How to design a current sense to Boolean circuit? - Page 1
Forum section: electronics / projects

author: sparkydog
author_rank: Regular Contributor
timestamp: October 05, 2023, 06:47:07 pm
content: I'm trying to design a non-intrusive¹ sensor to detect if an AC device is powered. (I'm not entirely sure how much power it uses; the data sheet mentions "7.5 VA" nominal "power consumption" (with spikes up to 70 VA) and "heat dissipation 2...3 W". So I'mguessingnominal RMS current is... 20-25 mA?)IIUC, the usual way to do this is with a current sense transformer. However, what's not clear is how I can convert this to a digital (on/off) signal without an MCU.Naïvety says if I use a 50:1 transformer (e.g. 56050C), I should see ~3.4 V peak voltage (relative to neutral), though when I try tosimulatea circuit I only get ~2.8 V (probably transformer losses, exacerbated by having no idea what the simulation parameters for the transformer should be). Current out is proportional to current in? However, IIUC current sense transformers correctly, thevoltageout is supposed to be proportional to thecurrentin? (How does that work?)In any case, if I take this circuit (link above, also included image), slap a smoothing capacitor on the other side, and feed that into a comparator, is that a reasonable approach to getting a Boolean signal indicating when the AC device is powered? Or is feeding a FBR like this a bad idea?Also, IIUC, the value of the resistor is relevant; how do I calculate what that value should be?(¹ Running a wire through a transformer core is sufficiently non-intrusive. What Idon'twant is something that has to be wired inline with the load I'm trying to measure.)

author: Benta
author_rank: Super Contributor
timestamp: October 05, 2023, 07:19:06 pm
content: A current transformer needs a burden resistor to provide a voltage output (value normally specified in the data sheet).You'll get an AC voltage proportional to the main/primary current.But you say "non-intrusive."I hope you're aware that you need to split Live and Neutral and only feed one of them through the current transformer. So you'll at least have to intrude into the mains cable somehow.

author: Shay
author_rank: Regular Contributor
timestamp: October 05, 2023, 09:13:53 pm
content: As far as I understand you're trying to detect if a certain measured current is above a threshold, it should output a signal (high or low).For measuring AC currents you can use a shunt resistor or current transformers. Since you're trying to use mains current, you can use a current transformer and benefit from isolation from mains.As previously stated, a current transformer outputs a current. You can convert the current to a voltage with a resistor in parallel.The exact value can be calculated if you have the specifications of the CT or you can use a decade box/potentiometer.After you have a AC voltage that corresponds to the current, you can rectify it and smooth the output and get the average current minus the rectifier VF, and utilize a comparator or zener to get a Boolean value.You can also use a precision rectifier and some filtering to get a RMS value if needed (although the accourcy depends on the form factor.)

author: sparkydog
author_rank: Regular Contributor
timestamp: October 05, 2023, 10:48:19 pm
content: Quote from: Benta on October 05, 2023, 07:19:06 pmThe data sheet says "Terminating Resistance to Produce 1VOUT/1AIN: 50 Ω". If my goal is, say, to get 2.5 V at 25 mA... that would need ~5k? And I guess Ohm's Law is governing this, which suggests I'd probably better not rely on being able to draw anycurrentoff this thing. Which suggests I'd better feed a transistor rather than a FBR to charge the capacitor. Okay, well, that actually has some advantages, anyway....but is the transistor going to blow up if it sees -5V? (I...think"no"?)All right,hereis a revised attempt. Does this look reasonable?Quote from: Benta on October 05, 2023, 07:19:06 pmRight, sorry, that's not a problem. I'm actually sensing the input of a contactor, so I'm already dealing with separated wires. I mainly don't want something that has to be wiredin linewith the circuit I'm sensing.

author: wasedadoc
author_rank: Super Contributor
timestamp: October 05, 2023, 11:02:55 pm
content: What exactly is the design spec? If you just want to know if AC mains is applied or not to the piece of equipment, an easy, safe and non-intrusive way for a one-off project is to have a phone charger or similar AC mains in to DC 5 Volt USB out on the same mains feed. 5 Volt Boolean logic signal right off the bat. Add two resistors if you need less than 5 Volt level, eg 3.3.But you would want a different approach if designing a product that is going to sell more than a handful.

author: sparkydog
author_rank: Regular Contributor
timestamp: October 06, 2023, 12:32:33 am
content: Quote from: wasedadoc on October 05, 2023, 11:02:55 pmSpace and idle power consumption are significant considerations, especially as I need several of these and they need to fit inside a box with a bunch of other stuff.This is actually fallout from a larger design change replacing DC relays (each with their own PSU) with AC contactors. Essentially I need a secondary mechanism to detect if the relay/contactor coil is energized. In the previous design, Ididjust tap the PSU for that signal (well, slightly more complex, but close). Getting rid of the PSUs was part of the objective of the original change, so in a way, I'd already considered and rejected your suggestion. 🙂 (It's a very reasonable suggestion, though!)AFAICT, Ishouldbe able to pull this off with a 56050C (which is a lot smaller than any PSU), a capacitor, a comparator, a couple transistors, and a handful of resistors and a TVS... and half those components are actually pulling double duty. IthinkI have a complete design that I'll probably post as a new topic; here, I'm just trying to figure out if I've got my head properly wrapped around how to properly use a current-sense transformer in a useful circuit.To that end:Is the circuit in the previous message going to charge the capacitor? CircuitJS thinks "yes", but only to ~1.6V. (I can work with that.)Is it going to toast the transistor in the process?Given the lack of current specification of the thing I'm trying to measure, are my guesses based on VA and W numbers reasonable?Is 5k the right value for the burden resistor? (Depends also on the previous question!)

author: Konkedout
author_rank: Frequent Contributor
timestamp: October 06, 2023, 03:05:55 am
content: The circuit in your .png might work as a current detector, but I think that a bipolar transistor would probably be better than a MOSFET in this situation. The current transformer (I do not have any specification for it) wants to put out current according to the detected current and the turns ratio, but it really does not want to put out much voltage. And a BJT base will generally work with lower voltage than will a MOSFET gate.Transistor could be MMBT3904 (200 mA max collector current) or MMBT4401 (600 mA collector current and higher base current rating) which are widely available general purpose transistors. The MMBT4401 is probably the safer choice if you are unsure of current levels.The best might be a voltage comparator which can sense at ground level; there are single versions of LM339 (that is a quad) which would probably work OK. Then your voltage threshold can be even lower, such as 100 - 200 mV.

author: Terry Bites
author_rank: Super Contributor
timestamp: October 06, 2023, 05:39:32 pm
content: don't make what you can can buy for a couple of bucks.https://www.aliexpress.com/item/1005005675543015.html?spm=a2g0o.productlist.main.1.16a84Sr84Sr8wS&algo_pvid=5860e4d4-829f-4564-b6e0-ce3e93ceeba3&algo_exp_id=5860e4d4-829f-4564-b6e0-ce3e93ceeba3-0&pdp_npi=4%40dis%21GBP%219.62%211.3%21%21%2183.07%21%21%4021038edf16966138454237582e68d1%2112000033978441413%21sea%21UK%210%21AB&curPageLogUid=kakaqvf520Y9

author: sparkydog
author_rank: Regular Contributor
timestamp: October 06, 2023, 07:40:27 pm
content: Quote from: Konkedout on October 06, 2023, 03:05:55 amI've mentioned a couple times I'm planning to use a 56050C. The first search result for "56050C" leads to theproduct page, from whence thedata sheetis readily available. 🙂Apparently an NPN can charge the cap to a higher voltage? Curious... so much about electronics I don't understand!Quote from: Konkedout on October 06, 2023, 03:05:55 amIn a sense, yes, but... I think that would add three resistors (two to make a divider for the reference voltage, and a pull-up) and I think I'd still need a transistor to charge the capacitor?On second thought, it looks like I can use the same reference voltage, so only one additional resistor if I use an LM393. (Why would I use an LM339 and add six pins that I don't need?) Since that's the difference betweenknowing(and being able to control) the threshold voltage and having whatever threshold the transistor happens to have, I'll take it.

author: schmitt trigger
author_rank: Super Contributor
timestamp: October 06, 2023, 07:47:40 pm
content: Quote from: Terry Bites on October 06, 2023, 05:39:32 pmThis happens to me very often. I have an idea for a module to realize a particular function. I perform a preliminary design, only to find out a complete module in a nice enclosure, fully assembled and tested, for less than half the cost of just the components in my design.

author: sparkydog
author_rank: Regular Contributor
timestamp: October 06, 2023, 07:59:44 pm
content: Quote from: Terry Bites on October 06, 2023, 05:39:32 pmTo start with, I'll overlook that that isn't sensitive enough for my needs.It's still big. Perhaps you actually read the thread and noticed that size is a concern? The "detect AC" function is also onlypartof the board I'm designing.Ifthat was usable, it would replace a whole five components on my board... and I have a sneaking suspicion there's a relay in that doohickey. That'sprobablynot fatal, but adding the delay for a relay to switch before I get a signal isn't ideal.Anyway, the point of this thread wasn't "how can IdoX?", it was "how can Idesign a circuitthat does X?". Giving me a fish doesn't help me learn to fish for myself. 🙂

author: Benta
author_rank: Super Contributor
timestamp: October 06, 2023, 09:34:23 pm
content: Let's go back to your OP first and nail some figures.Supply voltage is 120 VRMS(this is an assumption based on your location).Active apparent power is 7.5 VAThis means active current is around 63 mARMS"Heat dissipation" in the load is uninteresting here.Also, I presume an on/off behaviour in load current: either 63 mA or 0 mA.Can you confirm these figures? Or are you on a 240 V line?Setting a "slicing level" at around 1/3 is sound, so your 20...25 mA selection is ballpark as RMS value.

author: sparkydog
author_rank: Regular Contributor
timestamp: October 09, 2023, 05:14:48 pm
content: Quote from: Benta on October 06, 2023, 09:34:23 pmYes... sorry 🤦.Quote from: Benta on October 06, 2023, 09:34:23 pm"7.5 VA" is the number I could find, but feel free to perusethe data sheetyourself. I don'tthinkI'd regularly see lower current under normal operation (i.e. I assume that number is the nominal holding power), but do mind the inrush. (AFAICT, slapping a TVS across the output should be fine to deal with that.) Otherwise, the load is basically an electromagnet, so about as simple as it comes in terms of only being "on" or "off".

author: Benta
author_rank: Super Contributor
timestamp: October 09, 2023, 07:57:05 pm
content: IMO, it can be done much simpler.You have a 1:50 current transformer, so your RMS output current at nominal load is around 1.3 mARMS.Plenty for controlling a BJT (eg, BC547 or 2N2222).The attached circuit will simply give you a 50 Hz square wave when current is drawn, and VCC when no load is present.Converting the square wave "OUT" to DC is a piece of cake.You can also invert whole thing by using a PNP (no current draw = GND).Concerning 70 VA transient load: pffft. The BJT and diode will accept that easily. It's just 12 mA.

author: phil from seattle
author_rank: Super Contributor
timestamp: October 10, 2023, 05:27:27 am
content: Why not just select a contactor (relay?) with an additional pole and use that? seems too easy...

author: sparkydog
author_rank: Regular Contributor
timestamp: October 10, 2023, 04:19:47 pm
content: Quote from: Benta on October 09, 2023, 07:57:05 pmThat seems identical to my second iteration except for three differences:Mine included the capacitor that handles your "piece of cake" converting the square wave to "DC" (not perfect DC, but close enough).You used a BJT instead of a MOSFET, as previously suggested by Konkedout.You use a basic diode instead of a TVS.So, on the one hand, that suggests I'm on the right track! Yeah, my current design uses a comparator, but it's almost "free" since I already need the IC and voltage divider for another part of the larger circuit. (Using the otherwise-unused second unit does add an extra pull-up to the BOM, but that's worth it IMHO for being able to control the threshold with much greater precision.)Is the diode really necessary? Is the BJT going to go boom if it sees -6V at the gate?Quote from: phil from seattle on October 10, 2023, 05:27:27 amIt seems too easy because itistoo easy. 🙂The whole point is to detect contact welding. For that, I need to know if the coil is on (or supposed to be on, anyway)independentof what the contacts "report". If a main contact welds and happens to leave the relay in a state where the aux contact is closed when the coil isn't energized, that's a critical fault that I wouldn't be able to detect if I rely only on the relay contacts to determine whether it'ssupposedto be closed.(I suspect this circuit won't detect a relay that'sopenwhen it's supposed to be closed, but that's anon-critical fault, so... 🤷)

author: Ian.M
author_rank: Super Contributor
timestamp: October 10, 2023, 04:58:07 pm
content: Why are you stuck on sensing current through the contactor coil rather than simply sensing voltage across it?You can get multi-channel 120VAC to optoisolated logic level input boards off the shelf. Pay the premium to get one from an automation specialist rather than a cheap no-name Chinese one that has a high probability of having compromised isolation (usually tracks run under the optocouplers breaching UL's creepage and clearance distance requirements).

author: Benta
author_rank: Super Contributor
timestamp: October 10, 2023, 08:19:28 pm
content: Quote from: sparkydog on October 10, 2023, 04:19:47 pmYes, the circuits are similar. But it's interesting to look at current vs. voltage.Your circuit focuses on transforming a current into a voltage that can then be processed further.My circuit doesn't even deal with that, but simply uses the current from the CT directly (D1, Q1). Apart from the VBEand the diode drop, it's a short-circuitedcurrenttransformer, which is the ideal situation.Will it blow up without D1? No, but presenting a somewhat symmetric load to the CT prevents possible "flux-walking".The on/off 50 Hz signal at the collector of Q1 can be turned into a logic signal with a diode, cap and resistor.EDIT: just placing a cap in parallel to the resistor will give you a "digital" output.

author: sparkydog
author_rank: Regular Contributor
timestamp: October 11, 2023, 03:58:23 pm
content: Quote from: Ian.M on October 10, 2023, 04:58:07 pm...for the same reason I'm not using an off-the-shelf current sense module, and because I want something non-invasive? (Youdidread that requirement from my original post, right?) I can wrap a ~8 cm³ current sensor around a wire without otherwise affecting it. Unless I miss something, using a voltage sensor would require taking splices off the wire, yes? Those splices are already near or exceeding the volume of the current sensor, and that doesn't even count the board space to connect high voltage wires. At the very least, it's eight wire connections (two to the sensor, three for each splice) to sense voltage versuszeroto sense current.Quote from: Ian.M on October 10, 2023, 04:58:07 pm...and I'dstillneed my own boards to handle the rest of my logic. Unless you know a pre-made device that:Runs on 5VDC or 12VDCMonitors the coil (120VAC) and NC (board voltage) contact of a relay/contactorHas some sort of "output" that detects contact weldingKnowing whether the coil is powered or not is just one of the inputs. Building it into my own board also lets me pile on connectors in a way that minimizes the number of wire splices I need, by using the board to make the splices.As far as I can tell, sensing current is more space efficient than sensing voltage since the mains voltage never actuallyconnectsto the board.

author: IanB
author_rank: Super Contributor
timestamp: October 11, 2023, 04:25:08 pm
content: Quote from: sparkydog on October 06, 2023, 12:32:33 amIf the contactor coil is powered by AC, there is presumably something that switches that energizing current on and off? Is there any way to take a secondary output from whatever provides that control and use that as the on/off signal?

author: sparkydog
author_rank: Regular Contributor
timestamp: October 11, 2023, 05:18:42 pm
content: Quote from: IanB on October 11, 2023, 04:25:08 pmShort version: alas, no. 🙂Slightly longer version: ...only by turning the AC into DC some other way. (Already had that conversation earlier in the thread; trivial solutions are... very space intensive.) I need a few of these, and some of those unfortunately don'thavesecondary outputs, or any way to add such other than something that is ultimately driven off the primary AC output. Thus I am stuck trying to "tap" the primary AC output in as little space as possible. (Others I could maybe do that, but that would mean either designing two boards instead of one, or one that can work in either mode, so...)

author: sparkydog
author_rank: Regular Contributor
timestamp: October 11, 2023, 05:59:06 pm
content: Quote from: Benta on October 10, 2023, 08:19:28 pmAh! I see now. That... doesn't actually save me any components 🙂 (the resistor turns into a PNP, at least with my current extrapolation... unless I'm missing some trick, which is very possible!), but I see what you mean about the improved simplicity....and a few hundred mA (base to emitter) isn't going to blow up, say, an MMBT4401?Quote from: Benta on October 10, 2023, 08:19:28 pmI'm not sure where the diode is needed? (But it sounds like maybeyoualso decided it isn't...)Ultimately, I need the logic signal tochargea cap... so I don't necessarily need to smooth it at the input, but Idoneed to turn the BJT output into a switch, somehow. Unless I miss something, the way to do that is with a pull-up (which you already have) and the aforementioned PNP. So, I think that would look something likethis?The complete circuit measures two inputs, 'A' (the current sensor we're discussing) and 'B' (the switch on the right side of what's shown here). The goal is to do nothing if either input is present, but to do something if both are absent for an "extended" period of time. This is accomplished by charging the capacitor if either input is present, with a resistor to slowly discharge the same. Not shown, a voltage comparator watches for the resistor to drop below a trigger voltage. However, since both inputs must be absent for "some time", it's sufficient to only charge the capacitor during half of the AC cycle, i.e. for current sense output to be a square wave rather than plain DC. (The remainder of the circuit, mostly not shown, is therefore something roughly like a missing pulse detector.)I don't see anobviousway to lose the PNP while maintaining the ability to charge the capacitor via the right switch?

author: IanB
author_rank: Super Contributor
timestamp: October 11, 2023, 06:28:41 pm
content: One thought that occurs to me is a resistive opto-isolator. Basically one side of the isolator is connected directly to the AC circuit you want to detect, and the output of the isolator changes from high resistance to low resistance when the input is activated. The input can usually work with hundreds of volts, and the output just needs something simple to turn the changing resistance into a logic signal.You could make a home-brew version of such an isolator with a neon lamp fixed to a CdS photocell (or many variations on that theme), but the manufactured ones obviously come in a proper package and may have an LED inside them.I think something similar with a photo-transistor could directly produce an on/off signal when light is detected.

author: Benta
author_rank: Super Contributor
timestamp: October 11, 2023, 08:45:17 pm
content: Quote from: sparkydog on October 11, 2023, 05:59:06 pmYoustilldon't understand.You keep thinking in terms of normal mainsvoltagetransformers. That's a mental block I can relate to, but needs to be overcome.It's a CURRENT transformer!The short-circuit secondary current is the primary current divided by N. Loading a current transformer with 0 ohms is actually ideal, but pretty useless.There's no way the CT will ever provide "a few hundred mA" unless your primary current is high enough. At 5 A it would be 100 mA at N = 1:50.The dangerous scenario is when a CT hasnoburden resistor. In high-power installations, it leads to catastrophic failures.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: October 11, 2023, 09:32:28 pm
content: Quote from: IanB on October 11, 2023, 06:28:41 pmThere are AC driven optocouplers, that can do this simply.However, you only would only really need an optocoupler, if the current transformer did not have good enough isolation ratings.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: October 11, 2023, 09:39:22 pm
content: Quote from: sparkydog on October 11, 2023, 05:59:06 pmI think you mean NPN ?If you do not mind moving the transformer to +ve rail, you can use a single PNP transistor.Enough current/voltage turns on the PNP, and a RBE resistor can set the trip point.Some series base resistance could be a good idea, to protect against capacitive coupled spikes from the mains.

author: IanB
author_rank: Super Contributor
timestamp: October 11, 2023, 10:23:23 pm
content: Quote from: PCB.Wiz on October 11, 2023, 09:32:28 pmSure, but I am thinking about something that is both small and uses minimal components. The current transformer seems to need some extra components on the output to get the logic signal, whereas an AC driven opto-coupler or opto-isolator could almost produce a logic signal on its output pins directly.

author: sparkydog
author_rank: Regular Contributor
timestamp: October 11, 2023, 10:58:51 pm
content: Quote from: Benta on October 11, 2023, 08:45:17 pmRight. Um... on second thought, I don't know where I got that number. I seem to have either misplaced a decimal, or was looking at a sim with an absurdly high input current. On the other hand,willit stand up to that much current, if for some reason 25 A goes through the other side?Quote from: Benta on October 11, 2023, 08:45:17 pmUh... didn'tyouomit the burden resistor? 🙂Quote from: PCB.Wiz on October 11, 2023, 09:39:22 pmI'm pretty sure I meant PNP. That's what's in the circuit I simulated, anyway, and seems to work as desired. Note that the BJT collector is pulledup, making that point high when the BJT is "off". But I want to charge the capacitor only when the BJT is"on". Thus, I need a "switch" that's closed when its input islow... a PNP.Quote from: PCB.Wiz on October 11, 2023, 09:39:22 pmSorry, did not follow this. Sketch? (Modify the sim I posted?)Quote from: PCB.Wiz on October 11, 2023, 09:39:22 pmIn series with the output side of the transformer, you mean? Something likethis?BTW, can you explain the mechanism by which this helps with spikes? I'm failing to visualize how it would work, and I'd like to learn.

author: Benta
author_rank: Super Contributor
timestamp: October 11, 2023, 11:11:51 pm
content: Quote from: sparkydog on October 11, 2023, 10:58:51 pmNo. It's D1 and Q1 VBE. Basically a short-circuit except for 0,6 V offset. For CTs, you need something to tame the secondary voltage. It does not have to be a resistor.

author: sparkydog
author_rank: Regular Contributor
timestamp: October 11, 2023, 11:22:09 pm
content: Quote from: IanB on October 11, 2023, 10:23:23 pmWell, partly I don't care, because the logic isn't just the output of the current sensor. Mostly, though, it's a matter of space.I'm trying to avoid wires soldered directly to PCBs. Actually, I'm using screw connectors basicallyeverywhereso I don't have to deal with connectors, limited insertion cycles, etc. (Yeah, I know I still need crimped terminals on the stranded wire ends, but at least those are still "loose" wires.) And when it comes to AC wires, I'm usingbigterminals. (A lot of what I'm doing is fairly over-built, TBH.) If I were to plug mains directly into the board, I'd have to argue why I shouldn't be using the same 796740-2 that's directly upstream, not to mention I'm planning to use WAGOs for all the splices. So the real question is whether the optocoupler — or, more accurately, the entire setup I'd have using it — issmaller.Also, isn't an optocoupler (unless paired with a transformeranyway) going to use a lot more power? (Okay, ~100 mW isn't exactly a huge quantity, but I think the current sense approach only needs ~1 mW?)

author: PCB.Wiz
author_rank: Super Contributor
timestamp: October 11, 2023, 11:27:47 pm
content: Quote from: sparkydog on October 11, 2023, 10:58:51 pmYour SCH has a NPN transistor has a P-Channel MOSFET.You can delete the P-Channel mosfet, and use a PNP transistor with emitter to 12V. Coil then connects to 12V and PNP base resistor.Quote from: sparkydog on October 11, 2023, 10:58:51 pmYes, your added 2.2k B-E can change to set some trip level, based on VBE, and the 1k limits the base current.If you have a fast 1kV spike on the mains, even coupled through a small capacitance, a series resistor helps limit the peak current.

author: sparkydog
author_rank: Regular Contributor
timestamp: October 11, 2023, 11:31:41 pm
content: Quote from: Benta on October 11, 2023, 11:11:51 pmAh, so the problem is when you have aproperdead short? Okay.That said, I noticed that adding a resistor in parallel with the diode (as in my most recently posted revision) lets me "tune" the sensitivity somewhat, so that instead of triggering on basically anything that isn'texactlyzero, I can make it only trip on ~10 mA or more.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: October 11, 2023, 11:34:02 pm
content: Quote from: IanB on October 11, 2023, 10:23:23 pmOh, you meantinsteadof a transformer ?That violates the 'nothing inserted' requirement, and using a opto coupler for current sense, is non trivial, as the IR diodes inside a optocoupler need ~1V to turn on,If you have a 5A load, that's now 5 watts in a dropper, or an added isolated power supply added to power the LEDs.

author: Benta
author_rank: Super Contributor
timestamp: October 11, 2023, 11:37:05 pm
content: Quote from: sparkydog on October 11, 2023, 11:31:41 pmNo. A short on a current transformer is no problem, it's normal operation.Your mindset is still 100% stuck on voltage transformers. Try to break out.

author: Benta
author_rank: Super Contributor
timestamp: October 11, 2023, 11:41:14 pm
content: This would be the PNP implementation, meaning positive output voltage when current flows in the CT primary:

author: sparkydog
author_rank: Regular Contributor
timestamp: October 11, 2023, 11:59:42 pm
content: Quote from: PCB.Wiz on October 11, 2023, 11:27:47 pmD'oh. 😳 Yes, I meantP-FET, not PNP BJT. Too many acronyms that I haven't yet internalized sufficiently.I'm still not seeing how I can lose the FET without affecting operating characteristics, however. At least in simulation, every time I've tried, it results in the capacitor only charging to ~3-4V instead of 12V. I'm not convinced I'm following your attempts at a textual explanation. A drawing would be appreciated! (Seriously, it's not that hard to modify the sims I've posted; give it a try!)Quote from: Benta on October 11, 2023, 11:37:05 pm🤦 Right.To be fair, the problem wasn't that I was "stuck on voltage transformers", it's that all the talk about burdenresistorshad me thinkingresistanceis necessary for some reason. Actually, that made it worse, because Iwasthinking about how reallyhighresistance (no such thing as "infinite" resistance, especially in atmosphere) is bad.So, okay, let's try this again. It's not that you need a burdenresistor, it's that you needsomemoderate-resistance path for the current, right? And, obviously, a diode and BJT provide such a path. That makes a lot more sense...

author: Benta
author_rank: Super Contributor
timestamp: October 12, 2023, 12:25:15 am
content: Quote from: sparkydog on October 11, 2023, 11:59:42 pmSpot on.You need to look at voltage and current transformers as the inverse of each other and treat them as such.VT:Input: voltageOutput: voltageOpen output: goodShorted output: bad (current is excessive)Resistive load: goodConstant current load: goodWinding ratio: N:1CT:Input: currentOutput: currentShorted output: goodOpen output: bad (voltage is excessive)Resistive load: goodConstant voltage load: goodWinding ratio: 1:NVery simplified, but hopefully helpful.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: October 12, 2023, 01:25:11 am
content: Quote from: sparkydog on October 11, 2023, 11:59:42 pmSee #34.With a single element, the HFE will matter, and the cap R may need to increase.I'd also swap the shunt resistor from VBE to across the coil itself. That reduces a divider effect, and means the current set point is mainly set by one resistor.

author: IanB
author_rank: Super Contributor
timestamp: October 12, 2023, 01:46:01 am
content: Quote from: sparkydog on October 11, 2023, 11:22:09 pmSo here's my thought experiment, and maybe it's crazy, in which case it does no harm if you reject it. But a neon lamp (and series resistor) with insulated flying leads that connect to screw terminals where voltage is to be detected. The neon lamp is pushed inside a black plastic tube. Pushed into the other end of the tube is a phototransistor with two leads connected to wherever the logic input is needed. If the circuit is energized, the neon lights up, the phototransistor turns on, and the phototransistor drives the input to the logic gate (it pulls the input down or up as required), and detection is accomplished.Now, of course, you may not want to fuss about making such an assembly, or with attaching the neon leads to the screw terminals, but it is an idea, and it is about as compact as you could make it.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: October 12, 2023, 01:47:13 am
content: Had a quick play (could not find a way to spin a part) : A tuned version of #34, PNP HFE is set to 300, so appx BC857Chttps://tinyurl.com/yttx3c2gVarying the peak current by adjust of 10k test R, gives15k -> 1.48V pk on cap14k -> 5.62V pk on cap13k -> 11.972V pk on capimage of close to the limit (PNP is on for a few degrees) below

author: Circlotron
author_rank: Super Contributor
timestamp: October 12, 2023, 02:35:30 am
content: Quote from: sparkydog on October 05, 2023, 06:47:07 pmThe current output of a CT is proportional to the current input and the turns ratio. The voltage output of a CT is proportional to the secondary load. Zero volts for a S/C secondary and theoretically infinite volts for an open cct secondary. CTs are quite happy with a S/C secondary. Open CCT secondary can destroy them from over voltage, as has been mentioned.

author: Shay
author_rank: Regular Contributor
timestamp: October 12, 2023, 03:22:05 pm
content: Here is something that is close to what you're requesting. The goal of the following circuit is to indicate if one or more of the three phases have failed.The working principle is super straight-forward.Once the AC voltage at the input is high enough for the TVS to conduct (TVS are like back-to-back zeners), it passes enough current for the AC optocoupler leds to light up and produce a boolean output using the R+C to smooth the output.Maybe this principle can help you with your goal.

author: sparkydog
author_rank: Regular Contributor
timestamp: October 12, 2023, 03:49:14 pm
content: Quote from: Circlotron on October 12, 2023, 02:35:30 amSo... if youonlytoss a resistor across the CT, then youdoconvert current to voltage (per Ohm's Law). This can be sensed by e.g. an analog input pin of an MCU. I suspect the source of my confusion is that all the literature I encountered (including data sheets!)expectsyou to use this, and only this, configuration. (To be fair, I expect thatisthe typical usage.)Quote from: PCB.Wiz on October 12, 2023, 01:47:13 amYes, that's indeed a limitation.Someparts (mostly, ones with only two connection points) you can just drag the points around "however". Others take their orientation from how you first draw them and have to be recreated if you want them oriented differently.Quote from: PCB.Wiz on October 12, 2023, 01:47:13 am...except I chose the 2k AC load resistor to approximate my actual load. I guess your intent is to show what happens with other inputs?Also, fiddling with the burden resistor changes the sensitivity. My current iteration has a 3.3k, which gives a "detection threshold" of about 10 mA. As you note/show, this isn't an exact science, as near the threshold the BJT will let the capacitor charge to a lesser voltage. (If I cared, that would actually be an argument for keeping the arrangement where the BJT controls a FET.)Anyway, thanks! I might post the complete, updated circuit (probably as a new topic) at this point.