lukeweston
commented
11 years ago Thanks for the feedback.
This is a preliminary experimental design and hasn't actually been tested, so your concerns are likely to be quite valid.
It's probably worthwhile to consider increasing the current sense resistor a bit, maybe 10 milliohms. The limiting factor here is going to be the power dissipation in that resistor at the highest-case load current.
Currents of 16A or so are not unusual in some high-power domestic appliances in 120VAC countries such as the United States, and ideally I wanted to design something for worldwide compatibility. If we suppose that we design for 20A as the absolute maximum current, then a 10 milliohm resistor will dissipate 4W which is probably OK. Yes, that means you'll limit measurement resolution when measuring <10A typical currents in a 240VAC country, but I think that's an acceptable price to pay. You're also limiting the resolution when measuring the voltage signal in a 110VAC country.
Basically, as you understand, the offset and gain in both the voltage and current analog circuits (well, the divider ratio in the case of the voltage) should be designed such that under the "maximum possible" conditions, such as 20A in and say 260VAC in then we get waveforms into the microcontroller's ADC that are as large as possible without clipping outside the microcontroller's usable measurement range.
There is also an issue with the capacitive-dropper transformerless power supply... it's just not practical for this sort of application where we're trying to supply 100mA or so to run the relay coil, the XBee module (which is quite a power-inefficient device) and the AVR simultaneously. If you try and build it you'll likely find that the R14 input resistor in the RC dropper power supply circuit just has way too much power dissipation and heat. I think replacing it with a non-isolated high voltage buck switcher such as the Power Integrations LNK306 is the way to go.
On Mon, May 27, 2013 at 11:14 AM, Roger Clark notifications@github.comwrote:
Hi Luke,
I'm not an expert, but I modeled the current amplifier in LTSpice and it doesn't appear to have enough gain.
You are using a 0.005 Ohm shunt resistor, so at 10A this will give you 0.05V AC on the SHUNT input to that module.
The feedback 13k / 22k loop appears to give a gain of approx 2, hence the output is 0.1V AC offset to 1.65V
But you appear to need around 1.5V AC offset at 1.65V to get decent resolution on the ADC in the micro-controller.
You appear to need to replace the 13k with something more like 330K to get enough gain.
Also. In LTSpice, the input cap C2 appears to be a bit too low to get full transfer of the voltage across the shunt. It looks like 330n would be a more appropriate value.
And... I think you need to put something in to prevent the input of the opamp being destroyed by power spikes. (I've blown several opto isolators on mains circuits due to spikes with other equipment turning on and off).
Adding a resistor on the input to C2 may be enough (1k seems to be OK). Or you could also have 2 back to back diodes after the resistor.
But, as I've said, I'm not an expert, so I may be talking rubbish
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synergie7
Hi Luke,
I'm not an expert, but I modeled the current amplifier in LTSpice and it doesn't appear to have enough gain.
You are using a 0.005 Ohm shunt resistor, so at 10A this will give you 0.05V AC on the SHUNT input to that module.
The feedback 13k / 22k loop appears to give a gain of approx 2, hence the output is 0.1V AC offset to 1.65V
But you appear to need around 1.5V AC offset at 1.65V to get decent resolution on the ADC in the micro-controller.
You appear to need to replace the 13k with something more like 330K to get enough gain.
Also. In LTSpice, the input cap C2 appears to be a bit too low to get full transfer of the voltage across the shunt. It looks like 330n would be a more appropriate value.
And... I think you need to put something in to prevent the input of the opamp being destroyed by power spikes. (I've blown several opto isolators on mains circuits due to spikes with other equipment turning on and off).
Adding a resistor on the input to C2 may be enough (1k seems to be OK). Or you could also have 2 back to back diodes after the resistor.
But, as I've said, I'm not an expert, so I may be talking rubbish