Comment summary #39
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Comment 381 ... 390
| date | article | author | comment |
| 07 mei 2015 02:24:38 | Arduino wattmeter | Freddy | The Arduino Wattmeter doesn't need a negative voltage. If you want to use a negative 12 volt for testing than connect the positive terminal of the battery to "COM" and the negative terminal to "V". The wattmeter shows a negative voltage reading of the battery. |
| 06 mei 2015 16:27:11 | Arduino wattmeter | salim | how to generate Vcc-(-12) using battery of 12V |
| 03 mei 2015 17:12:34 | New wattmeter project | cesar lopez torres | como me inscribo a este foro |
| 03 mei 2015 13:42:46 | Theory and Definitions | Freddy | P=Vrms*Irms is only valid for sinusoidal signals with no phase shift. In the Pav1 equation, replace Vrms and Irms by the RMS equations above, and replace v(t) and i(t) by a sinusoidal function sin(2*pi*t). Because p(t)=v(t)*i(t), you can simplfy the equation to the one given as Pav4. |
| 29 apr 2015 18:21:04 | Theory and Definitions | Carl000 | Hello, please, may you provide me the demonstration of your formula: Pav1=Vrms*Irms=Pav4 Thanks |
| 14 apr 2015 12:48:25 | Arduino wattmeter | salim | hi tugla, I am working on a project similar to yours So I advise you use current transformer that will transform the current from the line to 1A then after that use shunt resistor or current to voltage transformer (op amp) to avoid the problem of impedance matching. |
| 14 apr 2015 12:14:26 | Arduino wattmeter | Freddy | Yes, you're right about the voltage difference calculation. If you are using a current transformer then it must be terminated with the recommended termination resistance. If the voltage drop across his resistor becomes larger than half the reference voltage than a voltage divider is required. If you're supplying the Wattmeter with the AC line voltage then it is better to use a 9 VDC adapter due to safety reasons. |
| 14 apr 2015 03:52:46 | Arduino wattmeter | Tulga | Thank you for your reply. What i'm confused about is Voltage is difference of two potentials right? So you have a analog signal (for example) -12V to +12 V from B to Com. If you make com +12V will it be 0 to 24V ? As i mentioned above i'm using current transformer to measure current so I think i should use voltage divider with coupling capacitor or opamp adder something like that ? And i'm going to feed my circuit from the AC line using rectifier and regulator. Is it affect my accuracy or something ? Sorry for my bad electronic knowledge. |
| 13 apr 2015 16:44:56 | Arduino wattmeter | Freddy | The analog and digital ground on the controller must be externally connected together. But from the analog ground pin, the connections may only made to the analog section of the circuit. The same applies for the digital ground: from this pin the connections may only be made to the digital section. Thus, the analog and digital grounds are only connected at the controller pins. The COM point may never be connected to any ground connection. The COM point has a voltage of half the reference voltage. The ref. voltage is made by a voltage divider supplied from the 5 V. This is done so because the ref. voltage output can't handel loads such as the resistive voltage divider. Because the controller can only measure positive voltages between 0 and Vref, and measured voltages can be positive as well as negative, the COM point is fixed on a half way voltage in order to make it possible to measure negative voltages. The difference between the measured voltage and COM is the measured voltage. |
| 13 apr 2015 03:51:42 | Arduino wattmeter | Tulga | Thank you for your work Freddy :D I'm gonna try this project with an Current transformer and 230/12 V transformer on a AC main line. And i'm using Atmega16 i guess it's not very different from arduino nano :) Only difference i've found out so far is internal Vref. I'm confused about analog & digital gnd. On atmega analog and digital gnd's are internally connected i assume ? is it has to be the same or analog ground must be connected to com ? and i don't understand the way that you are offseting your signals. Would you mind explain a little more about it to me ? Sorry for my bad english. Thanks in advance |