Alternative to this Circuit? - Page 1
Forum section: electronics / beginners

author: Bluestreak66
author_rank: Regular Contributor
timestamp: June 13, 2013, 03:03:06 am
content: I'm looking for a way to measure power consumption and came across the attached circuit. It meets all the requirements for what I need except Id like to measure the voltage that comes in on the board that goes to fan speed taps. This circuit uses an external transformer but Id like everything to be on the board and a transformer take up lots of space. Is there any other way to do this? I have a current clamp already worked out and the voltage Id like to measure would be between 100 and 250.

author: ataradov
author_rank: Super Contributor
timestamp: June 13, 2013, 03:20:25 am
content: To measure power consumption you need to know current as well as voltage. There are circuits without transformers, but they are dangerous if you don't know what are you doing.

author: Bluestreak66
author_rank: Regular Contributor
timestamp: June 13, 2013, 03:30:51 am
content: Like I said I'm measuring current with an external current transformer and I can assure you I'm more than capable of working with high voltage I do it everyday. This is for a project controlling Hvac fan speeds. I also need to mention that the circuit needs to be Isolated which is the main purpose of the transformer, I just don't have room for a transformer. I've look at circuits with resistor divider network but they are neither isolated nor can deal with there kinds of voltages with any sort of usable resolution. This will be 3 volt logic not 5 and in the diagram also. Thanks

author: Paul Price
author_rank: Super Contributor
timestamp: June 13, 2013, 03:31:26 am
content: They are dangerous even if you do know what you are doing.

author: ataradov
author_rank: Super Contributor
timestamp: June 13, 2013, 03:39:17 am
content: Well, for galvanic isolation there are no other options. Most often used schematic uses capacitor and resistor.

author: Paul Price
author_rank: Super Contributor
timestamp: June 13, 2013, 03:40:03 am
content: There are true RMS ic's that can measure the power use on AC mains. The idea here is to put this circuit on the HV side, powered by a small power supply, and then optically couple the RMS results to your MCU in some type of serial bit stream.Another idea is to make a small current transformer by winding a few turns of insulated wire on a small toroidial core. Make sure the wire insulation can handle the AC mains voltage that connects to your MCU circuit after amplification. This circuit gives you isolation and is inserted in series with the fans. This transformer will yield a small amplitude signal that is directly proportional to current used by the fans. This current sense signal is amplified and with some calibration you can calculate the current readings and since you know the fan supply voltage, you have what you need to measure power consumption accurately.

author: ataradov
author_rank: Super Contributor
timestamp: June 13, 2013, 03:42:04 am
content: Here is a typical example:http://www.electroschematics.com/5678/capacitor-power-supply/. You obviously don't need rectifier part of it.

author: Bluestreak66
author_rank: Regular Contributor
timestamp: June 13, 2013, 04:55:02 am
content: I'd like to stay away from the RMS ic that would add another ic and from what I can tell they are about $6 per chip in low quantities. I have an external current transformer that will plug into the board to measure current, that parts already finished. Can you elaborate on how I would use the circuit linked to above to calculate voltage?

author: ataradov
author_rank: Super Contributor
timestamp: June 13, 2013, 05:05:54 am
content: Capacitor and a load resistance form a voltage divider. Sort of like described herehttp://en.wikipedia.org/wiki/Voltage_divider.So you need to put a load register instead of bridge rectifier and measure voltage drop across the load resistor. Since capacitor reactance depends on the frequency, you'll have to adjust for the mains frequency and waveform. Definitely not the best solutions for a measurement device.

author: notsob
author_rank: Frequent Contributor
timestamp: June 13, 2013, 05:15:14 am
content: Read this app notehttp://www.analog.com/static/imported-files/application_notes/an564.pdf

author: Bluestreak66
author_rank: Regular Contributor
timestamp: June 13, 2013, 05:43:27 am
content: The more I look into it the better an external ic looks. I found thishttp://ww1.microchip.com/downloads/en/DeviceDoc/22011B.pdfwhich outputs a pulse for for both real power and a high frequency pulse for instantaneous power. So all i would have to do in software is count the pulses.

author: ptricks
author_rank: Frequent Contributor
timestamp: June 13, 2013, 10:42:29 am
content: You are measuring voltage so it doesn't have to be a 1:1 transformer or high current, 50ma transformers are tiny , you can use any AC transformer and multiply the ratio to get the input voltage. That is really the only easy way to get the voltage and remain isolated. You can do other methods like optical isolation but it starts to get a lot more complicated. For about $6 you can get transformers about 1 inch square or less.

author: Bluestreak66
author_rank: Regular Contributor
timestamp: June 14, 2013, 12:22:37 am
content: I had looked at transformers but even a small one inch square transformer takes up alot of space in comparison to a handful of discrete components or small ic. I found thishttp://datasheets.maximintegrated.com/en/ds/78M6610+LMU.pdfIm thinking of using it or something similar. It uses i2c to interface with the micro which is fine im using i2c for the real time clock as well. Would this count as isolation though? The only reason I'm concerned about isolation is the board will be communication over the network and i dont want any chance of high voltage making it onto the network. I will most likely add movs at the power input and all 24v signal lines for voltage spikes.

author: ptricks
author_rank: Frequent Contributor
timestamp: July 02, 2013, 12:54:37 am
content: That Maxim chip requires a transformer for the input side. Transformer is really the only way I can see you getting isolation in such a small PCB area.

author: lgbeno
author_rank: Frequent Contributor
timestamp: July 02, 2013, 02:05:36 am
content: Wrt isolation, someone talked about doing a small primary side power supply to power a few components then you could do some type of ADC on primary side and set a bit stream over an opto. Maybe a sigma delta modulator?Current transformer is a decent way to go, I think these parts are awesome too:http://www.allegromicro.com/en/Products/Current-Sensor-ICs/Fifty-To-Two-Hundred-Amp-Integrated-Conductor-Sensor-ICs.aspxI would still have a opto isolation on this allegro part since it won't have sufficent creepage and clearance. Make sure to observe this in your layout too.

author: Paul Price
author_rank: Super Contributor
timestamp: July 02, 2013, 01:31:26 pm
content: Here's a very simple and safe way to make the voltage measurement you need.Construct a simple bridge rectifier using any tiny diodes you have available. Connect the output of the bridge directly to the LED input pins of a transistor output optoisolator such as a PC817(that you can salvage from just about any junk switching power supply.)Connect one side of the AC hot connections on the bridge directly to the AC voltage you want to measure, connect the other AC side to the bridge through a 27K 2-W resistor to the other hot side of the HV AC.On the isolated side of the optoisolator you connect a 10K or so (value may need trimming for best dynamic range) to the collector, other side of the resistor to +3.3 or +5V of your MCU circuit, emitter of opto output to ground. This meas. circuit needs to be calibrated first with a known safe AC voltage source to relate the voltage output of the opto to the actual values, pretty simple to do, though.You can feed the proportional to the AC mains V in magnitude AC output of the opto to a single op-amp configured as a full-wave rectifier to then feed into and A2D input and let the MCU do the math.

author: JackOfVA
author_rank: Unknown
timestamp: July 02, 2013, 02:06:04 pm
content: I've found that cheap - and some no-so-cheap - power transformers are designed so that the peak of the AC cycle drives the core into moderate saturation, thereby flattening the sine wave crest.When using these for voltage sensing, therefore, some error will be introduced in the secondary waveform. Error in the waveform peak value may have a disproportionate effect upon overall power calculation accuracy. Whether this error is within your planned error budget is something you will have to determine.There are special transformers made to accurately follow the input waveform, designed for measuring AC mains voltages with lower voltage instruments - they are called "potential transformers" or "instrument transformers" but be prepared to pay much more for them than a standard power transformer.It might be possible to use - assuming you are in an area with 120V standard supply voltage - a power transformer designed for 240V primary instead of one rated for 240V. In theory the 240V transformer operated at 120V should have less waveform distortion, but it's something I've never measured or looked at in detail so this assumption could well be untrue in practice.

author: Bluestreak66
author_rank: Regular Contributor
timestamp: July 07, 2013, 04:12:39 am
content: Quote from: ptricks on July 02, 2013, 12:54:37 amI ended up optically isolating the chip with its own power supply, transformer may or may not have taken up less room but I now have more functionality at less cost than with the transformer. Thanks for the input everyone.

author: SeanB
author_rank: Super Contributor
timestamp: July 07, 2013, 05:25:32 am
content: Simpler than the instrument transformer is to buy 2 small 2VA transformers and run the primaries in series, with the secondaries in parallel. You get half the voltage but with no saturation of the core. Of course you can only draw a low current from this, as otherwise you affect the voltage accuracy as resistive losses are quite high in this use but are constant and can be calibrated out.