I was wondering if anyone has or knows where I can find a simple, cheap current sensing circuit that I could use to sense AC currents. I am currently working on a power control project that will monitor 120 VAC outlets with a total maximum current of 15 amps. We are currently working with a current sensor that outputs an AC voltage (actually a varying DC voltage since it does not drop below zero) from an AC input current from 0-50A. When the input current is zero the output is about
2.5V. At the full rated input current of 50 amps the sensor output has a peak voltage swing of about 2.5V which ranges from zero to 5V at the full 50 amps (swings around the 2.5V q-point).
The current sensor works great, but the problem is that we need to find out what the actual current is and this poses somewhat of a challenge since the output of the sensor is a varying DC (sinusoidal) voltage. We've tried some software techniques to capture the peaks, but it takes up too much processor time and is therefore not feasible.
Anyone have any ideas or circuits I might be able to try? Any help will be greatly appreciated.
"Patrick" schreef in bericht news: snipped-for-privacy@g44g2000cwa.googlegroups.com...
Patrick,
What about the electronic skills of your team? A capacitor is enough to separate DC from AC although it might be a huge one for 50/60Hz. If that's not applicable a simple opamp suffices to subtract the 2.5VDC from the signal leaving the AC-part for further processing. Of course you can find complete measuring systems on the net. Tektronix sells some great ones but I bet they are pretty expensive in relation to your budget. Current sensing devices with Hall effect sensors often has a half Vs offset. So if you buy one you will have the same problem. AMPLOC makes them for instance. Some of them however have integrated electronics to remove the offset.
I'm not sure I understand your problem. However, take a look at the AD636. It is a device which will give you a DC output equal to the RMS value of a waveform. Thus, pass the 2.5V through a capacitor, then run it into this thingy. You'll need to calibrate it, but it should give you a good view of the current when coupled with your current sensor.
Hi, Patrick. From your description, I would guess you might have an Allegro ACS750SCA-050, which has a zero ohm current shunt and attached Hall effect circuit to detect current going through the shunt. When no current is going through the shunt, its output is Vcc/2, or 2.5V. When current is going through the shunt, a voltage is superimposed on this DC level of +/-40 mV per amp up to +/-50 amps (which gives an amplifier output of 0.5V to 4.5V). It has a -3dB response at AC current up to
13KHz, and a combined guaranteed accuracy of +/-2% at room temperature.
You have been very unclear on exactly what you want. You've said you're using some kind of PC, SBC or uC to read the voltage, but the repetitive readings are taking up too much processor time. You'd like a simple, inexpensive means to do what you're doing now, but you haven't been clear on just what you're doing now, how simple, and how inexpensive. "Measuring current" can mean just measuring peak current, measuring average current or RMS current. You haven't indicated exactly what information you need, or in what time frame you need it. Do you need information every half cycle, every cycle, a certain number of times a second? Oh, and by the way, what kind of accuracy do you need? Your error budget is already over 2%. Do you need a combined accuracy of 2.1%, 3%, or 12%? The key to a good answer is a well-asked question.
Your lowest cost solution for doing just what you're doing now would be using a PIC with built-in ADC. You can read the voltage output off the ACS750 many times for every AC cycle (generate an interrupt off the power supply transformer secondary?), and then either figure out the peak current (if that's your pleasure) or add 'em up to integrate total current. There are many ways to communicate that information to whatever you've got as a processor. And there's no law against having two processors in a piece of electronics.
If you only need information on peak current on only half the cycle, you're only taking a few readings a second, and you can live with several percent accuracy, the simple and cheap answer is to use a schottky diode feeding a cap with a bleeder resistor, like this (view in fixed font or M$ Notepad):
By using software compensation for the diode drop, you can achieve accuracy within several percent. The obvious flaw here is that when the current decreases, it will take half a second to get an accurate reading. If that's OK with you, that's the simplest. You can juggle settling time and decay time by changing resistors to get something closer to what you want.
There are a lot of solutions to your problem. The Analog Devices chip mentioned in another post being an elegant one chip solution, and while not the least expensive, will give you the best, most accurate answer. The precision rectifier is a good answer which solves the issue with the voltage drop of the diode above. There are many other answers. But you could give a little more information if you want to get the best answer.
dragging up my limited analog knowledge ... how about demodulating the signal, and smoothing/processing the resultant dc level ?
- use a second ac signal as a reference, using it to control a two-way analog mux, using that to route either the true or inverse (via inverting opamp) ac signal. No doubt there are suitable circuits available on web. hth Neil
I know people are tired of hearing me ask, but "What are you trying to accomplish?" If you have a purely resistive load and a pure voltage sinewave, you can measure about any parameter of the current waveform and know all the rest.
Problem is that most applications are not even close to that ideal. In most cases, you have to actually measure the parameter of interest. Some typical parameters might be peak, average, RMS values. Phase may or may not be important.
If it's a current sine wave and you can determine the zero crossing with a comparator and you know the frequency, you know exactly were the peaks are and you can sample there.
If it's not a current sine wave, you can peak detect and subtract off the 2.5V either with an op-amp or software.
What do you want to do about noise? How fast do you need your loop response?
All comes back to, "what do you really want to know about this current waveform?" mike
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I agree that a simple scheme may work properly. One thing to factor into the discussion is that a microprocessor pin is a good way to zero a peak detector. He can use a shottky diode and cap, like you suggest, but skip the drain resistor. Before each sample, just use another pin to drain the cap. Then, make the pin an input, wait a few cycles, and take the reading.
Another thing to consider is that some micro ADCs want fairly low input impedance; the PIC 12 and 16 series, for example, want 10k.
Yet another issue is that this won't get him the right value if that
2.5V center voltage drifts around. He can fix that by using a big cap to make the 5VAC signal be centered at zero. Then, his answer will be between 0 and 2.5V, minus the drop of the diode.
If he wants to add a couple of cheap opamps, he can eliminate that diode drop as well by building a simple peak detector. However, that will involve more external hardware.
None of this really works, however, if the current isn't a sine wave.
Will the leakage through the big power Schottky be a problem here? The current through the diode is going to be very small, microamps. The datasheet for the 1N5817 says that the reverse leakage current is around 100 uA at 4 V though. My simulation doesn't rectify; the diode drop is negligible, in both directions... (but of course it works fine with a 1N4148 or something, or I guess you could make C bigger and R smaller).
I think R*C should be bigger anyways, because with a not-leaky diode there's still about 0.3 V of ripple on the output for a 1 V amplitude input. I think the ripple is roughly given by
dV/dT*T = (I/C)/f = ((Vout/R)/C)/f = Vout/(R*C*f) = 3/(1meg*0.1u*60) = 0.5 V which is kind of close
(assuming that the current through R is constant, and assuming that the capacitor spends almost all the cycle discharging and hardly any time charging)
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