# Tag Info

17

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.

7

The physical value that is important is the current that something will need to conduct, and the maximum power it might be asked to provide. Let's briefly talk about what all these abstract mathematical concepts actually represent physically: Real power ($P$) is the rate at which energy is consumed to do work of some sort (often simply in the form of ...

6

Your first circuit has a 3:1 voltage divider. Your second circuit doesn't. The op-amp input impedance is so high that the 2k resistor makes no difference. It sees the full 1.5 V of the battery.

4

Assuming the power is constant for every 15 minute interval. You get $$E[n] = P[n] * 0.25~\text{h}$$ where $P[n]$ is in kW and $E[n]$ in kWh $$Total Energy = \sum_{i=1}^n {E[n]}$$ Where n is the number of 15-minute readings in a month (n will vary from month to month)

3

Can you do it? Yes. Is is safe to do so? No! Thermostat wire is not rated for 120VAC but rather for 24VAC and also "wire sharing" is not allowed. The wire used to power something must have sufficient capacity to handle the power needs of the circuit by itself. Generally the smallest wire used is 14 AWG on a 15 A circuit.

3

250uW is the correct power in the first schematic. There is 0.5V across R1. The second schematic shows a voltage follower with 1.5V across the R1 so again the reading is correct (ideally it would be 2.25mW but the op-amp is not ideal).

3

No, for an ideal op-amp, no current flows into the terminals so for a buffer circuit there is no current flowing through the resistor, therefore no power in the resistor. In fact, there is no voltage drop across the resistor at all because there is no current flowing through it. I think you mistakenly thinking there is a virtual ground on at the inverting ...

3

How can I calculate an exact resistor value to my LEDs to get equal brightness? You can only calculate this if you have bought LEDs that have been 'binned' (selected, sorted) to particular brightness bands. Randomly bought LEDs can cover such a wide range of brightness that all you can do is adjust the resistors until they look matched.

3

The trick is not in matching resistors but in matching current, since even the same model of LED has differents Vf among parts and the resistors are computed over that. This, of course, if you have the same model of LED. If these are different (like different colours) you need to look up the curves for the light emission. If you are lucky your leds come from ...

3

Your dissipated power numbers are the "Total Device Dissipation @ TC = 25°C" from the datasheets. This is not a power you can achieve in practice. It is like saying "the Bugatti Chiron can go at 500km/h". Yes, but you need a special road for that, which is not easy to find. To achieve "Total Device Dissipation @ TC = 25°C" you ...

3

When you crank a stepper motor, it generates AC. Your capacitors look like a short circuit to AC, so all the current flows through the capacitor and none through the LED. This is also the reason it gets hard to crank. The short circuit allows a lot of current to flow. You have to crank hard to generate the high current. At a closer look, both of your ...

3

The metal plates are heatsinks for, and terminals of, the rectifier, The thing with 50 written on it is a press-in diode.

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

Look at the formula: - It's energy multiplied by frequency and that energy is the energy dissipated by the clamp circuit per switching cycle. In other words Power = energy x frequency. And, that energy is $\frac{1}{2} \cdot V_{CLAMP}\cdot I_P\cdot \Delta_t$ We also know that power x time = energy hence, Power = $\frac{1}{2} \cdot V_{CLAMP}\cdot I_P$ And ...

2

The first one is telling you that it is not likely to be practical to remove so much heat from the device even with 25C ambient (and I question whether you can rely on 25C ambient... is this in an air-conditioned room?) The second one tells you that even with Rjc of zero, you cannot dissipate enough heat for the device to survive. You would actually need to ...

2

So as you said, the gain of the 4 BJTs combined is huge and results in a severely unstable circuit, irrespective of the local compensation. Trust me, I have tried. When you simulated the circuit, you must have connected the input of the voltage regulator to a pure DC source, which has absolutely no ripple. Due to this, your op amp just has to maintain a ...

2

This can regulate V1 or V2 regardless of which one is higher, depending on how the FET's are switched. It could also regulate current with appropriate feedback. And it can regulate irrespective of power flow direction. So, for example, V1 could be a battery. V2 could be the output from a buck or boost converter. The load current could be regulated. I just ...

2

In your first circuit, there is voltage division due to the two resistors. That voltage division is 3 to 1 and that means the power is reduced in R1 by 9 times compared to it not being divided. So, if I divide the power from the op-amp (2.241 milli watts) by 9, I get 249 micro watts into R1 i.e. pretty much exactly what the first circuit is reported to ...

2

I believe you could do it with a SPST (or SPDT) relay IF you can tolerate having 3.3V appear momentarily at the 5V output. If not, you'll have to put in a diode on the 5V supply but remember that it will drop 0.7V. Edit: on closer inspection you'll need D1 to prevent the 5V supply from powering up RLY1's coil. This also prevents the 5V supply from shorting ...

2

There is no "one-size-fits-all" answer to this question. You need to contact the manufacturer and get their accelerated lifetime test data. It is likely that if you are using the device within its maximum specifications that there will be no data regarding small variations in supply voltage. Otherwise, you will need to do life testing yourself. ...

2

You are looking for a power sequencer. There are many brands that make them: Panamax, Furman, Pyle...

2

A common way to increase drive output current is made with a fairly simple modification shown below: - Resistor R is used to tie the BJT emitter to the negative rail and the output is taken directly from the emitter to drive a load connected to 0 volts. This class A amplifier will suffice in many cases but, depending on what the load actually is and what ...

2

You appear to have connected the regulator incorrectly. From the datasheet: You have 5V from the USB adapter going to pin 1. That's "Vout" according to the datsheet. Connect your yellow wire to the right most pin of the regulator instead of the left most pin.

2

In an introductory book, the examples usually are very simple to make it easy to understand a new concept, so they migh only have loads that are resistive so there is no phase difference between current or voltage so they are not mentioned. Just like the transformer is assumed to be ideal with no losses. There are no two loads as the ideal transformer does ...

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

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

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

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

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 ...

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