Calculating the Rt for a 1:100 20 Amp Current Sense Trans..

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Hi, I am setting up an Aluminum anodizing line and I have three 20 amp heaters and one 20 amp circuit available. I tested the on time vs off time for one of the tanks and it only needs 12 sec ON and 30 sec OFF so I built a sequencer to enable all three to be plugged in at one time. I am using 3 20 amp relays and one SSR (18 amp) and a tiny 45. I also have a sense transformer so that I know if they are on. Each heater has it's own thermostat. The plan is to switch SSR off, switch a relay on and then turn the SSR on, test for current flow, wait 10 seconds if on, wait 1 sec if off and then go to the next relay. My thinking is the 0 crossover SSR will give much more life to the relays and have a smooth turn on and off.

I tested the voltage output of the sense transformer and I only see 1.8mV. I calculated the resistor to be 20 Ohms to give me a peak 2.7VDC.

Quote:
3. To calculate value of terminating resistor (Rt) use the following formula:
Rt (Ω) =VREF * N / (Ipeak_primary)

Using N = 100 and Ipeak_primary as 20Amps I get:

20 ohms = 4VDC * 100 / 20Amps

At 16.6 amps for a 2000 watt heater I should see 3.32 V (minus 0.6V for the diode = 2.72V)

The circuit is connected like this. D4 is a bat85 diode. R2 is 20 ohms and the transformer is a PA1005 ( digikey link http://search.digikey.com/script... )

Where am I going wrong?

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That transformer is not for 50/60 Hz operation.

Check Digi-Key for the AC1030 current transformer, which is a 1000t unit designed for line frequency operation. If all you care about is detecting whether there is current or no current, a diode in the secondary circuit may be ok, but is unsuitable for making actual quantitative measurements.

Tom Pappano
Tulsa, Oklahoma

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I wondered what those frequency numbers were for :oops:
I guess these were for power supplies or something.

looking at the AC1030 I am confused. Do you just run one of the 120V lines through the center of this, wrap it around it? My off peak meter uses a big one like this but the datasheets don't show how to actually connect or wire the 120V.

I am also confused at the primary and secondary windings. I understand a transformer with a 1:1 gives the same voltage and 1:2 gives twice the voltage and 2:1 gives 1/2 the voltage but they directly run the 120V through the primary. Could you explain a bit how these specific sense transformers work?

I also see they have lower cost ones with 1:1000
http://media.digikey.com/pdf/Dat...

One of those would give 3.32V with 200 ohms, why would one use 1:1000 or 1000:1 ?

I will continue to read about transformers and await your reply. Also ordering some of each so I can better understand how they work.

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What you are wanting to use is a Current Transformer (CT), these have a specific turns ratio as Secondary:Primary.

You have to look at them as transforming current and that they must terminate into a specific load resistance to develop the desired voltage corresponding to the maximum current. The AC1030 data sheet has an example at the bottom showing 30 Amps through a single turn primary resulting 30 mA on the 1000 turn secondary developing 3 Volts across a 100 Ohm load resistor. Either the hot or neutral line can be run through the CT, running both through would cancel each other out unless they were wound in opposite directions. I'm a bit fuzzy as to what constitutes a single turn here on the primary, I suspect that a loop through the opening such that the incoming/outgoing leads form a straight line.

For industrial applications where high power equipment is being monitored shorting switches are provided to allow disconnection and calibration of the monitoring device. If a CT were to be open-circuited from it's load during operation the results would be quite disastrous for equipment and people.

Read wikipedia entry Current transformer and its links.

Stan

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Quote:
The AC1030 data sheet has an example at the bottom showing 30 Amps through a single turn primary resulting 30 mA on the 1000 turn secondary developing 3 Volts across a 100 Ohm load resistor.

Yes that is what is confusing me. The AC1030 turns datasheet shows 1000:1 and is called "TRANSFORMER CURRENT 30.0 AMP"

The other one I mentioned is 1:1000 called a "TRANS CURR SENSE 1:1000 20A" That makes sense as the primary:secondary because the single line of the AC would be the 1 in this case.

Why is the AC1030 called a 1000:1 there is not 1000 turns on the primary or is it notated differently secondary:primary. Perhaps it's made in Europe where they drive on the "wrong" side of the road :)

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By definintion the current transformer is sused in such a fashion so as to sample the current and give us acces to only a miniscule portion of it.

It is hardly ever ( i have never seen the CT beingused either without a load resistor or in reverse) used any other way. Therefore the ratio specification is not importat in the sensewhihc is primary which is secondary as the core in general only has the secondaru wound on it and the primary is implemented by looping the high current conductor through the core at the time the current transformer is installed.

Operating the CT witout load resistor would resuot in excessively high voltages on the secondary of the CT.

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If you really want to make life easier, take a look at these:
http://www.allegromicro.com/en/P...

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I wonder what those SO8 leads will do while waiting for a 30 amp circuit breaker to trip during a short 8-)

Tom Pappano
Tulsa, Oklahoma

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Hey K5HJ I have samples of ACS755SCB-050-PFF they are now obsolete how about that. i thought they were only for DC, can I run AC through it?

I had found a circular inductor coil in my stuff and I put the AC line through that coil wrapped around twice and I get 0.078V at 1200 watts so i was going to use an op amp to detect the voltage but if the allegro device will work heck, I will use that. Looks like I will get 1/2 the wave when it goes from pin4 to pin 5 on the device and nothing on the second half of the AC sin wave correct?

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metron9 wrote:
Hey K5HJ I have samples of ACS755SCB-050-PFF they are now obsolete how about that. i thought they were only for DC, can I run AC through it?

I had found a circular inductor coil in my stuff and I put the AC line through that coil wrapped around twice and I get 0.078V at 1200 watts so i was going to use an op amp to detect the voltage but if the allegro device will work heck, I will use that. Looks like I will get 1/2 the wave when it goes from pin4 to pin 5 on the device and nothing on the second half of the AC sin wave correct?

Yes, they will work with AC. The output of the device rests at 2.5V and swings + and - from there.

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Quote:

I wonder what those SO8 leads will do while waiting for a 30 amp circuit breaker to trip during a short

What are you shorting to? How much current are you then drawing (during this short) through your wires and other devices? How long are you going to short before the trip?

The short answer is that if you apply that SO-8 device in the type of circuit that is appropriate for that device -- nothing bad will happen, within a reasonable set of conditions. IIRC that is about a 20A or 30A max device, and withstand current is something like 5x for some milliseconds.

OP is somewhere around the max for the ACS712 that was suggested. Didja see the nice big tabs on the one the OP pictured? The answer there is a definite "nothing will happen".

You can put lipstick on a pig, but it is still a pig.

I've never met a pig I didn't like, as long as you have some salt and pepper.

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The data sheet for the ACS712 says 100A for 100ms. I have seen P&B T92 (dual pole) 30 amp relays *completely* blown to pieces when the load on a 30A breaker "protected" 240V circuit went to dead short. Their internal construction is a little more substantial than a surface mounted SO8, so it just makes me wonder about the long term survivability of such a setup.

Quote:
Didja see the nice big tabs on the one the OP pictured? The answer there is a definite "nothing will happen".

Of course, that one isn't the SO8 8-)

Tom Pappano
Tulsa, Oklahoma