New answers tagged power
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It is probably working because 5V is still within the input voltage's tolerance of the component (probably a voltage regulator) interfacing with your adaptor. Without a schematic of the radio, it is hard to say what is going to happen.
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You have to match two things in a power supply:
the voltage. The voltage rating of the source and the load should be matched, for example, 5 V or 12 V;
the current. The current rating of the supply should be more than the load uses, For example a 5 V, 5 A power source will be able to supply a 5 V, 1.2 A load.
In this example the power supply is a \$5\ V\...
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Between the +5V and the -12V outputs, you have a total of 17V, definitely not what you need. Connecting the two together will damage the power supply. Not good either.
The proper way to get +7V is to use a buck regulator to regulate the +12V down to +7V with good efficiency. This is usually easier than to convert from a negative voltage.
The Alibaba web ...
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If the MCU has an ADC to measure some sensors, then you should assume the ADC ability to reject VDD trash is zero.
That is, during most-significan-bit decisions, a 100 millivolt spike into your +5 regulator will probably produce 100mV output, because LDOs just cannot work fast enough to prevent input spikes from becoming output spikes. And your ADC likely ...
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I am hoping to use 9V batteries, and realize I might have to run two or more in series?
This is how to make bipolar power supply using two batteries.
And for reverse engineering the board - check to which connector terminals goes pin4 and pin8 of the IC. Then connect batteries correspondingly +Vs to Vcc+, -Vs to Vcc- and GND to GND.
Image taken from ...
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If the voltage source has low output impedance, its output voltage will not drop a as much as it continues to supply higher levels of current, so the load will continue to try to draw a higher current. This can cause the voltage source to overheat and damage itself.
But if the voltage source has a high output impedance, it's voltage will droop as it tries ...
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The wiring of the power depends on the layout of the strips, if they are connected exactly like in the image, i would switch build a ring supply (powering both sides of the strips.
Please be aware of the following:
5V, 96A is no joke in terms of security. If the wires touch, there will be a huge arc.
5V 96A requires large wire (like 25mm2)
Each strip of 60 ...
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Here is how you can wire up the two power supplies to the single switched power entry module.
Make sure to check the actual terminal assignments for your specific power entry module as the lug locations may vary from manufacturer to manufacturer.
Also note that I have shown some double crimping at the female spade lug terminals. If this is not acceptable ...
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You can connect two power supplies from a single switch if it is rated to handle total power of both supplies.
I don't know how to connect them in a single IEC C14 fused plug switch.
Wire inputs in parallel.
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In a large power system one large generator is dedicated as master and responsible for the frequency control and is governed(synchronised) to run at the system frequency.
Loads and generators are now added to the system. The load supplied by each generator is determined by the power source (steam turbine.) The whole systems runs at the same frequency.
If ...
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Not mentioned elsewhere it is even quite possible that two different devices are powered over longer cables in the same building and then could share grounding through some common signalling cables.
The use of small diameter wires for POE and 48V drive implies that there will be voltage drops across the cables. Such drop is proportional to the current draw ...
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How to connect two power plant (11 KV)located at 15 KM away from each other in parallel?
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Rather spend your time understanding the Laws of Thermodynamics.
If they could patent the stuff they would. One cannot patent a perpetual motion device without a working model.
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I don't have the spec in front of me but I've done a number of PoE designs. As I recall the PoE isolation spec is to meet safety requirements and avoid ground loops. If your device takes the 48 Vdc and uses it within a fully enclosed, non-conductive enclosure, with no other cables, then the 1500 Vac power isolation (typically done with a flyback circuit) isn'...
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To make it simple...
With pure resistive loads, the current being in phase with the voltage (V & I being +ve or -ve simultaneously), real power would be (V * I) Watts.
With pure inductive loads, the current lagging the voltage by 90 degrees, at the instant of peak voltage the current would be at zero and vice versa. The power (V * I) would alternate ...
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If you want to call it PoE, you need to stick to IEEE 802.3af-2003 or IEEE 802.3at-2009.
These standards call for isolation. This answer could end here: no, you cannot build a PoE device without isolation.
This is also assuming that the data lines are already isolated, so the only potentials without isolation is the power lines.
You also will have a ...
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The current shunt of an electricity meter is in the live conductor. This means the low voltage circuit reference is the live conductor.
The voltage circuit of the meter is as you explain connected between the live and neutral circuits, with the volt drop resistor connected to the neutral circuit.
All shunt electricity meters use this principal. They are ...
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Have a look at this slideshow explaining AC power and energy from first principles.
(Use Acrobat instead of Google viewer to display the animations.)
In concrete terms reactive power does not exist. Only active- and apparent- powers exist. Maybe the best way to explain reactive power, \$Q\$ is:
\$\hspace{4cm}Q= \sqrt{V_{rms}^2I_{rms}^2-P^2}\$
This means ...
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Your diagrams are a little confusing. The effort is appreciated, but the critical bit that's missing is how the fuse blocks are wired.
Consider the following two schematics:
simulate this circuit – Schematic created using CircuitLab
The only difference is the fuse position. From the perspective of the power supply (what you call "live") the fuse is ...
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Since I don't know what you know I will explain several related concepts related to single phase sine wave electricity.
The short answer is that the imaginary part of the power arises from the phase shift between the voltage and the current. If they are in phase, the imaginary part is zero. If they are 90 degrees out of phase, the real part is zero. In ...
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I quess some images can be useful. The wanted formula is at the end.
V1 is 100V peak sinusoidal voltage f=50Hz. There's 100 Ohm resistor R1 and 0.5H inductor L1 as load. The load current is about 530mA peak and its phase lags from V1 as it should be with inductive load.
In the next image the red curve is the momentary power from the source to the load i.e. ...
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When energy goes "back and forth" it ends up in different places (magnetic vs. electric fields in inductor/capacitors, or kinetic vs. potential energy in pendulums and springs, etc.) at different times.
Phase is a measurement of how different the times are between the different places (with respect to some frequency or rate of the back-and-forth oscillation)...
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A mathematician would tell you that with complex numbers, the imaginary part is a way to track a +/-90 degree component. Or in correct parlance a QUADRATURE component. Caveat: this works when dealing with a single complex conjugate (ignoring the other one) which is a trick for using the imaginary component to tell you what happens in the real world where ...
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If you have a simple circuit with an AC voltage source, a resistor, and a capacitor in series then the 'apparent' power will be composed of a 'true' part and a 'reactive' part.
The true power is consumed by the resistor and dissipated by heating it up.
The reactive power is alternately absorbed by the capacitor to charge it up, and then contributed back ...
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The imaginary part is energy that is bouncing back and forth between reactive components without being dissipated. Essentially, you can think of it as energy that gets stuck in the wiring due to the existence of capacitance and inductance.
Now to your actual question:
I am trying to understand how the mathematical expression Imaginary(s) is what you just ...
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Reactive power is when X amount of REAL power flows to the load, does no work, and then X amount of REAL power flows back to the source. The reactive power in that case is X.
That means that in order to view reactive power you have to observe over some time span, since if you observe a sufficiently short time span all you will see is real power flowing one ...
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For 2., the lower inductance means lower loss.
There are two main sources of losses in high power transmission lines. One is the I^2*R (resistive) loss, and the other is the loss caused by the self inductance of the transmission line. Hence reducing the inductance lowers the loss.
Note - eliminating the inductive losses is a big advantage of DC ...
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Any time you transmit/transfer/send power (through wires in this case), not only is your transmitted power not all received/used at the end, but the voltage is reduced as well. So you have both a power loss and a voltage loss.
In order to transfer a given amount of power (let's say 100W), you can use any combination of voltage and current to get that much ...
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A simple example may help.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) Direct transmission at LV. (b) HV transmission.
In Figure 1a we drop 5 V on the feed and return wires giving a 10 V drop in voltage at the load. (These are all just rough numbers. The real voltage drop will be a little less because the current will be ...
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Ignoring the difference between reactive power and resistive power for a minute...
Power P = V x A
So power in a load is the product of the voltage across it and the current flowing through it.
So if you want to dissipate 1 W in a resistor, you could deliver it in the following ways:
Apply 1 V to a 1 R resistor, drawing 1.00 A
Apply 5 V ...
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Because the losses are due to the current squared times resistance.
So as the voltage increase, so the current reduces for the same resistance.
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This figure is used to compare power efficiency of different processors. The lower the number, the better is power efficiency.
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In a few papers I found a figure of merith called power density and defined as Pdens = uW/MHz, so basically it is energy.
"Power density" in this case is just energy required to do one cycle of computation. If you multiplied by how much work per cycle the processor did, you would have computational efficiency.
The thing is I don't understand what it ...
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The higher the frequency, the more the power dissipation will be. All else equal, a processor that consumes 10 W at 25 MHz will need 20 W at 50 MHz*.
The power density (I'm not familiar with this name for it, but I'll assume you have it correct) is roughly constant for a given processor, so if you have a power density of 0.4 W/MHz, you can predict how much ...
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For the generator to deliver 50 Hz, it would need to operate at 5/6ths or 83.3% of the original speed. To deliver the same power, the torque would need to increase by 20%. The prime mover may or may not be capable of that. In addition, the generator excitation would need to be increased to supply the same voltage at a slower speed. The generator may or may ...
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A solar panel is, fundamentally, a photodiode*. If you forward-bias that diode, current will flow through it.
To elaborate a bit: You may have used photodiodes for circuitry that has to respond to light. When used for light sensors, they're generally reverse-biased and operated in the photoconductive mode, but they can also be used in the photovoltaic mode, ...
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If your schematic is correct then you are using the DA2032 part. From the datasheet it appears to be a 3A part.
Taking a look at the datasheet of the LT3757: they recommend taking 20% of the specified current sense (100 mV minimum).
On top of that the RC filter at the sense input lowers this sense voltage more.
Based on this: you should work with 20% of CS ...
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Your friend is right, and so are you.
Power is the derivative of Energy over time. Since the field energy in your capacitor changes over time, there's obviously power needed to effect that.
However, when you look at what happens over a full period (or multiples thereof), the capacitor stores as much energy as it releases; the average (over an arbitrarily ...
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Perhaps it's this?
simulate this circuit – Schematic created using CircuitLab
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simulate this circuit – Schematic created using CircuitLab
EDIT:
You get oscillations, which is the correct scenario. Look this video, the turbine valve has to move very slowly, additional deflector is used for fast rpm control, not the valve , it would tear off the entire mountain.
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A critical piece of information is missing here: battery voltage. Though it is probably 12 V lead-acid, you should state it clearly just to be sure.
Assuming this is correct, then you have approximately 173 Wh of capacity. Your router presents a 6 W load and your monitor a 24.7 W load, so a total of 30.7 W. Dividing one by the other, you would get about 5.6 ...
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Not sure about your "the USB-C wall adapter", but if it is compliant to USB Type-C specifications, you will NOT have any power. Per Type-C specifications a Type-C port will NOT output the VBUS until a cable-device connection is made. If there is no Rd on consumer side, the port (charger) will not sense the connection and will not power the link.
More, not ...
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As far as wire specs go, you can use any "magnet wire" catalog as a reference. I like https://prod.essexwire.com/sites/essexwire.com/files/2017-08/Essex-Wire-Engineering-Data-Handbook-EN.pdf for general wire physical properties, but you will need to check each of the many insulation materials to determine the breakdown voltage for a specific insulation ...
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If this is just a one or two layer PWB, then it really doesn't matter which approach you take.
Any interference/EMI is going to come in broadside to the board (probably), and the power and return (ground) traces are not going to do anything to protect the signal trace, so matter how they are placed relative to the signal trace.
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Except in the most extremely extreme situations paint or similar is NEVER used for insulation.
Any conductive case, chassis, panel, handle, other metal thing, ...
would normally be grounded.
NOT being grounded is liable to be a shock hazard and possibly a noise source.
Any such parts which accidentally become electrified during operation should activate ...
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Paint is not a reliable barrier for safety. To meet UL and other types of approvals, the line side is isolated from the low-voltage side by reinforced insulation. This means physical distance and / or a rated insulating material to prevent contact between line and chassis.
When a power supply has a primary-side chassis connection, it will make this only ...
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Its not a question of „who should be next to who“.
If you want a design that is robust concerning EMI (I assume that‘s what you‘re after), then you have to defeat 3 enemies:
1) inductive coupling (loop to loop). Current only flow in loops and loops carrying current affect each other. So you must figure out all loops and find a way not to interfere. ...
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If you have an impedance with a negative real part, it delivers power, yes. But remember: no actual passive component has a negative value! To build that negative resistance, you need active circuitry that do consume dc power. The very fact you mentioned (infinite power loop) is the concept behind designing oscillators. They consume dc power to be able to ...
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If your watt meter is crappy, it won't take into account power factor (but then I wouldn't really call it a wattmeter anymore since the whole point of a wattmeter is to see through the power factor). I kinda doubt your drugstore humidifer is drawing 640W.
You can't calibrate it. Voltage and current have to be in phase for real work to be done. When they ...
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Will I then risk something by touching/soldering the wires after shutting it down again?
Hm, the output side of your supply should generally be safe for touching, because it's low voltage.
But:
When cutting multiple wires at once, you might create a short on the output, which might discharged a charged capacitor in very short time. That can damage tools (...
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