AC Current Source

Discussion in 'The Projects Forum' started by MVIBoss, Jan 21, 2007.

  1. MVIBoss

    Thread Starter Member

    Jan 21, 2007
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    Hello,

    I need to build a simple, inexpensive, variable AC current source for motor overload relay testing. I want to source 0-100 Amps, 60 HZ

    I plan to simply backfeed a Current Transformer with a 120 Volt Variac, taking all necessary safety precautions. Then pass one or multiple turns of wire through my CT primary to feed my motor overload relay.

    I am curious if anyone has suggestions to improve this technique, such as CT Ratio, burden rating of CT, applied secondary voltage (more or less than 120 VAC), window versus bar type CT, or anything else that comes to mind.

    Thank You!
     
  2. mrmeval

    Distinguished Member

    Jun 30, 2006
    833
    2
    Do you mean you will output the voltage from the variac to the primary of the CT?
    And with that you'll create a secondary and use that voltage as a sense line to the relay?

    If so it should work though I'd suggest a current meter in the primary loop and perhaps a fuse or breaker matched to the limits of the lesser device.
     
  3. MVIBoss

    Thread Starter Member

    Jan 21, 2007
    13
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    My intent is to apply the variac voltage to the CT secondary, and pass a loop, or several loops of short circuited wire through the CT Primary. I will insert the overload relay that I want to test directly in series with that loop of wire.

    I am attempting to create a variable AC current source, to test current sensitive relays used for 3 phase motor protection.
     
  4. fanie

    Active Member

    Jan 20, 2007
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    Hi MVI

    I have a different solution that may be a nicer and more elegant solution. Do have a look at Allegro's hall effect current sensors, you get some 150A ones and overrange specs is great too. I've used some and I am very impressed. The mains is isolated from the electronics and makes it easy to use any kind of electronics with it, in my case I use pic micro's. If you use triacs to switch the mains power on / off the overcurrent can be switched off as fast as one mains cycle when overcurrent is detected. A contactor is usually somewhat slower.

    Almost forgot... CT's suck ! Had lots of troubles with them... oh, lotsa reasons.
     
  5. MVIBoss

    Thread Starter Member

    Jan 21, 2007
    13
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    Thank You for the reply, Fanie.

    I am not an electronics design engineer, but it sounds as if you are suggesting I build a Solid State power supply, using Hall Effect sensors as the feedback element for regulating my current output.

    At this time I need to stay with a magnetic solution, centered around the current transformer. Thanks for the information on the Allegro products. Very interesting web site!
     
  6. fanie

    Active Member

    Jan 20, 2007
    63
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    Mmmm... fear for the unknown. It's simpler than you think, but there are electronics involved. There may however be someone in your area that could help you with it ?

    If you say so, you're the boss !
     
  7. MVIBoss

    Thread Starter Member

    Jan 21, 2007
    13
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    I can assure you that it is not fear of the unknown. As stated before, I have a cost constraint.
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I'll preface this by saying that I have never used a current transformer. The following is just based on my understanding of transformer theory, and what I have learned over the years by reading about current transformers.

    A current source is characterized by high impedance (infinite, ideally). A transformer's output impedance is N^2 times the source impedance. A current transformer works because, in the normal direction, there is (typically) one turn in the primary and many in the secondary. For example, if there were 100 turns in the secondary, the impedance ratio from primary to secondary would be 1:10,000. If you turn that around, 1 amp in the primary would give you 100 amps in the secondary, but if your source impedance were 100 ohms (about the most you can tolerate in order to get 100 amps out), your output impedance would be 100/10,000, or 0.01 ohms. This is a voltage source (1.2V max), not a current source. If you want more voltage out, you have to lower the source impedance, which will further lower the output impedance. The output current will be highly dependent on the load impedance. Maybe that's OK, if your load impedance is constant. Just be aware that you are making a low-impedance, low voltage source with high current capability, not a current source in the classical sense of the term.
    NOTE: The inductance of the multi-turn winding may affect your source impedance.
     
  9. MVIBoss

    Thread Starter Member

    Jan 21, 2007
    13
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    Hello Ron H,

    You really have me thinking on this one .... :)

    First, in addition to not being an Electronics Engineer, I am also not an academician. ;)

    CT ratings are expressed as the current ratio of the primary to the secondary. For example, I constructed my high current test set today by obtaing (for free) an 800:5 ratio Bar Type CT. 800 amps input results in 5 amps output. From the transformer equation:

    N1*I1 = N2*I2
    where N1 = primary turns, I1=primary current, N2=secondary turns, I2=secondary current.

    N2 = (N1 * I1) / I2 = (1 * 800) / 5 = 160 Turns

    So you are correct in stating there are more turns on the secondary, but to be consistent with the transformer equation, there is less secondary current flow. And indeed, this is true in practice. This is contrary to your assertion that "1 amp in the primary would give you 100 amps in the secondary"

    In regards to circuit impedance, "real world" current sources are modeled as an ideal current source in parallel with its source impedance. (The dual to the ideal voltage source in series with its "real world" source impedance) The closer the real world current source is to ideal, the larger the source impedance value. My understanding is that the CT's capacity to source current into incrementally larger loads is a function of the quantity and quality of the core iron. Magnetic saturation results at a critical load, where increasing the primary current no longer increases the secondary current.
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I was referring to the 160-turn winding as your primary, since you are swapping the original primary and secondary. I was struggling with how I should refer to them. Primary generally refers to input, secondary to output.

    A current transformer normally works with the secondary (many turns) terminated into a very low resistance - a short circuit, ideally. You may have seen warnings that dangerously high voltages can be generated if the secondary is open-circuited. You can also see That this is consistent with the equivalent circuits you mentioned.
    You need to study transformer theory to understand the impedance ratio. You might then understand why it is that a current transformer's "normal" output is a high voltage through a high resistance (the Thevenin equivalent circuit), whereas if you turn it around as you propose, you will get a low voltage in series with a very low resistance. As I said, this may be OK for your application. The impedance transformation ratio (within the "linear" region of the B-H curve) has almost nothing to do with the core characteristics, and almost everything to do with the turns ratio. The maximum current before saturation does have a lot to do with the core characteristics, as you stated.
    BTW, I am an electronics engineer, and have been for 40 years.
     
  11. MVIBoss

    Thread Starter Member

    Jan 21, 2007
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    Thanks again Ron H for your thoughts! I think we are in agreement now.

    With my old 800:5 CT, I was able to source over 450 Amps with the primary short circuited, when excited with 4 Amps (28 volts) applied to the secondary terminals. I am guessing that the reason I did not achieve an output of 4 * 160 = 640 Amps is that the impedance of the wire used to short circuit the primary was too large. I intend to graph the appied secondary current against the output current on the CT primary, to see through what range of input currents my output remains linear.

    I now have my 100 Amp current source for testing my motor overload relays!
     
  12. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Congratulations!
    Keep in mind that you should monitor the relay current when testing, as it will be highly dependent on the relay impedance. You will only get a few volts at most across your relay. Do you know how many volts it requires?
     
  13. MVIBoss

    Thread Starter Member

    Jan 21, 2007
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    The relay I am testing is current sensitive, and has an impedance of only a few ohms. These style relays are designed to use with CT's, and the relay's "burdens" are carefully matched with the CT, so that the CT can source linear current flow throughout the useable range of the relay.
     
  14. fanie

    Active Member

    Jan 20, 2007
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    If I understand this right, you have a relay you set (contacts closed) so power can flow through the contacts and it is latched, - then if a certain current through the contacts and a seperate low impedance (high current) winding is exceeded the relay goes open circuit ?
     
  15. MVIBoss

    Thread Starter Member

    Jan 21, 2007
    13
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    Very close, but not 100% right.

    A motor overload relay does have a normally closed set of contacts. The contacts are wired directly in series with the Motor Contactor coil, which switches on and off the motor. (A Motor Contactor, sometimes called a "Starter", is nothing more than a relay with large power contacts)

    The CT is used to sense the amps the motor is drawing. If the motor operates in an overloaded condition, it will draw more amps than indicated on the motor's nameplate. The Motor Overload relay (in its simplest form) has a calibrated resistor in series with the CT secondary. The resistor heats a "bi-metallic element" inside the relay. When a critical temperature is reached, the Bi-Metallic element will move, opening up the Normally Closed contact, thereby turning off the Contactor.

    As you point out, the contact is latched, so that first, the bi-Metallic element must cool down, then a reset button on the overload relay must be pressed, that will once again close the contact, permitting the motor to be re-started.
     
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