Comment summary #40
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Comment 391 ... 400
| date | article | author | comment |
| 09 apr 2015 14:53:02 | Arduino wattmeter | Zak | How would you modify the voltage divider circuit above if you wanted to measure 3 phase voltage? Would you use a ground from the source and connect that to the analog ground? |
| 06 apr 2015 23:56:41 | Magnetic permeability | Freddy | There is something odd with your hysteresis curve. For a cgs emu unit system I would expect a Gauss vs Oersted graph. Another thing that doesn't add up is the given symbol and unit: M (emu), with a capital character M, it is the symbol for magnetization with the unit emu/cm^3. The conversion to SI is: 1 emu/cm^3 = 10^3 A/m. On the other hand the unit emu is associated with magnetic moment with the symbol m (small character). The conversion to SI is: 1 emu = 10^-3 J/T. The conversion for the field strength H is: 1 Oe = 10^3/(4*π). When using SI units the permeability can be calculated as: μ = B/H. To calculate permeability for a specific area in the hysteresis curve, the difference in field strength and magnetic flux of a small portion must be taken from the curve. See also the article Magnetic hysteresis |
| 06 apr 2015 18:35:52 | Magnetic permeability | Mak | I have a hysteresis i.e. [M (emu) vs Hc (Oe)] curve of nickel filled thermoplastic and want to calculate the magnetic permeability. How I can get from hysteresis curve? |
| 19 maa 2015 20:52:38 | Shunt resistors | Freddy | Maybe you will find your answer in the article diode measurements . There is a chapter that explains the measurements of diode leakage currents. |
| 19 maa 2015 12:02:16 | Shunt resistors | Jay | How can i use a dual measurement DMM to measure V-I characteristic of an power SCR to find leakage current? |
| 11 maa 2015 00:31:50 | Arduino wattmeter | Freddy | It is not the harmonics that are causing troubles. If you look at the effective voltages on the primary and secondary side, the transformation ratio will change with the applied voltage. Especially when the voltage is around the working voltage. To avoid this kind of problems you need a transformer with a higher primary voltage. Or you take two similar 230 V / 6 V transformers and connect the primary is series and connect the secondary in series. In this way you get an 460 V / 12 V transformer. You have to do this only if you want to make high accuracy measurements. If you accept a certain amount of deviation you can use a single 230 V / 12 V transformer. |
| 11 maa 2015 00:10:48 | Arduino wattmeter | salim | so i ll use transformer of 240v/12v then after that i ll use the dividers as they are in the circuit above. what do u suggest to avoid harmonics?? knowing that the freq of our country is 50 Hz so to prevent my circuit against harmonics i ve used low pass filter with cut of freq of 80 Hz (is it correct??). |
| 10 maa 2015 19:25:33 | Arduino wattmeter | Freddy | Yes you can use transformers to isolate the voltage and current inputs. But, keep the voltage dividers in place. These act also as input protection for the Arduino. Be aware that you can measure only AC voltages and current when transformers are applied. And another thing: Normal mains transformers balance on the edge of saturation. So, they will not behave as linear as you expect. |
| 10 maa 2015 19:00:55 | Arduino wattmeter | salim | thanks for those information . i have more questions. if instead of using voltage divider i ll use a potential transformer to step down the voltage, and a current transformer to step it down ? |
| 04 maa 2015 00:19:01 | Arduino wattmeter | Freddy | You didn't mention if the 220 V is AC or DC. I assume it's a AC mains voltage. Be careful with this kind of voltages. You can easily hurt yourself or damage a computer connected to the Arduino. Read the safety section. The chapter "Voltage & current range" explains how the resistor R1 and R3 are calculated. If the voltage is 220 VAC, the top voltage is sqrt2 higher: 311 V. Because voltage are subject to fluctuations and voltage spikes can occur, It is recommendated to make the voltage range 400 V. The resistance R1 will be according to the equation 7.26 MΩ. A practical value is 8.2 MΩ. Assuming the current is a sinewave, the top value is 45 A. With a little room a 55 A will be suited. The resistor R3 will be according to the equation 0.001 Ω. |