Analog signal conditioning from Current Transformer Output

[jnnewton]

I a have a current transformer, that runs on 120-480vac, 0-25A. on the secondary side i want a 0-5 V output. I've designed and simulated a circuit where I have one leg of the ct going to a voltage divider 10k/10k on 5V to 0V. This gives me a signal centered around 2.5V, which now that i think of it i should bump that up to about 5V, to let me operate without a negative supply, or change the burden resistor to output less mV/A.

So then i started with the rectification / filtering and get a signal out that is 2.5V out @ zero A input. This is the schematic that I've posted.

The current transformer datasheet is here: https://www.mouser.com/ProductDetai...=sK90oOEhQaVS76c2z9pE3BH5d3VZpkTntGf143uYsGE=

The output is 110mV / A, so i want to design for the full 30A in case I go over. That would put the output at 110mV / A * 30A = 3.3V and less than the max output of 4.0Vrms. I could change that output by changing the burden resistor, but not sure how that would change the signal chain after it.

What I need help with:
1. I can't seem to work out how to work a difference amp into the circuit and get it to work. Maybe I just can't figure out the correct resistors.

2. is there a simpler way to do this, seems 3 op amps is more than should be necessary.

 

[MrAl]

Hi,

Part of this would be based on what kind of accuracy you are after. Many applications simply use one 'burden' resistor and two diodes using a center tapped secondary. You could even use one diode for half wave rectification if your load is always symmetrical.

The burden resistor goes by the turns ratio and the required output, and the limit is set by the number of turns and the core size.

For example, with 1000 turns on the secondary and 1 turn on the primary, with 1 amp in the primary you'll get 1ma in the secondary so a resistor of 100 ohms would provide an output of 0.1 volts per amp primary current. If you switch to 200 ohms then you get 0.2 volts per amp primary current. It's that simple, but there are a couple catches.
One is that the limit of the core can not be exceeded. This is where the core starts to saturate because the output voltage is too high. The makes the output response non linear with input amplitude near the high end of the scale. For good linearity a relatively low value resistor is used.
The second is if you use a diode or diodes the output is somewhat non linear near the low end of the scale. An op amp or two for rectification can help this situation but that's if you need good linearity near the low end...if you dont you may get away with just diodes. This works well for a lot of applications because it's more rare to need to measure very low currents when very high currents are normally measured. Also, the burden resistor is connected after the diodes to improve the linearity.

 

[jnnewton]

I'd like 1% accuracy, I could deal with 2% accurace, most importantly on the low end (Bottom 2%). The limit of the ct is 30A input 4Vrms output. Does that help to clarify my problem?

 

[MrAl]

Hi,

I took a look at the data sheet and i see they already spec the max output as 4vrms. So they recommend 60 ohms so that at 30 amp input the output is 3.6v so that's well enough under 4v. So you are sort of stuck with 60 ohms unless you dont need to go up to 30 amps on the input side. For 3v output (30 amp input) you'd use a 50 ohm resistor. If you only need 15 amp on the input you can go to a 100 ohm resistor. A 1/2 watt unit is usually used for ruggedness for lower inputs and for power handling in some cases.
For that kind of accuracy you are probably better off using op amps rather than just diodes, but it doesnt hurt to try.
You might also have to incorporate some calibration routine into the process.
They seem to be cheap enough if you have to use two.

 

[jnnewton]

Yes, all of that seems correct to me. And you're right, It doesn't hurt to try the diode route, but I'd rather be safe than sorry. I'd hate to get close to the accuracy I need with diodes, and then it fall out of tolerance 6 months from now. I'm going to stick with the 60 ohm resistor for now for the 30A input, and will increase that for lower current ranges It is a 1/4W right now, which I could bump up, and will do based on your suggestion. My real problem is after that portion (CT / burden) is done, which I think it is, barring the upping of the wattage. I can rectify and filter, but am not enough of an analog expert to shift that signal down so that it's in the range of my microcontroller without using a negative supply. Right now, with 110mV / A and a 2.5V starting point, I run outside of not only a 3.3V mcu analog input, but a 5V one too. I haven't decided on the mcu, but this is a problem

 

[MrAl]

Hi,

What do you mean by a 2.5v starting point? What is this and what causes this?
Oh is that because you use a voltage divider?
If you use diodes you wont need this.
If you want to use op amps, you can go half wave (symmetrical load). This also affects speed of response which im not sure what you need. Many apps dont have to be super fast and are symmetrical too.
There might be another solution that does not required a voltage divider, or a voltage divider has you have it but with slightly different op amp configuration that allows a 0 to 5v output for a symmetrical 2.5v about ground input signal. Once the signal is rectified with the op amp(s) a stage that provides offset correction can be used and that should take care of it. This would be a simple addition or modification.
For example, it looks like if you run Vref to ground (0v) instead of a voltage divider you'd get a full wave rectification of the input. Not sure why you have that running to 2.5v anyway.

One other thing you should note as shown on the data sheet, is that the response drops off rapidly after 20 amps. At 30 amps the response is down to 70 to 80 percent of nominal. You might want to think about this. If you are using micro controller then you can use calibration data to correct for this, also for the physical unit you actually get. It might be a good idea to make some measurements and plot the input vs output and use that data to program the uC for the calibration.

The response should not change too much over time just because you use diodes, although there may be a small amount of temperature drift.

For a more exact output value, the secondary series resistance has to be considered. That's probably why they are saying 110mv typical rather than 120mv.

 

[jnnewton]

MrAl,
Thanks for the response. To answer your questions:

2.5V starting point is due to the biasing of the transformer and is the cause.
Yes, it's due to the voltage divider.

And to respond to your comments:
I'm not interested in using diodes.
I do want to use op amps.
I'm not sure what you mean by symmetrical load half wave. I'm already slowed down quite a bit due to filtering, and the rectification seems to work fine. I don't see any issue with the rectification and filtering in my tests, or in your comments.

You mentioned a slightly different configuration to give me 0-5V for a symmetrical 2.5V input. This is what I'm looking for. This is the problem I can't solve.

I have it running @ 2.5V center due to wanting to keep the signal above 0V. I'll probably need to bump this up to 5V or so for the full range, or change the burden resistor.

I'm aware of the drop off, and I know how to handle that.

I hope I covered everything. Hope that helps clarify.

 

[MrAl]

Hi,

What do you mean the biasing of the transformer? Your circuit shows the 'transformer' connected one side to ground. That's not biased. The only biasing i see is the 2.5v divider, which can be eliminated. Connect that Vref to ground instead. That gives you a rectified output out of the first op amp stage. For a symmetrical 2.5v peak input you'll get a 2.5v peak DC output, then filtering will bring this down a bit due to the averaging effect.

You dont have to keep the input signal above 0v. The precision clamp takes care of the negative signals without the need for bias, of course this is without the 2.5v divider as mentioned above. I guess i thought you realized this already and that was the reason for using this kind of rectifier circuit.

So assuming your current transformer secondary is not connected to anything but this circuit, what you would do is connect one side to ground, the other side to the input of the circuit. You also remove the 2.5v Vref divider and ground the point labeled Vref. Your final output will then be the average of the 2.5v peak wave or something close to that, which would be around 1.6 volts DC. If you want higher than that we'll have to adjust the gain a little bit.

By symmetrical with respect to a half wave output, what that means is that if the load draws a symmetrical AC current (as many loads do) then a half wave rectifier can work ok in many cases where speed is not a huge concern. But it's important that the AC current in the LOAD be symmetrical otherwise the half wave rectification does not adequately represent the true load current.

 

[alec_t]

Here's a possible circuit, but even though it uses a 'precision rectifier' approach it's error is > 2% for low current inputs.
The current transformer's burden resistor comprises two halves, R3/R4, to allow full-wave rectification without the need for a dual-polarity supply. There are two peak rectifiers; one for the positive half cycles and the other for the negative ones. Separate peak detectors allow for asymmetry of the measured current as MrAl mentions above.
D3 and D4 prevent the non-inverting inputs of opamps U1, U2 from going negative by more than ~ 0.6V (which would cause unwanted output signal inversion). Peak voltages on C1, C2 are averaged by R9/R10/C3 and the average is scaled by opamp U3.

 

[MrAl]

Hi,

Alec, he doesnt really need a new circuit he needs to know how to use the one he's got. That circuit should work just fine.

 

[itshayhaysturn]

try to open your circuit design and i just get a blank page i would really like to get it. I am working on a current sensing circuit i need it to monitor 120v ac 60hz using a current sensing transformer, the load can be anywhere from .5 amps to 12 amps. This is how i need it to work i turn the load on whatever the load may be at that time and I simply turn a pot until on green led comes on now the load is good as long as the green light is on. lets say the some coponent on the load burn out now a red led should light on the circuit notifing me that something has changed. I find and fix the bad component and the green led comes back on. all components on the load will not have varying amp draws they will be constant. However one component might have .5 amp draw and the other might be 150 ma or another might be 1 amp. so i just wanted the circuit to sense 100 ma change no component will draw less. for instance the load is lets say 5 amps so the voltage out of the current transformer will have a value and i just compare that value to a comparator and if ever in the future current sense transformer value changes then the red light lights. I really em completly lost on the electronics and the math and a/c d/c frequency all of it, I have blown about 30 components a power strip and one breaker and a whole lotta leds. i just purchsed a cst 1015 being shipped to me but I understand the output will be a/c do i need to convert that to d/c if not dont know how ill compare the voltage from it to the supply voltage for my chip when the voltage changes.

 

[atferrari]

Could you separate that horrible mass of thext in paragraphs?

When you speak, you do it that way?