# Tag Info

31

When you're looking at an AC source in isolation such as in your question, indeed there's no polarity and you can connect the wires either way round. When combining two or more AC sources in series or parallel, the relative phasing is very important.

16

Clearing up some misunderstandings will probably help. First, your data is a 16-bit value. There's no "overflows" and "actual data" -- the 16 bits are just divided into two 8-bit pieces (bytes). To get the right binary value, you need to concatenate the bytes. In C, you can do it by starting with unsigned values and using bitwise operators, like this: ...

15

As long as you are dealing with a closed system (like a transformer secondary winding + bridge rectifier) then no, AC is not polarized. However, when dealing with outside power (like what comes out of the wall) we do consider AC-carrying cables to be "polarized". There's the hot wire carrying the juice and the neutral wire carrying the return. The hot wire ...

14

You need to know the power factor, which is highly depedant on your application. Then you simply use this formula: $$P = S * PF$$ where P = actual energy S = apparent energy PF= Power Factor This answer tells us, that power factor in general public households can be expected to be greater than 0.9.

12

You are confusing amp-hours and amps. A 1.3 Ah cell can supply a LOT more than 1.3A - try 13A or even more if the screw is stuck - albeit for a lot less than a whole hour! In this case, theoretically 6 minutes (0.1 hour) but practically, probably a bit less, depending on the condition of the battery, the temperature and so on. So the problem isn't that the ...

10

If you have a 100W electrical load and you drive 100W plus efficiency losses, say 110W, into the generator, things will be in a state of equilibrium, with 100W being converted from mechanical input power into electricity, and the other 10W of mechanical input power being eaten up by losses. Now suddenly put 1kW of mechanical power into the machine; at that ...

9

"10 amps" and "2.16 kilowatt-hours" are not the same units and can't be directly converted. An amp is current, a kWh is energy. However, "10 amps at 240 volts for 1 month" is energy. Google will happily convert the units, if you ask it "(10 amps * 240 volts * 1 month) in kwh": 1,753 kWh. Coming from the other direction, since kWh is energy, it doesn't ...

9

I'm understanding this question to mean that you're trying to run a three-phase motor off a single-phase line. If you're trying to run the motor directly off the AC line, the phase angles involved will make it difficult to get the motor started, which is part of the reason three-phase exists in the first place. Single-phase motors usually have motor start ...

9

Answers given will either not work in the real world versions of what you describe or are far from lowest cost. Steven's LM2623 datasheet is a reasonable choice and will start on 1.1V and run on 0.9V but the IC costs about 60 cents. If you genuinely want lowest cost then a properly engineered version of the Joule Thief is a good candidate. I use that name ...

9

Energy must be conserved. You cannot get out more than you put in. Assuming you are having a constant power draw, the power in is going to equal the power out. Therefore, [Amps in] x [Volts in] = [Amps out] x [Volts out] So, if you are stepping up 12V 100A to 100V, you are going to get 12A (less with losses) out.

9

Assuming no losses, yes. 120V x 10A is 1200W. This could be converted to 12V x 100A, or 24V x 50A, or any other similar combination. However, losses do enter into it, as do maximum current ratings, etc. Also, you can only use a transformer with AC. If you want to convert DC it becomes a more complicated problem.

8

The power fed into the generator equals the power taken from the generator. The latter includes heat or mechanical energy (e.g. vibration). So any additional power is converted to heat or mechanical energy which may lead to destruction of the generator. But consider, that it is not so easy to put more energy into the generator, because if there is not ...

8

Yes, devices of the same voltage can be run from the same supply. For this you would need a (preferably decent quality regulated) 12V supply that is capable of sourcing more than the total (max) current draw of all your devices combined. If you have other devices at e.g. 9V you would need a supply with a 9V and 12V output. The required info re current ...

8

Not aware of any tool that will automatically do the conversion. You could do it by hand or write your own script. Here is a list of common SystemVerilog to Verilog-2001 (or vice-versa) Easy conversions: always_comb --> always @* always_latch --> always @*, may want to add a synthesis directive for latch always_ff --> always int --> integer or reg signed [...

8

Your PSU seems to be a common one and the necessary pins were determined by members of an external forum devoted to radio control enthusiasts found here. The solution seems to have been discovered by trial and error, and the poster claims that shorting pins 31 and 34 (which should be labelled) allows 12V to flow to the rail. Server power supplies don't tend ...

8

A simple model to show how an ADC behaves is this: simulate this circuit – Schematic created using CircuitLab There, you have a source ($V_{\text{Analog}}$) and its impedance ($R_{\text{Source}}$). As well as the internal model for an ADC with a switch which is open and close every sampling period, and the ADC's impedance which is a combination ...

8

An analog signal that is sampled is called a discrete-time signal. An analog signal that is quantized is a discrete-valued signal. A signal that is discrete in time and amplitude is called a digital signal. ￼

7

Sounds like the following string.toCharArray(buf, len) is what you want. as exampled String stringOne = "123456789"; char charBuf[50]; stringOne.toCharArray(charBuf, 50) ; Serial.println(stringOne); for (int i = 0; i < 9; i++) { Serial.print(" 0x"); Serial.print(charBuf[i], HEX); } Serial.println(); yields: 123456789 0x31 0x32 0x33 0x34 0x35 0x36 ...

7

From the datasheet: The table in the datasheet specifies the module resistance to be 1.9 to 2.3 Ohms. it gives a ball park figure. I say ball park figure because the current cannot be predicted exactly, as it depends on the temperature difference between the plates too. The graph assumes one has constant current source. In OP case, i assume it is constant ...

6

The Dash Button The [Amazon] Dash Button operates at 3.3 V boosted from the battery’s nominal 1.5 V, drawing 200–300 mA from the battery when on and 2.3 μA when in sleep. This means the ~1200 mAh battery should be able to power the device for at least four hours while on and decades while in sleep. Since the button is only on for a few seconds when ...

6

There are two possible answers to this as a processor can be bigendian meaning that the byte with the lowest address is the most significant or littlendian meaning that this byte is the least significant. I'll assume the the bytes are stored in an array unsigned char a[4]; and we want the result in unsigned long retval; If the system is bigendian ...

6

Call the base "N". Then write the equation in base conversion form: The first one being: 4*N + 2 - 3 = 3*N + 6 Then solve for N In general, digits in the second column (from the right) are multiplied by N, digits in the third column would be multiplied by $N^2$, digits in the fourth column would be multiplied by $N^3$, etc.

6

Since your microcontroller doesn't support floating point, then you need to scale the numbers up and then divide back down. To do so, you will need to use long (32-bit) arithmetic. For the ADC, 4 ma = 0 and 20 ma = 16384. Therefore the difference 16 ma is also represented by 16384. If we take the full scale reading of the bar graph, 25, and divide by ...

6

While experimenting a satisfactory solution was found. (1) Connect track 31 with 34 (to boot) (2) Install a resistor between track 32 and ground see http://pa0fri.home.xs4all.nl/Diversen/DPS-800GB%20A%20Server%20sypply/DPS-800GB%20A%20PSU%20eng.htm

6

Apart from noticing that maybe the integral in the average inductor current at the end of the second time period should (maybe?) run from D1 to D2 rather than 0 to D2 (but I might be wrong), I have an observation ... ... this sort of algebra-heavy approach may be correct, but I don't think it's useful. My approach, which may be too approximate for some ...

6

First convert it to binary: 0001 1011 0000 1000 1100 0010 0010 1011 Then map it to its fields: 0 00110110 00010001100001000101011 S exponent -------mantissa-------- $S=0$ so the value is positive. The exponent is stored in excess-127 notation, which simply means they wanted negative magnitudes to sort smaller than positive magnitudes when the sort ...

6

If something uses N kWh in a 30-day month, has an average consumption of $$N \frac{1000\text{ W/kW}}{720\text{ h/month}}\text{ watts}$$ In other words, to convert "energy" to "power", divide by "time". Just as an example, suppose you use a 1500 W toaster for 3 minutes to make toast for breakfast. That would be a total of 1500\text{ W}\frac{3\text{ min}}...

6

If energy consumption in one month is 305kWh the mean (i.e. average) power consumption during that month is $\frac{305kWh}{30 \times 24h}=\frac{305000}{720}W\approx 423W$ (assuming a 30 day month).

6

The symbol represents a common mode choke. They look like these: (Source) There are SMD versions as well. They are used to block common mode noise i.e. noise incident on both lines of power supply. In the picture above (Source), you can see that common mode noise is coming on both lines in the same direction. At the same time, power is also coming on ...

5

This is actually a very common and legitimate desire, for example to run quasi-industrial machine tools equipped with 3-phase motors in a home workshop. Three approaches are generally available: use capacitors to accomplish a phase shift to manufacture at least enough sense of a third phase to tell the motor in which direction you wish it to start. This ...

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