New answers tagged

2

Beside the inductive energy which is realtivelly small, there is a kinetic energy of rotating mass. In case of induction motor reversal, the whole kinetic energy is dumped in the rotor, i.e it would glow or burn. ad 1.) The energy is transformed into mechanical energy or heat. ad 2.) There is no large kickback, since there is a free magnetic path in the ...


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There would be arcing at the contactor contacts, due to 'inductive kickback', every time an induction motor is switched off. When a running motor is reversed, there would be a finite time between the motor switching off in the forward direction, suddenly stopping and then starting in the reverse direction. Hence the same arcing would result. Such a reversal ...


2

Fuses are intended for short-circuit protection only. A standard mains filter would take care of electrical noise. Heat dissipated by a load is inherent and cannot be avoided by a series resistor but only taken over by it.


0

Assume that the core is of infinite permeability (ideal magnetic material). g=0,113mm


1

We could just plug numbers into the formula, but it's more instructive to do it from first principles. Then if you forget the exact form of the formula, you can re-derive it. You've already suggested some numbers for turns and pole area, so let's do it for a specific airgap as well, say 1 mm. Inductance is inversely proportional to it (assuming perfect ...


0

It depends on the effective diode impedance at some reverse voltage < Vbat and leakage current rating @ some -V and the ambient temperature as well as the diode type e.g. Schottky diodes leak more than Silicon. Consider at some temp the Fairchild Semiconductor FDV303N has a reverse leakage of 0.1 microamp @ -1V. That would be equivalent to 10MOhm, thus ...


3

For ideal components, the circuit is described by ill defined mathematical equations. The ideal diode when reverse biased will present infinite resistance. The ideal voltmeter also has infinite resistance. So the voltage division formula will be \$V_{\text{voltmeter}} = \frac{\infty}{\infty + \infty}\$. For realistic components, the volt meter would read ...


1

Data centers use 3 phase because utilities will insist on it for the amount of power they use. If they were to give them single phase, it would cause a severe imbalance of the 3 phase utility transmission lines. Single phase service drops are generally limited to 167.5kVA (around 800A) maximum, most data centers use thousands of amps.


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The real problem is that the stresses involved with achieving say 10% ripple voltage with split phase is that the charge current can only operate during 10% of the time so its decay period of 90% of the time is the actual discharge load current while the charge current is the load plus the charge current to reach diode peak voltage. this results in the ...


3

We were offered 120V or 208V at at one US data center, so three phase is available. Six pulse rectifiers are efficient at making DC, but powerful DC circuits are hard to manage once you exceed the working voltage of your typical arc welder. Ordinary switched mode power supplies start with rectifier followed by a a power-factor correction circuit, these will ...


0

you don't mention it, but this vonverter is bi-directional, energy (and net current) can flow either way, this is a good feature as it allows regenrative braking to recharge the accumulator. Neither switching scheme is ideal. the second implementation doe not conduct current directly from input to output this means that all current flows through all four ...


2

This TEC is 58% efficient with 170W in and 50’C difference. Efficiency goes down with higher temperature difference needed to reverse the process and may in fact be not equivalent. I suggest if you input a 50’C temperature difference and load the TEC with some MPPT regulator to match the source impedance to the load, you might compute the thermal resistance ...


0

TL;DR datasheet Your IC is designed for 38 to 55V batteries with 35 to 55 Vinputs. My explanation is for Laptop chargers. The same philosophy may apply to Lithium secondary batteries but not Lead Acid which can tolerate 2.3V per 2V cell in continuous charge voltage ( with temp compensation for glass Matt types.) So the hysteresis may be as small as you ...


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Here's the logic. And here's the timing diagram. Relay 'A' would set and reset on alternate pulses of the reed switch.


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I solved the issue. The pads of one of the switching transistor were lifted. I repair these pads and SMPS worked fine. Thanks for the support. @winny @RohatKılıç


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There are two effects that cause ghosting in LED matrices like the one you designed. The first the the switching time of the row transistor you use, and the second is the parasitic capacitance of the LEDs themself. Every transistor takes a finite amount of time to turn on or turn off. If you turn off Row A at the same moment you turn on Row B, the reality is ...


4

It is a gas discharge tube (GDT) in a glass package. In some power supplies they are shunting one of the common mode choke coils, so this is not an unique design. It makes somewhat sense that it seems to shunt high voltage transients so that they bypass the common mode choke neutral coil.


3

My speculative answer: You wouldn't put a unidirectional device across an input CM winding for clamp suppression. It would conduct every half-cycle, after all. A bidirectional TVS or a gas-tube would work for clamping - in both cases, should sufficient voltage be present to cause breakdown, the TVS or gas tube would clamp the voltage. (This sort of event is ...


0

I think its a Flyback Diode. When the circuit is turned off after reaching a steady-state, The coil has back EMF which is reversed in polarity. This will result in high voltage, potentially creating an electric arc. The flyback diode is reverse biased when the circuit has power, but when turned off it provides a path to dissipate power, avoiding an electric ...


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The formula for calculating allowable current through a trace is published in the IPC-2221 standard section 6.2 and you may find this on-line calculator helpful. Best to check with another source before deciding on a value to use ...


1

The type 3 compensator can be built around a variety of amplifiers like an operational amplifier (op-amp), an operational transconductance amplifier (OTA) and a TL431 to cite the most popular options. The type 3 offers the following characteristics: a pole at the origin offers a high open-loop gain in dc (for \$s=0\$) and helps minimizing the static error (...


0

A half bridge is a buck-derived topology, and peak current mode doesn't work without some interesting tricks. So we'll assume voltage mode. You would place a pole at the origin for high DC gain. Then place a pair of zeroes at the resonant frequency of the L-C filter to cancel the complex poles there. Then a pole to cancel the ESR zero of the output ...


1

I agree that using the \$\omega_m\$ to produce a time-varying observer is a valid and intuitive approach. The "theoretical correctness" of it will mostly depend on two aspects, and both will require some reading and research on your part. (1) the criterias you want to guarantee for the closed-loop system. This mostly concerns stability guarantees ...


4

is it possible? No. 5 V < 8 V so no charge will flow into the battery. is it safe? No. Lithium batteries are explosive. any other way? Yes. Buy a proper charger. This will monitor the battery's state of charge and limit voltage and current to a safe value to prevent fire and explosion. ... lithium iron battery ... Are you sure it's not lithium-ion?


2

No, you want to hook up the pot as a variable resistor, not a potentiometer to ground. Remove the connection from the end of the variable resistor that you have going to ground, and hook that to the wiper of the variable resistor like this: The way you have it drawn puts the "bottom" resistance of the pot in parallel with the lower feedback ...


3

I'd go straight for a single stage dual phase converter like this: - One for each motor should do the trick. Be warned though, this is not something to be undertaken without good tools and PCB design skills. You will certainly find more designs available if you split the power requirements by four and concentrate on a single regulator per motor load. ...


1

With a 100% efficient power conversion from 3.7 volts to 10 volts, any current taken from the boosted 10 volts will be seen as 10/3.7 (2.702) times as much on the battery so, it's probably better to just consider the battery capacity as unchanged but, in these circumstances it will be dispensing 2.702 times as much current. Given that the efficiency of a ...


0

The battery capacity does not change, it will stay 750mAh. It's the current taken out of battery that will change, so that changes how long it can drive the LEDs. 9V×30mA is 270mW load. Assuming the boost converter has 100% efficiency (it won't), there is 270mW taken out of the battery. At 3.7V that is 73mA so 750mAh/73mA is only about 10h.


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I am using IRF3205 to switching GND, and the gate is controlled by the microcontroller for the delay, so when the connector attaches to the load circuit, the GND is already disconnected, so the spark not happens. Solution: - When the power In signal detected by the microcontroller, it turns on the gate after a few seconds to avoid spark. (Note:- Here I can ...


3

The FOC already does all of this. If the motor's actual velocity is higher than it's velocity setpoint, then you will have a generator mode. You can even control the torque with Iq setpoint. For example: if n>0 // positive direction Iq= -Iq_set; // negative torque (braking) elseif n<0 // negative direction Iq = Iq_set; //positive torque (braking) ...


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I don't think there is a strict or definitive way to define it here, so the measurement goal should be the capabilities of the controller, if you want a more powerful mppt controller you want better components, they may be thicker and bigger. So the weight will increase accordingly.


0

You need relays with more robust contacts for a reactive load such as those on the front end of laptop chargers. The arc occurs when connected with contact bounce or disconnected with an inductive filter and storage cap to filter RF. The down side is the surge current derating on silver oxide contacts is HUGE often from 10A to 0.3A for miniature and 10A to ...


0

I finally got my device to turn on without using a battery source. I have charged the capacitor to a certain voltage, then had a switch to connect to the load when above a certain threshold. To do so, I needed to break down my calculations to Energy (Joules/sec). My device can turn on for at least 2 minutes before shutting down and having the capacitor to ...


3

You are welding the contacts of the relay by the surge current, probably caused by capacitance in the power supply front end. One approach would be to use a physically bigger relay with a better surge current rating. Probably a TV-5 rated relay would be acceptable, though the rating is UL/CSA so it's limited to 120VAC, so make sure the relay itself is rated ...


0

I'm having some trouble understanding how the diode-connected device M3 maintains a fixed voltage difference between VDD (source) and node X (drain)? In other words, why is it called a diode? The operation of the "diode-connected" M3 can be explained by the help of the negative-feedback principle as follows. The output drain-source voltage Vds3 (...


2

Can't Vgs be greater than Vth for a diode connected deivce?, so then the Vdd to Vx difference will no longer be a constant Vth. That's not the point here. The point is, for a given voltage difference between the gates of M1 and M2 there will be defined currents through M1 and M2 (sharing Iss). This current through M1 sets the \$V_{gs}\$ of M3 and thereby ...


2

If a pin would be externally held to the positive or negative rail, it will use the same trivial amount of power when configured for analog mode as for digital mode. Likewise, there would generally be negligible extra power consumption from enabling internal pull-up of a pin which would either be externally pulled high or disconnected would draw no current, ...


0

No you can't, because your neutral is floating, so any imbalance between L1 and L2 loads will cause imbalance between L1 and L2 voltage. Perhaps it would work with a transformer tha has 230V input and center tapped 120V-0V-120V output, if you can get the grid tied inverter to turn on without a grid.


15

According to the reference manual: This apparently can reduce the typical current draw from uA to nA. It's not clear if they make assumptions about whether the pins are tied high, low or floating for these comparisons.


1

My suggestions, MOSFET based inrush current limiter: https://github.com/msglazer/Anti-Spark_Switch (Generally used in octocopter) Anti-Spark XT60, XT90 connectors explanation: https://youtu.be/X71Suakve6A (Video) In Anti-spark connectors, they are using small value high power resistors approx 6 ohms. This Resistor helps in reduce inrush current due to ...


1

You can get a spark hooking up 12V jumper cables between cars because you are making and breaking a circuit in the many-millisecond time it takes to connect. The inductance in the cables or receiving circuit is surprising in its ability to keep current flowing in a collapsing magnetic field when a circuit is broken. It's nothing to worry about. If you want ...


1

Your design has excessive inrush current because it has no protection for a bulk capacitive load. Thus the surge current is I=V/ESR for ESR includes cap and contact resistance. You could consider a PTC rated for the max Joules of charge energy but that adds a minor cost and heat. Even 2.4A USB power to Apple iPad products creates a black carbon strip in the ...


1

An electric spark is an abrupt electrical discharge that occurs when a sufficiently high electric field creates an ionized, electrically conductive channel through a normally-insulating medium, often air or other gases or gas mixtures. - Courtesy Wikipedia. A spark could be caused by breaking current flow through an inductor (resulting in a high back emf ...


3

This is a very common occurrence if your load contains a large capacitor bank. The problem is that if the voltage over a capacitor changes fast (as when you connect the load to the power supply), this results in very large currents. This can be seen by the following differential equation for the capacitor current: $$I_c(t) = C \frac{dV(t)}{dt}$$ ...


0

Battery charging in buck mode battery charging in buck mode (S1: pwm, S2: off, S3: off, S4: off) No, it's like this: - In other words it's a synchronous buck regulator as per this: - Battery charging in boost mode battery charging in boost mode (S1: on, S2: off, S3: off, S4: pwm) Nearly right (you have assumed a non-synchronous boost circuit). A ...


0

Your diagram is missing any way to sense, and therefore control, current. As drawn, it a voltage source, which is not good, strung as it is between two low impedance batteries. While which battery gets charged would be determined by energy state of the batteries, and which mode of buck or boost would be selected by which side has higher voltage, the actual ...


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