detected , the battery is locked out from drain. This is usually called UVLO and you may choose to create an early warning threshold should your design choose to increase your broadcast interval. You must get smart enough to design your load demand to always be less than supply ...

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We can't answer your question (except in general) without knowing the specifics of your system. In general with these kinds of systems the battery can supply the load both while charging and when there's no wind (which is the point of the battery). If the charging current exceeds the load current then the battery will continue to charge until full. If ...

0

How critical is the timing and the amplitude? My experience with these types of devices is that they need higher voltage (and hence higher current) just for the initial magnetic pull-in. Once the the brake is engaged, drop the voltage to minimize heating. Similar to Russel's approach, I'd start with a large electrolytic capacitor. Since the sustain voltage ...

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A solution could be using a 555 timer. Use a 555 in monostable mode generate the 200 ms pulse for the 24V supply. Benefit of a 555 is that it can make relative large time pulses and you can easily change the timing. The brake signal has to be converted into a break pulse in order to have the 555 timer reset again (after 200 ms). This edge detection is ...

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"Mickey mouse" and only sort of right solution. On switch over Vout is Vin + a charged cap, decreasing to Vin - a diode drop. Vin could be higher with a series resistor, if desired. Cap size is based on the desire for about 5 J or energy. Use: 0.1F supercap with a diode across the capacitor, non conducting when charging. With switches as shown the ...

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You could use 2 relays. One a single pole relay, the other a double pole relay. On the single pole relay, use an output from the micro controller to pull in the coil. Then wire the 12v source through the NO contacts on the single pole relay as well as a NC set of contacts on the double pole relay then on to the brake. On the other pole of the double pole ...

0

Line reactors help protect the VFD, Load Reactors help protect the motor. NEITHER of them can be effectively justified for the sole purpose of "reducing harmonics" in my opinion, because they do NOT reduce the harmonics enough on their own to be of much value. Load Reactors in fact have ZERO relationship to harmonics. If you are concerned about harmonics, a ...

1

Why is 3% on the line side and 5% on the load side recommended? 3% is recommended for the line side because a 3% voltage drop in addition to the "typical" voltage drop that would be seen without the reactors is estimated to be the maximum that is tolerable. 5% is recommended for the load side because that is the maximum that is estimated to be tolerable ...

2

Q: the grid-side current feeding stage takes the source AC into consideration. How is this possible? A: the condition is a grid fault (0V sc.to gnd) with the voltage source now appearing as a current sink to the 400V charged microgrid capacitor. Q2: Also, the first and second stage is described as a natural response and the third stage is described as a ...

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The fault characteristics can be divided into three stages as described because "the VSC...interfaces to the ac side through an inductor (Lac)." In addition, there is inductive impedance on the DC side in series with the fault. That impedance controls the timing of the response such that it can be analyzed in three stages.

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Note: I have noticed that some people use the term "open" for conducting, which is in direct conflict with "open" as in "open circuit" (not conducting) as we use in North America, at least. It would be better to use some other term. Threshold voltage is specified at a given Id. So if you apply a constant current (with limited voltage) or simply connect a ...

1

The waveform from $\pi /6$ to $\pi /3$ is just the symmetrical reflection of the waveform from $0$ to $\pi /6$. Therefore, the average value will be the same over those two periods.

0

how would C1 discharge since only I1 will be passing through the mosfet? Because it had been charged when the mosfet was not conducting as shown in your edited question, picture 6. In picture 2, when the mosfet is open, and C is being charged through D1. This is simplified in picture 6: the current $i_1$ is charging C1 and $i_1 = i_{C1}$. When ...

1

In a switching converter this is desirable because it means that it lets the current in the inductor never fall to zero which lets you run in continuous mode instead of discontinuous mode which produces less ripple and noise (I think...it's something like that). In a motor this is desirable because you don't throw away all the energy put into building up ...

1

You typically choose a PWM frequency high enough to limit the ripple in the current through the motor to prevent torque oscillation and audible noise, and so the current during the off period won't decay to zero except at very low duty cycles. At 50% duty cycle, the motor will be running at about half full speed, and the current through the freewheeling ...

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Is not the lift controller that determines if motor is motoring or regenerating. This task is accomplished by the VVVF (variable frequency variable speed). For two quadrant operation (this is the most common elevator lift, but there is also four quadrant inverter) the inverter senses the DC BUS voltage. If it increases above a preset level (perhaps 800 VDC ...

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The low side sink current switch causes a high release voltage after clamp to 0V.

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This is an inductor current graph, not an output capacitor voltage graph, so it's difficult to tell from it whether the output capacitor discharges through the load (it does). The current between $DT_s$ and $T_s$ is not constant, but the derivative of the current is proportional to the voltage difference between the inductor terminals, which during "off"...

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It sounds to me that you are running an open loop "Volts per Hertz" algorithm similar to what you could run with an induction machine. The problem you are seeing, I think, is that a permanent magnet synchronous machine has very little damping (contrary to induction machines), and your rotor is oscillating pretty wildly. When your currents are high, this is ...

3

You have quite a few questions. No the cap across the transformer secondary does not short it, maybe it reduces some higher frequency transients from external sources or from the diodes switching. Pretty sure it will work about the same without it. The current through the probes is AC (ignoring a brief bit when first turned on) so long as the series ...

2

The switch allows for PWM output, which I have seen can be filtered to an analog output voltage - however I have not seen this done on high current inductive loads. Is this a feasible solution? Probably not. The pump's built-in motor controller probably has a large filter capacitor across the supply to damp voltage spikes caused by lead inductance, so ...

0

The internal motor controller should limit the starting current. If it doesn't, the selected input voltage controller appears to be capable of limiting the current. It appears to be capable of withstanding the typical stalled current of a motor with the 4.6 amp current rating. There does not appear to be any reason to regulate the output of the selected ...

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I found a solution better suited to my application: https://www.nxp.com/docs/en/data-sheet/MC33981.pdf

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This is all the protection you need simulate this circuit – Schematic created using CircuitLab D_Freewheel prevents the output of the inductor from going more than 1v below ground. It's got to be able to handle the full current that the motor takes. Even 'slow' diodes like 1N540x are quite sufficient, they're only slow on turnoff, they turn on ...

2

L1 should have a freewheel diode across it .Any fast diode will work .It is not good to allow negative backswing on the fragile highside driver chip .The freewheel diode also ensures that M1 will not see much more than 12VDC .

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From my comments, with credit to @Sean and TimWescott (Since they have not yet answered). If you only want to knock the peak voltage down by ~100V, there is no need for a super high voltage regulation system. Typical modern linear regulators really only require a "reference" for ground since they often sink far less then 1mA. Look for low "Quiescent Current"...

1

A Google search reveals that they fail because they overheat. The Rds is very low, so that is unlikely to be the problem (assuming that it is fully turned on). It could be that the Vgs is too low and it isn't fully turning on. Or, it is used in a PWM circuit and it is switching too slowly. You would need to scope a working circuit to diagnose the marginal ...

0

The TPA6111A2 headphones amplifier is not designed for electret microphones as inputs because with enough gain then its input impedance is too low and the external compensation capacitor required will cut high audio frequencies. An LM386 power amplifier works fine with an electret mic since its inputs are a higher impedance than the other inverting amplifier ...

0

Core losses are resistive due to eddy currents that rise with f^2 and iron thickness, t^2 and and hysteresis losses due to nearing the L-10% saturation threshold. Eddy current losses are the result of Faraday’s Law http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html which states that, “ Any change in the environment of a coil of wire will ...

2

The synchronous motor/generator has two power ports, a mechanical power port at the shaft and electrical power port at the motor/generator terminals. Because mechanical losses and magnetic losses (core and stray losses) are (approximately) proportional to the motor’s rotational speed, they are often taken together and referred to as rotational losses. ...

1

Core or iron losses are not mechanical. They are mainly due to frequency and voltage, so they are constant. Eddy currents (batteries set up by impurities in the iron) and hysteresis (heat created by reversing magnetic field) are the main iron losses. Eddy currents are the main reason the core is made of laminated segments.

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(EDIT: Complete rewrite) The TPA6111A2 is a stereo speaker driver with an inverting input. To use it correctly, the source would need to have negative polarity. The datasheet shows that, labeling the inputs IN1- and IN2-. The LM386 is a single channel with internal biasing that accepts a positive input. Neither chip is really a mic amplifier: they don't ...

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Any sound picked up by a microphone can be assumed to come from some point distant to the microphone (maybe mm to several metres or more). Take a sound at 1 kHz at 10 metres distant to the microphone. 1 kHz has a wavelength in air of around 0.34 metres and this means that between the source of the sound and the microphone diaphragm, the signal has undergone ...

3

In some cases it will make no difference, as the microphone signal is an AC signal, and in many cases the inversion does very little, if anything, to the audio (there are cases, like sending differently phased signals to different speakers, can do bad things to sound). In other cases, it can make a big difference, often depending on the type of microphone. ...

1

You are asking for trouble; the drive frequency is 140 kHz and your primary series resonant tuning is precisely 140 kHz therefore, the L and C act as a short circuit at the switching frequency. You need to run the primary circuit either from an output stage designed to handle the series resonance or use a different approach - maybe add a current limit ...

1

In the given diagram it is assumed that the inductor in the load will try to maintain the current without much ripples..This inductor induced voltage keeps T1 and T2 on during negative half cycle

0

Depending on narrow-band versus broad-band design, your SNR becomes a key design variable because broad-band interfaces that have 50ohm Rsource and 50ohm Rload will cost you 6dB signal level. Yet you have the magic-incantation of "but boss, I matched it".

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There’s a difference. conjugate matching reduces the VAR LOSS to real power transfer. maximum power when matched results and 50% efficiency from a voltage source. Higher efficiency and gain means less output power but lower input signal when load is say higher impedance than source. At extremes Gain with no load is twice the voltage but no current so no ...

0

where does this relationship come from? It comes from the fact that if you have a forward travelling wave at $z=-l_1$, then it will travel forward a distance $l_1$, then reflect off the load, then travel in reverse distance $l_1$ again before you see the reverse travelling wave at your measurement point. Are there any assumption (for instance that ...

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