current measurement of AC coming into bridge rectifier powering circuit itself - Page 1
Forum section: electronics / beginners

author: djacobow
author_rank: Super Contributor
timestamp: November 20, 2014, 01:24:15 am
content: I have a circuit powered by a 24Vac transformer. The 24 Vac is used to drive some stuff through relays. I am also tapping off that voltage to power my circuit: rectifier and voltage regulators.I'd like to measure the current that my 24 Vac devices are consuming. Accuracy is not important. I'd like mostly to be able to distinguish between open, dead short, and looks-about-ok.I could use something like Hall effect, but cheap is important, so something with just a sense resistor and an opamp would be great.Thing is, I'm getting wrapped around the axle because the sense resistor is in series with the circuit before the rectifier and I can't quite figure out how to arrange the opamp. It's AC and neither high- nor low side, at least relative to "system ground" at the negative terminal of the bridge rectifier.If I could just get something that made peaks at 1.5 volts when there was about 1 amp of AC current, peaks at 3 volts when there is 2amps or more, and no peaks when open circuit, I could easily peak detect that and send it to the uC.But the high common-mode voltage of 24v across that sense resistor is vexing me.Any hints?

author: Paul Moir
author_rank: Frequent Contributor
timestamp: November 20, 2014, 04:13:12 am
content: That's a tricky one since it's naturally outside the rails of the op-amp!Here's two ideas that really don't answer your questions:1. Couple it down with a transformer like a little audio transformer (once used in every transistor radio) or one of those 600 ohm ones used in modems. Then it's isolated so you can just rectify and measure the DC signal. If you could find the right ratio, you might not even need the op-amp.2. (Left field): if accuracy isn't important maybe you could measure the unregulated DC out of the power supply, relying on the 24vac transformer's voltage drop to see it's load. 'Course that would be totally transformer dependant. To get a better reading, you may want to use a separate rectification circuit (I would imagine just a diode, small capacitor and resistor divider would be all you needed).

author: djacobow
author_rank: Super Contributor
timestamp: November 20, 2014, 06:50:35 am
content: Thanks, Paul. Yeah, it is tricky. I was on mouser and current transformers cost a few bucks and there really aren't that many to choose from that work on the order of 1 amp and 60 Hz. I know I can do it with a Hall Effect device, too. But I just feel like there has to be a way to do this on the really cheap. A few bucks matters because I'm going to make a lot of these.

author: Seekonk
author_rank: Super Contributor
timestamp: November 20, 2014, 12:54:43 pm
content: It takes quite a core to do 50hz. I have a case of small open frame power transformers that are wound on a split bobbin, instead of primary on top of secondary. I just cut out the primary with an exacto knife and wind one or two turns of wire. It is a lot of work.

author: rs20
author_rank: Super Contributor
timestamp: November 20, 2014, 01:01:51 pm
content: Quote from: Seekonk on November 20, 2014, 12:54:43 pmIt takes quite a core to have a voltage-input, high-power, high-efficiency 50Hz transformer. But this application requires a current-input, very low power, what-does-efficiency-even-mean 50Hz transformer that takes in 50 Hz and barfs out 3V to a Hi-Z uC pin. This doesn't not require an impressive core at all.

author: fcb
author_rank: Super Contributor
timestamp: November 20, 2014, 01:25:30 pm
content: Use a diff. amp to measure between the two potential dividers, something like an LM358 is adequate.

author: djacobow
author_rank: Super Contributor
timestamp: November 20, 2014, 03:34:24 pm
content: Are you suggesting that I take the center of those 22k dividers as the input to my diff amp? The problem I see is that the voltages there, relative to the ground as it is located at the rectifier, is about 15V. (I'm using 33.9V, 24*1.414 in my "sine"). So, if I power the opamp with 3.3V, that's way out of range. I could power the LM358 right off the main rectified voltage, but 34V is just outside the operating range of the LM358.Quote from: fcb on November 20, 2014, 01:25:30 pm

author: fcb
author_rank: Super Contributor
timestamp: November 20, 2014, 07:43:10 pm
content: Use an LM358 (dual opamp) to buffer the two voltages of interest (fit a zener, say 10v, in the Vcc - you'll only need a few volts of headroom above the half-supply point). You now have a nicely buffered pair of voltages that sit at approx. half rail (so about 17V in your application).Now you can either use another pair of potential dividers to drop this down to something your more comfortable with (i.e. feed straight into a micro's ADC and do the differential function in software if you've got the resolution). Or what I would do would be use a diff. amp (half an LM358) to provide gain and reference the output to the 50% point on the ADC.Probably use the spare half an LM358 to provide a buffered reference voltage for your differential amplifier.

author: Zero999
author_rank: Super Contributor
timestamp: November 22, 2014, 12:23:23 pm
content: What about a small mains transformer?Connect the low voltage winding across the sense resistor and take the output from the mains voltage winding. A 120V to 6V transformer will have a turns ratio of approximately 20:1 so a100m Ohm sense resistor in parallel with the 6V winding will give about 2VRMS per Amp.