Is the power rating of an SCT-013’s burden resistor known? It’s not in the datasheet tables.
What should I be concerned about, if I push the current limit, say 20A RMS with a -015 CT?
Actually damaging the CT/burden? Non-linearity?
Thanks!
Welcome, David, to the OEM forum.
I’ve never actually tested any of the SCT-013-xxx voltage-output c.t’s, so I’ve not looked inside and don’t know the size of the burden; but as they are the same basic design as the -100 50 mA output type (but with slightly fewer secondary turns), I think I can say with reasonable confidence that you’ll be running into saturation before you need to worry about the rating of the burden. The rms burden voltage of the SCT-013-000 flattens out at about 160% of the voltage at rated current, so about 2.5 times the dissipation.
My notes from contributions by other forum members say the secondary is 1860 turns and the burden is 124 Ω, so I make the dissipation at the rated current to be 8.06 mW, so unlikely to be ever more than 25 mW, and that’s below the smallest SMD resistor I can see in the Farnell catalogue at 31.25 mW.
I think you’re unlikely to damage either the c.t. or the burden unless you operate it at well over the 133% of the rated current that you mention, but it will read low and increasingly so as the overload increases.
I’m curious to know why you’d want to run a 15 A c.t. at 20 A, unless you’re worried about overloads or inrush currents.
Thanks for the reply, Robert!
Most circuits I’d be monitoring wouldn’t exceed 15A, but a few are rated 20A.
I’d like the greater output, higher resolution, when available.
Also, so I can purchase just one model CT.
I planned on rectifying and (slow, ~25mS time constant) filtering the output, reading directly in A RMS, or essentially that. While keeping within 0 to +3.3V ADC range.
The maximum realistic voltage for a 3.3 V ADC is 1.1 V rms with a pure sine wave, after you’ve taken all the tolerances into account, so at 133% overload, you’ll be clipping at the ADC input unless you add a second burden in parallel to lower the output voltage, about 560 Ω.
What sort of rectifier do you have in mind for your true rms output?
First off, I don’t claim True RMS conversion. I’m looking for a fairly basic current measurement.
I’d use a quad Schottky bridge package, followed by a 50k/100nF RC filter.
You might be better off using a true current-output c.t., with the burden on the d.c. side of the rectifier. The c.t. would still drive the current it wanted, overcoming the diode drops automatically. It should be linear to a much lower current with this arrangement.
Good point! Even with Schottky’s, diode drop is significant.
I can get SCT-013-000 for the same cost, but I think it’s a much higher turns ratio.
Rated 100 A : 50 mA, so 2000 : 1
A 40 ohm burden would get me back to 1V/15A and I can scale from there.
Not that much: 2000 vs 1860, about 7½% more.
You could also look at the Talema range of ring-core from RS, if you want a physically small device:
https://uk.rs-online.com/web/p/current-transformers/7754909 or /7754903 or /7754912 25 A / 40 A / 50 A, Secondary 50 mA, good to 1.6 V. (Ignore ±400 A Output on the website, it’s wrong.)
also physically smaller:
https://uk.rs-online.com/web/p/current-transformers/1243897 50A:50mA (good to 1.6 V)
https://uk.rs-online.com/web/p/current-transformers/1243896 25A:50mA (good to 1.6 V)
Huh… I’ve been going on the basis of calculating the turns ratio from the scale factor and someone’s value given for the burden.
I worked it for the SCT-013-015 as:
Output voltage =1V at 15A, with 10 Ohms burden => 100mA in the resistor.
Then, 15A / 100mA = 150:1
How did you arrive at 1860?
Is there a table available of the number of turns for the SCT-013 family?
The value of 1860 was reported by Bob Lemaire, a former contributor and confirmed for the -030 by a second contributor George Yundt.
I don’t know where they got the number from, but it’s not hard to take an SCT-013-xxx apart, remove the burden and measure the current ratio to derive a reasonably accurate value for the number of turns. In any case, it’s likely that the secondary turns will change by a small amount from batch to batch as the manufacturer endeavours to achieve the best overall accuracy over the specified range, due to changes in the properties of the core.
The SCT-013-015 uses a 124 Ω burden, according to my information. That number has probably been read off the burden resistor itself. I don’t know why you thought 10 Ω, because I’d already told you the value in post no.2.
Not that I’m aware of. We’re not really interested in the number of secondary turns anyway. Certainly In the UK, and probably in the present and former Commonwealth countries, current transformer ratios have always traditionally been specified as a ratio of rated currents, not turns. So the SCT-013-000 is 100 A : 50 mA, whereas a 2000:1 c.t has a 2 kA primary current and 1 A secondary.
If you’re buying a voltage output c.t, then this is specified as the ratio of rated primary current to the corresponding secondary voltage, again it’s the physical value of the input and output quantities, not dimensionless numbers.
And I must have missed that - sorry! That makes a lot more sense.
1V @ 15A → 15A/(1V/124R) = 1860
This is also reasonable in my mind, with all of the series using a “nominal” 2000 turns, and adjusted ad-hoc for the core characteristics.
Thank you for all the explanation - it really does help.
Dave
As a follow-up, I received two SCT-013-015 CT’s I had ordered.
Took one to work and clipped it into a test we were already running.
This was applying a 5A pulse, 40ms width at 2Hz to a test article.
Scoping the CT output, I read 330mV peak on the pulse rise.
0.33/1V=33% * 15A = 5A, as close as the scope trace allows.
Also, the parallel combination of the coil and 124 ohm burden reads 53.5 ohms (DC), for a coil series resistance of ~94 ohms.
Any guess on the coil’s inductance? My LTspice model of the transformer is currently 1 uHy for the primary conductor, or 4Hy secondary, which seems pretty high…
Interesting numbers. 
Clearly wound with the same wire as the 100 A : 50 mA one - no surprise there.
I’ve been meaning - almost ever since I joined OEM - to develop an accurate model of the SCT-013 series c.t. And guess what - we’ve moved away from them and it hasn’t happened yet.
What I do know about the inductance is the frequency response only just includes 50 Hz, and starts to roll off very soon below this. If you look at Robin Emley’s website, you can see what I did for him in an attempt to equalise the response a little to make the measurement of the first half-cycle a bit more accurate. It’s the Tech Notes page, Faster Control Algorithm Updated (March 2021).
Thanks, again, Robert!
ROM, what does the low frequency roll-off point suggest for inductance, or are there just too many variables to form an estimate?
Where would I find Robin’s website?
Dave
I can’t remember now - it was 2016, a good few years ago. I was really interested in the top end from the harmonics point of view. I know I looked below 50 Hz, but I didn’t record any numbers.
Puzzling over the mk2 PV Router
“surplus energy flows away to the grid and is of no benefit to the PV-owner”
Oooh! No credit for back-feed to the grid! Shameful.
We don’t get paid full price/kWHr, but it’s definitely something!
That was in the days of Feed-In Tariff. The same amount was paid for generated energy, whether it was exported or used on-site. So clearly it was beneficial to use as much as possible to avoid imports.
Most Ferraris meters here had a mechanical ratchet so that it wouldn’t ‘unwind’ when exporting (meaning the consumer was in effect paid the same for export as they were charged for import), but I heard of at least one person whose meter didn’t have the ratchet and did unwind. They were onto a winner - the supplier should have checked, but didn’t.
But there’s worse: I understand in Spain NO export is permitted, and the customer is penalised if they do. I don’t know the thinking, maybe grid stability?
My guess would be safety, if the grid is down, AND the inverter still runs, without sync.
Lineman could get a big surprise…
That’s hardly likely. As far as I’m aware, all inverters have to obey very strict anti-islanding rules for exactly that reason, and must shut down very promptly when they lose the grid connection, and aren’t allowed to start up until the grid is present. But I don’t know how they sense this.
I’ve been trying to think of a way to trick my micro-inverters into powering up against a small AC source.
It would be on me to disconnect from the grid, in that case.
This past winter, we bought a generator, and haven’t dared to try syncing up to it…