# Dumb "current transformer" questions

• posted

Let's say that I have a cable and I want to measure the AC current going through it. Up to, say, 100 amps.

I could use a current transformer, right?

If I have a say 200:1 current transformer, then on a 100 amp AC current it would want to produce a 0.5 amp current. Then if I stick, say, a 1 ohm resistor across it, it would produce 0.5*1 = 0.5 volts AC across the resistor.

Is that right?

i
• posted

Yes

AWEM

• posted

Yes, assuming that you haven't exceeded the primary or secondary current ratings for the transformer.

Furthermore, since you're probably looking to use it on your tig welder inverter, the transformer probably should be rated to be accurate for pulse work rather than just 60hz sine waves.

Here's more than you want to know about CT's:

Oh, and don't ever leave the secondary open with a load in the primary....

• posted

Thanks.

No, I will use it for my phase converter. For measuring current on the welder, I have a DC current meter on the welder's panel.

Thanks, that was all very helpful.

I have some CTs and I want to use them with some resistors and diodes to measure several currents in my phase converter, like input current and leg 3 current.

I would use this meter (cost \$19.99 on ebay):

My schematic would be

Diode with low voltage drop CT pole 1 ------------| Resistor Capacitor ~~ process meter CT pole 2 -------------------------------------------|--

I need the diode because this meter is for DC readings.

Since the voltage reading is screwed up by the voltage drop of the diode, I would need to find a diode with low voltage drop.

i
• posted

Wouldn't a hall sensor do a better job? I mean with all the square wave and changing frequency a current transformer seems to be not the best choice. And if he is in DC ...

Nick, just curious

• posted

How about this one? You just have to push the decimal point to the right. And it is exactly what you need.

Nick

• posted

The problem is way worse than that. Your circuit will measure the *peak* voltage of *1/2* of the sine wave (less the diode drop).

The reason is that the diode will only pass one side of the sine wave. But when it's passing that side, the capacitor will charge up to the peak value (rms * sqt(2)) For the nitpickers, yes, I'm assuming a well-formed sine wave.

The input impedance of the meter would also have an effect, but since it's an unknown, I won't go there.

So, 10 volts in would be 7.07 volts out.

The good news is that many times these process meters have scale and offset values that can be programmed. You could change the scale to correct for the 1/2 wave and cap and change the offset to correct for the diode. Not sure it's the best way, but if it tracked a clip-on ammeter, it may be good enough.

• posted

>

I see. I coud use a 4 diode bridge and subtract 2 diode drops.

Yes, it will be a normal sine wave.

It is a "known", actually, there is a table of impedances in the datasheet. 1 or 10 megohm depending on the range.

Yes, that's OK with me. This meter can be programmed to perform linear conversion y=ax+b.

All I have to do is bring various measurements (input current, input voltage) to similar scale, so that:

1) when I switch to voltage (say 240V), my meter displays "240", 2) when I switch to current (say 25A), the meter displays "25".

That means that I have to perform conversion of my voltages from the AC line voltage and from current transformers by bringing them to the same scale. It's a simple algebra problem and can be done with appropriate voltage dividers and also diodes.

Just as mine is.

I agree. I will mess around with it, hopefuly it is not going to be too difficult.

i
• posted

While this is correct it's not the best way to use a current transformer as a measurement device. This because most meters capable of reading a few volts AC FSD are messed up by the forward drop of diodes used to rectifiy the AC.

The trick is to use it as a true current transformer without an intermediate voltage transformation. Feed the output of the current transformer directly into a full wave rectifier - silicon diodes are OK because forward voltage drop is not important.

Short circuit the rectifier output directly through a DC AMMETER. The ammeter will then read directly the MEAN value (0.90 x RMS) of the secondary current. The voltage drop of the rectifier diodes will slightly increase the voltage drop at the current transformer primary but will not affect the current transformation ratio. For sine wave input waveform the equivalent RMS current should be read as 1.11 x the indicated DC value.

Jim

• posted

Assuming the transformer is capable of the current needed use ample wire so it won't heat. I wouldn't make the turns so high - but do as needed. You might have to adjust the high side a bit for calibration - but the easiest is to buy a 2 ohm 1 watt and adjust from 2 ohm to almost zero. Might want wattage higher if possible so if the value is low the current will be high.

The professional current transformers I have are 10 turns on a core and the user puts in the high current windings.

Martin

Martin Eastburn @ home at Lions' Lair with our computer lionslair at consolidated dot net NRA LOH, NRA Life NRA Second Amendment Task Force Charter Founder

Ignoramus29580 wrote:

• posted

I want to have a nice panel. I already have a clamp on ammeter.

i
• posted

Thanks, after some thinking I realized that it is excellent advice.

That's very nice. I looked at my current transformers today and realized that they are 200:0.1, not 200:1 as I incorrectly reported.

So, 50 amp current would translate into 0.025 amp current on the transformer.

With, say, a 50 ohm resistor it would amount to 1.25 volts across the resistor, or 0.03 watts of power dissipated on the resistor. I can get away with using a small cheap 1/8 watt resistor.

i
• posted

>>

Just get a standard 5A full scale AC current meter. There's a few problems with this diode stuff. If you want to measure upto 100A get a 100:5 current transformer. Connect it straight to the meter.

1) they don't less than about 0.6 volts though with one diode 2) using one diode like in the ASCII art diagram will magnetize the current transformer core. 3) the full rectifier will have a drop of 1.2 volts. You will not be able to read any voltages below this.

A 10VA current transformer can only output upto 2 volts at 5A. SO, with a 100:5 transformer and 50A though the wire though the donut, you won't even get a DC voltage out of a bridge rectifier.

• posted

>>>

I am confused. With a current transformer, I am guaranteed a 200:0.1 current even if I put a full diode bridge into the current transformer. Voltage drop should only result in extra power disspated on the bridge.

Current xfmr 1 ----_ / \ + - \ / Current xfmr 2 ----~

the current will flow in one direction between points marked + and

-. Right?

Then, I can simply install a resistor between + and - and measure voltage across the resistor (with a filter cap to decrease ripple).

Am I mistaken?

The reason for this is that I have to go with what I can get, and I could get a process meter with DC capability.

(datasheet and all)

I will check my current transformers again tonight, but I would think that with 200:0.1 ratio, they should hopefully be able to overcome the voltage drop. Perhaps I am completely wrong.

i
• posted

Then I would build a rectifier out of OpAmps that has no diode-drop.

So you want to have a R behind the xfmr and then a rectifier that rectifies that voltage across the R. Behind that a C and then into your DC V-meter.

Nick

• posted

Well, since it is Iggy that posted the question, he probably has a big bag of surplus CT's that he bought for \$.50 (:

• posted

No, I only have very few CTs, I need to use them for my phase converter measurements.

i
• posted

Never use a half-wave rectifier in a CT secondary. On alternate half-cycles it presents an open circuit to the CT secondary. If this happens, the CT secondary voltage will skyrocket to very high numbers on those half-cycles. You will very likely damage the CT or the connected circuitry from the high voltage pulses.

Randy

• posted

Ouch... Good catch.

• posted

That was not how the diode was supposed to be used.

My plan was to use a "burden" resistor on the secondary, and the diode/filter capacitor connected in parallel to the burden resistor, to accumulate a measure of average voltage across that resistor.

i

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