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Posts posted by Gazza

  1. Does current that is out of phase with voltage mean lower available power. I see inductors used everywhere in power supplies, but are they used for more than ripple reduction?

    It is all about the Power Triangle. So the answer is, it depends on what you mean by power. If the current and voltage are in phase, all of the power will be "real" so real power = apparent power or KW = KVA.

    If there is an inductor or a capacitor in the circuit, the voltage and current will be out of phase, therefore the real power will be the same but there will be a component of "active" power which has a phasor +/- 90 degrees out of phase with the real power. When you determine the "apparent Power" it will now be more than the "real" power or KVA>KW

    The long story short is study the power triangle.

    Audioguru, there are many types of AC power supplies, what about a UPS or a VFD?

  2. Alright. I took a 3:1 transformer, 34Vac input 10.3Vac output with 100 ohm load. That's 2.1 Watts peak power. I reduced the load to 25 Ohms. 34Vac input 6Vac output. That's 2.9 Watts peak power. The transformer is rated at 120Vac, and put's out 40Vac when plugged into the outlet. That's 3:1. A 100 ohm load keeps it a 3:1 transformer. A 25 ohm load makes it a 6:1 transformer, and the the current peaks at only 340mA. Well within current rating.

    A light load of 100ohms works. The energy is not available for the 25 ohm load. Else I would get 10Vac on 25 ohms for 9 Watts peak power.

    The resistance has NOTHING to do with the turns ratio of a transformer. What you are seeing is something totaly different.

    PIn=POut therefore if you have a transformer that is 5:1 with a primary of 120V and a 60ohm load.

    The secondary voltage is equal to 120/5= 24V
    The current is = 24/60 = 400 mA

    Now, if you reduce the load to 30 ohms

    The secondary voltage is equal to the primary voltage divided by the turns ratio, it does not change 120/5 = 24V
    Current = 24/30 = 800mA

    The secondary current is a function of the load, and is not used to determine the turns ratio.
  3. You can do it either way as long and you keep your signs consistent.

    Example (XT = Xc-XL)
    The reactance in the circuit is equal to the absolute value of Xc-XL

    If the value of XT is positive, the reactance is capacitive, if it is negative, it is inductive.

    If you reverse the formula, the opposite is now true, positive values are inductive and negative numbers are positive.

    do you add it or find the difference ( subtract) ?

    X= XL+Xc or is it this X = XL- Xc  or this X= Xc-XL

  4. A "travel adapter" is made to reduce the power to electrical hair dryers, not transformers.
    The LED lamp might also be burned out.

    Your transformer has a rated output of 12VAC, 1667mA. Get one with a primary winding to match the mains voltage in the UK (230VAC?) and a 12VAC, 2A secondary.

    Not to mention the frequency of the voltage is 50Hz in Europe and 60Hz in the USA.

  5. I think the voltage might step down okay with a light load. A power transformer has an iron core to keep the inductance not so high, which makes the reactance lower, which makes the current high. 120Vac can't generate as much current. 1 Watt on the primary can't provide 2W on the secondary. Bring the voltage up 240Vac, and you get 2 Watts on the primary, 2 Watts on the secondary.

    I don't think you know what you are talking about. Once you have the turns ratio which is equal to

    a=V2/V1 or a=N2/N1

    You can find the voltage on the opposite side of the transformer.

    If you have a 240 to 120V transformer, the turns ratio is equal to 120/240 = .5

    Therefore if you feed the primary side of the transformer with 120 volts, the secondary voltage will be

    .5= v2 / 120
    v2 = 120 x .5
    v2 = 60

    Since current is inversely proportional to the voltage and power in must equal power out, the opposite is true, if you have 1 amp of current at 120V you must have twice as much current at 60V.

    The iron core is to help mitigate losses in the transformer, I suggest you look into the equivalent circuit of a transformer.

  6. I have to disagree with Gazza.  Since the voltage ratio rockazella inquired about is the same for both input voltages and both are below the voltage rating for the transformer, it will be fine for stepping 120 volts down to 60 volts, as Ldanielrosa said.  You don't need and couldn't use a tap to do the job unless there was a 50% tap on both primary and secondary.

    The primary disadvantage would be excess weight due to more than the necessary amount of steel and copper in the transformer.  There might also be slightly greater fixed losses due to the larger volume of steel to be magnetized, but it's been a long, long time since I studied transformers so I may be mistaken on that.  If that is true, it would show up mainly in slightly greater fixed losses at low current draw.

    Even though you would be using the transformer at well below its power rating, you shouldn't exceed the rated secondary current because the wire gauge was selected by the designer for the design current.


    My bad, I read the question too fast. You are 100% correct.

    If the turns ratio is .5, the output voltage will be half of the input voltage. Which is the case here.

    My appologies.

  7. i was trying to determine the frequency of the a.c[ i know it is about 50-60Hz] but wanted to find it myself,

    The easiest way to do that is to use an oscilloscope, measure the number of divisions between the start and the finish of one sine wave. Multiply that by the S/div and invert :)

    and also to find the number of harmonics in a fixed length and mass[i got up to the 3rd harmonic at max. with .9m and .15kg]so i simply shorted it out, the thing got so hot that the plastic on which the windings were made melted, even though the room temperature was close to 0` Celsius.

    This doesn't make any sense, there will only be harmonics present if the load is non-linear. Even if there were any harmonics, the only way to "short" them out is to tune a filter circuit to the frequency of the harmonic, it will then act like a short to those frequencies. Even then, the filter exists between line and ground, not a dead short between the terminals.

    Shorting out the transformer's secondary with a wire is a sure way to set your transformer on fire.
  8. Hey Gruthos,

    You are right, you need to vary the frequency to change the speed of your motor. I would use a single phase v/hz drive for this.

    The cap on the top of your motor is a starting cap for single phase induction motors

    You motor is using approx. 1.5HP, (10.2 x 110)/746

    I would just buy one from www.automationdirect.com, http://web2.automationdirect.com/adc/Shopping/Catalog/AC_Drives_-z-_Motors/GS1_(120_-z-_230_VAC_V-z-Hz_Control)/GS1-22P0

    That one costs about $150, You just have to make sure you can single phase the output, I dont have enough time to look right now.

  9. wow I didn't read that first post properly, you can short circuit the seconday of a transformer for testing purposes but you have to be really careful about the current getting too high.

    However I have only ever heard of short circuiting a transformers secondary to find copper losses in the transformer.

    What were you trying to test? The resonant frequency of a wire?


  10. i was using a 1500mA, 12V transformer to find the resonance frequency of a copper wire. but the transformer was getting hot too quickly, within 15mins of its running it was too hot to touch. is it because the circuit was short circuited by directly connecting to the copper wire with nothing else? or is there a problem with the transformer?

    Can you post a picture of the circuit?

  11. Hi all,

    I would like to build a circuit to protect a 12v pump from running dry, or even better to change to a second pump once the first pump runs dry. The current the pump draws when pumping water is approx 1A and when running dry approx 0.3A.

    Any help or advice would be greatly appreciated,

    Thank you in advance


    Cool idea,

    If it were me, I would use a cheap PLC from www.automationdirect.com, you can get a cheep one for $129.
    I would probably monitor the water flow with some flow sensors and write a bit of logic to decide what pumps to use.

    Flow sensor - http://cgi.ebay.ca/Flow-Sensor-Air-Water-0-5-Vdc-Output-Omega-FLR1010_W0QQitemZ200071635353QQihZ010QQcategoryZ67003QQssPageNameZWDVWQQrdZ1QQcmdZViewItem
    PLC - http://web4.automationdirect.com/adc/Shopping/Catalog/PLC_Hardware/DirectLogic_05/PLC_Units
    Software - http://web4.automationdirect.com/adc/Shopping/Catalog/Software_Products/Directsoft_PLC_Programming_Software/Directsoft_Software - there is a free version

    I am a Power Electronics guy, so I tend to use a sledge hammer to solve these problems, I am sure one of the electronics guys will have a smaller idea :)

  12. hi gazza.

    thanks for the article. but it still doesnt let me extract out dc voltage for small signal usage from the mains.. theres a device i have in hand now that does the specific application. but i cant see the components used as it is covered with some kind of protective layer over the critical components.

    is it a rectifier?

  13. thanks for the quick reply.

    my purpose for this is to replace an existing wall switch with some RC control switching (infrared, wifi, etc). but.. those conventional wall switches has only a live wire coming in... and an output wire going out connecting to the load. i do not have the intention of connecting any neutral wire to the device... because that would require extra wiring to the existing switch...

    yeah, in my sketch it acts exactly like a switch and i would want to know if there is a way to draw power from the live wire and converting it to dc power without lighting up the load

  14. Filtering out large magnitudes not only affects the waveforms appearance, it affects the amplitude. I see it very good when I apply superposition theorem.

    You need to step back about 50 yards, harmonics are not just added to the fundamental.

    PT=(PF + P 3rd harmonic2 + P 5th Harmonic 2 + P 7th Harmonic 2.........)1/2

    I really don't mean to be rude but I think you are completely missing the point here, harmonics do not affect the value of the fundamental portion of a signal, they are additive but in a bad way. They are adding "distortion" to the signal.

    The fundamental frequency is the only part of the signal that you want to use. The harmonics are undesirable side effects created by non-linear loads.

    Here is a bad analogy:

    I have a rock, the rock grows moss due to the environmental conditions surrounding it, I scrape the moss off.

    Rock = fundamental
    moss = harmonics
    environment = type of load/power grid

  15. I'm thinking crystals are the main ingredient for generating harmonics. Producing perfect multiples of a fundamental does not sound easy at all without a crystal.

    You are missing the point. Harmonics are created in any load that is non-linear. There is no magic crystal. Just non-linear loads.

    I promise

  16. Why don't they just say the 60Hz has all sorts of noise.

    Hey Kevin,

    I am not 100% sure that I am getting your question. As far as this quote is concerned maybe I can help a bit. When you are talking about the power grid in North America, the fundamental frequency of the power is 60hz. If you have a non-linear load (i.e. a rectifier) harmonics are created. Depending on the nature of the load, you will get different harmonics.

    Here is an example:

    If you have a fundamental frequency of 60hz feeding a 3 phase 6pulse rectifier. You will have 5th and 7th harmonics present, they will have frequencies of 300 and 420 Hz respectively.

    If you try to tune a filter to 480 Hz, you will get rid of the 7th harmonic, however the waveform will still be distorted due to the 5th. If you could filter out all harmonics (which is very difficult and costly) you would see a pure sine wave at 60hz.

    If you are passing a purely fundamental wave through an inductor or capacitor, ELI the ICE man has all of your answers.

    When a circuit is inductive:
    Voltage leads current.

    When the circuit is capacitive:
    Current leads voltage

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