I would like to generate an AC current in the 100 amp range and unsure how to do this. I do not want to use multiple coil turns but a single conductor that can carry this current. I have looked at a few transformers and a fairly large one I have was rated for only 10 amps. I will have a low impedance load as I only need the current but dont need to drive a load. Can anyone please advise me on how to do this? thanks

## Generating High Current

get a stick welder...

It depends also on how many watts you need from the supply.

Perhaps is time for an SMPS. Low voltage output for low power, and current regulation instead of voltage.

Anyway, it looks dangerous, at least for the switching transitor/s.

Guillem.

Yes, the big question is "how many watts"? 100amps at 0.1V is only 10W; thats not so bad.

Of course, you have to be careful about wire size.

Another big question is where does the power originate? Mains? Battery? Is the source AC or DC? If AC, what frequency and is it the same at the load?

Jim

The power will come from a US wall plug 110VAC 60 Hz. I was actully thinking of making something that can be either AC or DC but the AC part is a priority right now. As far as a load, I dont really need a load, only a current source as it will be use to test current transformers (CT's). These CTs connect to a conductor and the output is a 0 to 5 volt signal that is proportional to the current going through it. This 0-5 volts is fed to a controller or ADC only for knowing the current in the conductor. This also means I dont require much power.

Warning - Dangerous Experiment!

Put a single turn around the core of a medium size transformer (see photo). This should really be a substantial copper strip, but since I didn't have any handy for a photo, I improvised with folded aluminum foil. You could use a strip of aluminum cut from a soda can, doubled if necessary.

current out = (number of primary turns) * current in

To discover the number of primary turns, wind 10 turns of thin wire around the core, fire the transformer up and measure the voltage across the 10 turns. Take care not to short it out!

Number of primary turns = (110 / measured voltage) * 10 turns = N

This is a fairly vital measurement so if you can't get a good voltage reading with 10 turns, try 20, or 50, or whatever it takes; but don't forget to substitute the right number in the equation above.

The voltage across a single turn secondary will be 110/N, call it "V", so now you can calculate the load RL for 100A -

RL = V / 100 ohms

This will be very small indeed, and very difficult to make, so instead put a 10 watt load resistor in series with the primary, equal to N*RL. For the best accuracy, measure the DC resistance of the primary and include that as part of N*RL. Now when you short the secondary turn, the primary current will be limited and hopefully nothing will catch fire. You can also measure the primary current and calculate the secondary current, for a check on your equipment under test.

## Attachment(s):

You won't have a current source, in the traditional sense, from a transformer.

There will be an open-circuit voltage (read by opening the secondary and reading the voltage). The current will then be set (mostly) by the total secondary resistance (including the winding resistance.

You should have a fairly repeatable current, even if it is not constant. However, at these high current levels, you will have to watch out for the non-repeatable resistance of the hardware used to open the secondary to put the CT on (if the CT is not split core).

Jim

**new2use wrote:**

I would like to generate an AC current in the 100 amp range.

**new2use wrote:**

I will have a low impedance load as I only need the current but dont need to drive a load.

**new2use wrote:**

...I dont really need a load, only a current source...

If you read the three above quotes, there is something intrinsically incorrect in the concept.

1. "...generate an AC current in the 100 amp range.", in itself implies a 100 anpere load.

2. "low impedance load... ...but dont need to drive a load." How can you have a low impedance load sucking 100 amperes of energy out of a power source but not be driving a load?

3. "I dont really need a load, only a current source." If you have no load, no current will be required.

I can't seem to grasp the mis-conception here.

I think a better understanding of the concepts regarding power, voltage, current, resistance, current sources, current transformers and possibly power supplies in general are imperative here.

If these concepts aren't well understood, 100 amperes represents very unwelcome circumstances - especially in the open secondary of a current measureing transformer.

Be extreemely careful about that, which you are about to undertake, your life may well depend on your understanding of these concepts.

Be warned!

Carl makes a good point and let me amplify on it.

The single turn winding >>IS THE LOAD<< (plus what ever is coupled into the loop by the CT). I think what the OP is suggesting is that no OTHER load (such as a resistor, etc) is needed in that application.

Core loss and low coupling coefficient may come into play here more than one might expect. Both will have the effect of adding to the apparent resistance of the single-turn loop such that the loop current is less than is anticipated by the simple open-circuit voltage and winding resistance. About the best that can be done is to build one and check the current with a pre-calibrated CT.

Further, it will be very difficult to design the thing to get a specific current. The OP asked for 100A. I would be hard pressed to start with any transformer in my posession (and I have more than a few) and pick one and add a "winding" and know that it will be anywhere close to 100A. Again, about the best you can do is build something, check it with a CT, then try again.

Jim

As a suggestion, a shunt resistor used to measure current may be the better alternative. They are about the same cost as a CT. In addition, a potiential transformer would be used following the shunt resistor.

The shunt resistor will output so many milivolts per ampere. The potiential transformer would then be used to safely provide isolation, Substantially smaller voltage potientials and, some small amount of scaling. From there, the AVR can do the rest of the scaling, if need be.

Again, if the end goal is to measure up to 100 amperes of load current using a CT in an existing system, consider that the CT may have a voltage "Step Up" ratio of several hundred to one. When the secondary winding (the low current side) opens, the voltage may be several thousand volts. In addition, the balast resistor (the resistor that develops the voltage drop to something tollerable in the CT secondaty side) will dissipate a fair amount of heat and may be of considerable physical size.

To create a CT from scratch, there is a good amount of magnetic theory and mathematics involved. In fact, I believe (its been quite a few years) there is some calculus involved somewhere in the process.

A shunt resistor and potiential transformer is limited to fairly simple algebra substantially smaller voltage potientials and lends itself to a "Less Elegant", but more "Robust Solution".

Hello,

I can appreciate all the advice given thus far. Is he building a CT or is he using one that is commercially available? I am unclear on this point. If he is using one "off the shelf" he should be able to test is with any steady state resistance device (space heater, light bulbs in parallel) without needing the dangerous 100 amps! How about something around 10 amps instead? THis should be fine for his calculations. Guaranteeing 100 steady amps would be very difficult, unless an AC arc welder was used (as suggested above), still, even this amperage would fluctuate with arc length. I would forget trying to get one hundred amps. Current KILLS!!!

John

The interesting thing is, it needen't be that complicated.

Check out the devices at the Allegro-Micro web-site.

http://www.allegromicro.com/en/P...

I have experimented with a 50 ampere device and it worked quite well.

My impression from the second post is that this is an attempt to test or calibrate CTs.

I dont really need a load, only a current source as it will be use to test current transformers (CT's). These CTs connect to a conductor and the output is a 0 to 5 volt signal that is proportional to the current going through it. This 0-5 volts is fed to a controller or ADC only for knowing the current in the conductor.

Jim

**new2use wrote:**

The power will come from a US wall plug 110VAC 60 Hz. I was actully thinking of making something that can be either AC or DC but the AC part is a priority right now.As far as a load, I dont really need a load, only a current source as it will be use to test current transformers (CT's). These CTs connect to a conductor and the output is a 0 to 5 volt signal that is proportional to the current going through it. This 0-5 volts is fed to a controller or ADC only for knowing the current in the conductor. This also means I dont require much power.

So, this is a power supply (on the form of an adjustable current source) that needs to develop 100 amperes, in order to test the CTs at the rated current.

That being the case, you will need a load capable of developing a 100 ampere load - there is no other way around it! In order to have current flow, you need a voltage source capable of delivering the required current load.

Current is a function of I = E / R

In order to develop 100 amperes of current flow, you will need a given voltage, say 100 VAC divides by 1 Ohm. 100 Volts / 1 Ohm = 100 Amperes.

10 Volts / 0.1 Ohm = 100 Amperes.

1 Volt / 0.01 Ohm = 100 Ampers.

But also consider that the power dissipated in the resistor is: P = I^2R so:

P = (100 Ampers x 100 Ampers) x 1 Ohm = 10,000 Watts

P = (100 Amperes x 100 Amperes) x 0.1 Ohms = 1,000 Watts

P = (100 Amperes x 100 Amperes) x 0.01 Ohms = 100 Watts

For a 100 VDC power supply, it needs to be capable of handling: P = E x I = 100 Volts x 100 Amperes = 10,000 Watts.

If using a shunt resistor seems to confusing, then consider the situation and realize that I have only touched the surface.

If you are still wanting to use a CT, then also consider that the resistance values called out in the equasion: I = E / R above, might be typical load resistances that will be required to develop a 100 ampere test current needed to verify the CTs that you want to test.

If this is a production situation, one where, you will be required to test CTs as a function of product accuracy, you will be required a "Real" 100 Ampere test current for accurate verification of system calibration.

I have brought a lot of confusion and doubt into this thread. Perhapts, that is because there is a lot of unclarity within your initial post as to what really needs to be accomplished.

While taking a short smoke break, the following came to mind.

1. The primary winding of a CT is the wire, whose current value is being measured and that current is represented by the voltage drop across the balast resistor across the CT secondary. The secondary output voltage is usually expressed as milivolts per Ampere.

2. You can, in fact test the CT at say, 10 Amperes or even 1 Ampere. But, if the measured secondary voltage is in error by say, 1%, the error at a "Real" current of 100 Amperes may now be off by as much as 10%.

But there is yet one other way to test the CT using say, 10 Amperes, or even 1 Ampere. But this approach isn't conducive to production testing. One adavangage here is that, rather then needing 1000 MCM wire for the CT test, 14 gauge wire can be used for the test simulation of 100 Amperes.

Again, the primary of the CT (the wire whose current flow is being measured) is passing thru the CT core, usually only one time. So, simply passing the primary wire thru the CT core 10 times will reduce the test current from 100 Amperes, to only 10 Amperes.

As you can see, 10 turns of wire thru the CT core won't be as burdensome in a production situation as say, 100 turns of primary wire and a 1 Ampere test current.

So, I have given you many things to think about. Most noteably, you need to figure out how to perform a full scale test function, using a current value that is as small as possible. Using high voltage at high currents, especially in a production situation, is not something you want to expose production personell to - at any time!

While I was having my smoke, I realized that I had to solve a smiliar problem about three years ago, just not at 100 Amperes. Rather, using a 30 Ampere CT, but needing to measure 5 (full scale) Amperes accurately. 6 turns of 16 GAa. wire thru the CT core worked perfectly.