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1

So my question is when we have a transformer (delta-wye) for example 220kV/110kV, so primary voltage is 220kV, but what voltage is it, voltage A to ground(VAG in video) or phase A (VAB)? The default voltage specified for a three phase system is line voltage irrespective of the type of transformer that connects to it. So, if it is specified as 11 kV then ...


1

Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab When analyzing a transistor we need to use the following relations: $$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$ Transistor gain \$\beta\$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$ Using KCL, we can write: $$ \begin{...


2

Simply use something like HCPL3700, which is basically your circuit in one package. You only need one input resistor and bypass cap to make it work. Note, that either your circuit or HCPL3700-based one are not actually "isolated", but rather "optocoupled", because you still have to connect wires to mains somehow, so full voltage is present at least on part ...


4

Half-wave rectifiers are mostly a relict from electron tube times. Making circuits with electron tubes is tricky as the cathode and anode are not symetric ; the cathode has a heating circuit which needs to be a cathode potential. For implementing a bridge rectifier, you needed seperate heating windings on the transformer for the high-side diodes. Also, ...


8

The 4 diode bridge is a full-wave voltage doubler and achieves 200% of the voltage and 50% of the current of the centre-tapped 2 diode full wave version. Thus the tapped version has lower impedance. Some users might consider the tapped 2 diode version as a 2 phase each half wave combined to make a full wave. But in fact, the secondary is only a split-...


1

But what would seem important to me for that would be to know the maximum absolute value of the instantaneous power, which would be the apparent power plus the active power (S+P, not P+Q). Is that correct? It is rarely if ever important to know the maximum absolute value of the instantaneous power. That would be an instantaneous value. When power is a key ...


4

The half wave rectifier uses one diode, does not need center tapped transformer. The full wave one uses two diodes and needs center tapped transformer. Efficiency for both cases are the same. At the same load with the same output filter design, full wave rectifier has less ripple. From another perspective, with the same ripple requirement, filter design is ...


4

With a full-wave bridge rectifier you have four diodes, resulting in two diode drops, and the capacitor value is roughly half what you need for a half-wave rectifier with the same ripple. If you have a center-tapped transformer you can use two diodes to get full-wave rectification with only a single diode drop, but the extra transformer tap and associated ...


1

A slightly off-beat approach: Figure 1. Intersection of the curves. Because they're sinewaves we can use some trig functions. At (1) the blue phase is at 0. Let's call this 0°. At (2) the blue phase is at 2400. That will be 90°. At (3) and (4), the intersection point with the other phases, the voltage is 1200 which is half the peak. We know that \$...


0

The easy way would be to put your vertical grid lines at 60 or 120 degree spacing. Then you'll see that the rising-edge zero-crossings of phase 2 are 120 degrees after phase 1, and the zero-crossing of phase 3 is 120 degrees after that.


1

Plexiglass is not a well defined term, 2mm of certain exactly defined and approved material such as polycarbonate is ok. I guess you mean polymethylmethacrylate (=PMMA) when you say plexiglass, not some resembling trade mark. It can be used as well in this place. Warnings 1) There should be also certain minimum route length along the surface of the ...


1

I need to regulate it to 1to 4ma maximum output without having much voltage drop... Can I just use a resistor in series? No. A resistor works linearly according to Ohm's Law. Resistance = Voltage / Current, so for a maximum (short circuit) current of 4 mA the resistance needs to be 12 / 0.004 = 3000Ω. However at 1 mA the resistor would drop 3000 * 0....


3

The short answer is "no." A capacitor is, fundamentally, an energy storage device. The energy density (joules per liter) of a particular kind of capacitor (film, electrolytic, whatever) is roughly constant, whether a particular module is optimized for high voltage/low capacitance or for high capacitance/low voltage. Transformers let us trade voltage for ...


2

Yes you can Place the cap bank on the end of a transformer or even an auto transformer .Remember that the windings must handle the leading reactive capacitor current .So if your Cap bank is rated for say 100KVAr then 100KW is your required transformer rating .This at power frequencies looks expensive .I have not seen this done .I have seen tertiery windings ...


1

There will be several of these detector circuit in a single pcb, the number will range from maybe 5-10 This means that you have to route mains voltage from several places to a single detection circuit - really bad idea! It makes more sense to have a tiny PCB that can fit next to each individual switch with low voltage signal going to the common location. ...


3

Low consumer quality boards are done like this. Since dust and moisture accumulation causes greatly reduced breakdown voltages down from 1kV/mm, it is far better to have an air gap milled in the board between primary and secondary of the Opto. However , your external schematic is nonsense with Line Neutral and switch going to LED.


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When Capacitor Reactive Impedance, Xc(f) rises to affect impedance ratio with R such that they are equal, you can measure many responses; the voltage amplitude divider ratio (gain) reduces times \$\dfrac{1}{\sqrt(1+1)} = 0.707 = -3~dB\$ the current phase shift changes by 45 deg, (=trig. angle with equal sides of impedance) which as already started to ...


1

You're creating a high pass filter between your capacitor and whatever resistance you're connecting to Vref with. In the first circuit, I can't see the impedance to Vref, but it's apparently not zero. In the second circuit, it's the output impedance of U_2; again, it's unknown to me. I might be able to estimate them if I knew the frequency of your test ...


2

You are passing what appears to be 60 Hz through a 100 nF capacitor. That capacitor will have an impedance of: - $$|X_C| = \dfrac{1}{2\pi f C}$$ Plug in the numbers and you get 26.5 kohm so, if you need very little attenuation of the AC signal the resistor value needs to be many, many times the value of 26.5 kohm. In your last scenario the equivalent ...


1

IN GENERAL the whole approach of setting \$D=E\$  is incorrect. It fails e.g. in the quite obvious example of \$A=D\$ and \$B=E\$. $$$$ The approach is only valid when D is constant and E is variable (or vice versa) the variable can become very small or very big with respect to constant $$$$ Let D be the constant and E the variable (think of E being ...


1

I was told that I must set \$D=E\$ No, because that would alter the magnitude of the input voltage \$v_1\$ and, unless the output magnitude were altered by the same amount then the ratio result would be wrong. Stick with finding the magnitude of the input and output signals as seperate entities then make the division.


2

Recall that displacement \$d\$ is the area under the velocity curve. For a sinusoidal drift velocity \$v_d\$ having radian frequency \$\omega=2\pi f\$ where \$f=60\,\text{Hz}\$, the magnitude of maximum displacement over one half cycle can be calculated as the integral of \$v_d\$ with respect to time, during the time interval \$(0 \le t \le \pi/\omega)\,\...


0

I just saw the answer to this question by user freecharly. They claim that the mean drift velocity of electrons is $$v = \dfrac{j}{n e},$$ where $$j = I/A$$ is the current density for current \$I\$ and cross-sectional area \$A\$, and the electron density is \$n\$. If I'm not mistaken, applying this gives us $$\dfrac{\frac{3.02 \times 10^4 \text{A}}{\...


1

This means the voltage drop across the equivalent resistance must be \$i_{c}R′_{L}\$ No, you forgot that at DC, the transformer primary has a DC resistance of circa 0 ohms and this is in parallel with \$R'_L\$. For AC signals of sufficient frequency that the primary inductance can be ignored (i.e. is much bigger in magnitude than \$R'_L\$), the AC ...


0

DC doesn't cause substantial voltage drop in the transformer, if it's properly designed. In idle state the voltage between the collector of the transistor and GND is =Vcc or more precisely Vcc - Ico x Rt where Ico is the idle state collector current and Rt is the resistance of the transformer primary. The signal makes Ic swing around Ico. Voltage between ...


1

You specify AC mains operation and a 5A sensor. This answer focuses mainly (but not solely) on the consequences of the internal shunt in the sensor IC failing open circuit (as can happen). My discussion below assumes potential fault currents far in excess of 5A, or indeed, in excess of the continuous 65A rated current of the ACS723. I've also mentioned ...


0

A simpler and perhaps more intuitive explanation... As the input moves up and down, the capacitor will charge and discharge via the resistor. If the input to the resistor moves up and down quickly (high frequency) then the capacitor will not have chance to charge very far and the output will be small. If the input to the resistor moves up and down slowly (...


0

You can look at this type of a filter as voltage dividers. $$ V_{OUT} = I*X_{C1} = V_{IN} * \frac{X_{C1}}{R_1 + X_{C1}}= $$ This circuit should be treated as a voltage divider where the role of the variable resistance is taken by the capacitor. At low frequencies the capacitor has a high reactance (\$X_C\$), so almost the entire input voltage is ...


0

So why does this need to be AC? Three-phase AC current makes the magnetic field rotate. Basically, what causes the induction? The rotating stator field induces AC voltage across the rotor conductors resulting in AC rotor currents. Those currents produce a magnetic field that rotates with respect to the rotor. The above is the intuitive explanation. The ...


2

Note: I am not trained in electrical safety. Do not rely on this post for safety advice. I know you are asking theoretically, but I feel like I should put this disclaimer here anyway. Is it because the waves may not be synchronized and could then cancel out each other? Or something else? That is the main reason, but the consequences are worse than just ...


0

Your premise is incorrect for experts and only applies to newbs. When you apply voltage sources in parallel, the impedance, voltage and phase must be known before connection to know the resulting power being transferred , if it will be stable, safe and performs expected task. This is why an experienced Tech. or Eng. should be consulted. This is done all ...


1

Danger to Wires With this type of setup, while your intention may be that it would only ever be used with "one cable's worth of electricity", there is no guarantee. So someone could use the exact same configuration to get double power - e.g., 20A from circuit 1 and 20A from circuit 2 in order to have 40A total. There are a bunch of problems with this. In ...


3

When two DC sources are paralleled, blocking diodes prevent back feeding from one supply to the other. The supply with the highest voltage will always supply current without being affected by the other. There is no comparably simple and effective provision for paralleling two AC supplies. Two AC supplies from the utility can be obtained with matching phases ...


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