# Tag Info

## Hot answers tagged power

18

An unloaded power transformer is a pure inductance. As you increase the load, the power factor improves. So, take a Variac, plug an incandescent light into it, and you should be able to dial up any power factor you like.

8

The cheaper LED lamps - the ones that use a capacitive dropper rather than a switch-mode power supply - can have a very poor power factor. The power factor could be as low as 0.2. Buy a bunch of them, and wire them in parallel to get a significant current draw.

5

Take a look at figure 6 on that page (I've added the red lines): - So, when the input supply voltage is constant and the output voltage and load current is constant (i.e. the circuit is operating in "steady state"), the two current changes (during inductor charge and discharge) are equal. This mode of operation is called continuous conduction ...

4

Here, "steady state" means periodic steady state. In other words, all node voltages and branches currents have waveforms of the form f(t) = f(t+kT), where k is an integer and T is the period. During the periodic steady state, the (ideal) inductor current must increase linearly when a positive voltage is being applied and decrease linearly when a ...

3

In this context, steady state means that the circuit has completed the start up phase, and if left undisturbed, it will continue to operate in the future as it does now. In other words, the behavior of the circuit will look the same now, 10 seconds from now, or 10 hours from now. Of course, a circuit will not in reality continue to operate the same way ...

3

You can buy inductors with a wide range of current and inductance ratings. An unloaded induction motor has a low power factor. If you use a single-phase motor, you can probably disconnect the capacitor if it has one. Unloaded, it should start with a little manual twist of the shaft. You may have difficulty with the distorted magnetizing current waveform if ...

2

Poor power factor in practice is usually inductance-based. On the other hand, inductors for testing purposes look rather bulky and expensive. If you are OK with capacitive load, the capacitors for asynchronous motors come in various capacitances and are rated above mains voltage. They are also rather cheap. You can combine them with incandescent lamps, space ...

2

In general, you can run this at 37V input and 380mA. This would still comply with the absolute maximum ratings where neither a maximum current nor maximum power dissipation are specified directly. But: as you might have seen, this device has some protection features, like short circuit and overtemperature protection. Those implicitly limit both the current (&...

2

How much current it can pass is one property, and how much it can heat up by the power dissipated is another property. You can input 5V and make 380mA 3.3V, and the regulator would heat up only at 0.65W, which is certainly doable. If you input 37V, and output output 3.3V at 380mA, the poor chip has to dissipate 12.8W, which is completely absurd amount of ...

2

Power = Voltage * Current I'm assuming that you have a continuous forward current of 400 mA. Note that these diodes are very small so they heat up very quickly. So even if the current is 400 mA for a short time, that would need to be really short like less than 1 second if you wait a long time (many minutes) before doing that again. Now look up the forward ...

2

It's not that 'a lot of commercial transformers like sinusoid'. It's simply that every transformer has issues (usually heating or vibration) with frequencies over the one it is designed for. So a 50Hz transformer would have to handle 100Hz, 150Hz, 200Hz and so on. The EN 61xxx standard (in EU, the US would probably have some FCC related stuff) mandate a ...

2

I'm confused though as on the datasheet there is no voltage mentioned, only a max phase current of 2.1A and same holding torque. But on the line right under the amps/phase of 2.1A, there's a line that says the coil resistance is $1.6\Omega$. $2.1\mathrm A \cdot 1.6\Omega = 3.36 \mathrm V$, so there you go. I know that voltage applied to a stepper, ...

2

In general, for a circuit where the output is switched on and off, how can we analyze the power factor? In general, the power factor in such a circuit is not the power factor defined as the cosine of the angle between the current and voltage but as the real power divided by the apparent power. In this case, the apparent power is the total RMS voltage ...

1

However when the micro-controller are being manufactured what determines their operational voltage? All ICs are fabricated in a certain manufacturing process. That process is the recipe to make the transistors and other components. That process together with the design of the transistors (large or small) also determines what the maximum allowed voltages are ...

1

Both cases are protected against reverse voltage. That's what Q1 is there for. That will not change just by swapping Q3. The n-mosfet is definitely more suited for your purpose, since the bipolar transistor requires some current to be polarized, if you want to have low quiescent currents, then the mosfet is the way to go.

1

You can replace it with an MOS transistors. No problem. The BJT had an internal resistive divider to increase the threshold where the switch "flips". I am wondering why was it implemented inside the BJT. The transistor is less versatile this way. But I also not see any reason why a BJT should be used here. The current it has to supply is very low, ...

1

An isolation transformer is just a normal transformer that has the same input voltage as output voltage. It is also usually fused or has a circuit breaker. You can make one by putting two transformers back-to-back such as 120:24 and 24:120. It's not usually a great thing to do unless your current requirements are modest and you have no other option. Good ...

1

All torque is proportional to current BEMF reduces applied V from velocity so holding torque reduces slightly with rising step rates. Voltage affects current from I= V/DCR(=1.6ohm) V=L(=3mH)dI/dt or I as the integral of voltage.

1

But why do some devices have a current limit and some a power limit? The current limit will be determined by the maximum allowable current density in the bonding wires and internal components of the chip. As current density increases so will heating and there has to be a physical limit to that. The power limit will also be due to thermal considerations but ...

1

You don't provide any details about your FTDI board. But if an USB to serial converter has a 5V pin, there is no reason to assume it is a power input. The 5V pin would be expected to give out the 5V USB supply provided by the PC. When that ia connected to the 5V node on the Nucleo it will power it up. Same way like connecting it directly to USB from the ...

1

"Off the shelf" requests for specific devices make for shopping questions, which are forbidden, but it's ok to ask what the name of such a device is. In this case, "Power Factor Load Bank", "Resistive/Reactive Load Bank" or separate resistive and reactive banks will work for you. They look expensive. A large inductor has a ...

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