How do you hook up a 120v to 12.6v transformer with a Bridge Rectifier?
143811
Aboslutely, provided the fans are for 12V service.  Leave the secondary center tap unconnected, and put each end of the secondary to the AC input of the rectifier.  The + goes to the red lead of the fan; the - goes to the black lead (traditionally speaking of course.) Now the supply will be very ripply, with a 120 Hz waveform which could be more than a volt.  If you put a 470uF 35V capacitor across the plus and minus of the rectifier, you will smooth out the ripple, but increase the voltage above 12 volts.  Make sure the fans can take the increased voltage.  Transformers are rated at the RMS output at the 3 amps in your case.  The peak voltage will be around 1.4*12.6, but in practice I have never seen it go above 15 V with a load.  Therefore at 1 amp, if you put a cap on the supply, it may produce excessive voltage.  Check it with a voltmeter-- you may luck out.\n\nI have been in a similar situation.  Get a 5 watt resistor at maybe 1 to 3 ohms.  Insert it in series between the capactitor positive and the fans wired in parallel.  When running, use a voltmeter on the fan circuit.  Increase the resistor till you get 12V or thereabouts.  The resistor will get hot, so leave the leads long and connect them to a terminal strip or block if possible, and mount it in the airstream.\n\nCheck out the LM7812, however, at 1 Amp.  Google it, or check out jameco.com for the part.  Put a 470uF at 35 V at each end of the regulator to keep things very smooth.  Your fan will run perfectly with the voltage regulator.  Be sure to mount the tab on a heatsink to keep it cool since you are running it at full capacity.
767776
The Power, P = VI, hence\n\nThe current at 120V is I = P/V = 65000/120 = 541.667 A\nThe current at 1200V is = 54.1667 A\n\nHowever, here the power is carried by TWO parallel lines, each with 0.1 ohms of resistance. That means the effective resistance is given by:\n\n1/Reff = (1/R1) + (1/R2) = 20 or Reff = 0.05 ohms\n\nTherefore, the Power Loss at 120V is\n\nP = I^2R = (541.667)^2(0.05) = 14670.15 watts\n\nThe Power Loss at 1200V is = 146.7 watts\n\nHence the power saving by stepping up to 1200V is\n\n= 14670.15 - 146.7 = 14523.45 watts
377734
n1/n2 = V1/V2\nFor step down transformer n1>n2\n\ntherefore, 11000/400 = 120/n2\n\nn2 = 400 * 120 / 11000\n   = 48/11\n   = 4.36\n   ~ 4 turns
905900
It depends on whether you simply want a boost in voltage value or a boost in signal power. In order to simply boost the voltage you need a transformer but if you want the signal the be boosted in power also you what you need is a simple amplifier circuit.Amplifier chips are available in IC patches
416476
You need some way to clip the signal, using a diode of some sort.  Simply R L and C cannot generate a square wave using AC voltage.
261103
10 amps if the voltage difference is 120 volt as usual
364172
First of all, you cannot turn a 240 V, 200 A service into a 480 V, 200 A service with a simple 3 phase transformer. In a transformer, power out (roughly) equals power in. If you put 83 kW in, you can't get 166 kW out. You will have your electric company run a new 480 V service for you. From "your side" of the service, you will have to have a licensed, qualified electrician make sure that all breakers, wiring, fuses, and other components are rated for 600 volts, so you may not be able to use your existing breaker box, or even the wire feeding it. All of your existing wire to motors and other 240 / 480 VAC devices should be of ample ampacity, as a motor draws half of the current at 480 than at 240, but again, it may be wired with 300 V wire that does not have insulation rated for 480 V circuits. Another investment would likely be a step-down transformer for your electrical outlets. You won't be able to go phase to neutral anymore, as that would get you 277 VAC (in a 4-wire Y system). Make sure that your qualified electrician complies with all local codes, as well as the NEC. Good luck!
379234
Hi! A transformer works on the principle of mutual induction which is valid only when the voltage (or EMF, to be more appropriate) on one of the coils of the transformer is changing with time, as in the case of A.C. voltage. As the magnitude of the voltage changes on one of the coils, the corresponding flux on the other coil also changes, thus giving rise to a voltage on the other side.\n\nAn AA battery provides a constant voltage, as a result of which there is no change in flux and no voltage on the other coil (120 V side, in this case).\n\nHowever, there will be a flux change for a very short duration of time as the current in the 24 V side rises from 0 to its full value after the battery has been connected. During this time, a voltage will be induced in the 120 V side and you'll get an A.C voltage on this side. But this would be for a very short time duration.\n\nSo, the answer: After some time, there is no voltage (A.C. or D.C.) on the 120 V side.
665401
hi,\n\nas i understand it, the way a DC-DC converter operates is in 3 stages:\n\nfirst, V_in is used to drive a pair of coupled transistors (or some other oscillatory circuit) that output a square wave or some kind of time-varying oscillatory voltage.\n\nsecond, that AC signal is used to drive a transformer which consists of two inductors.  the ratio of the number of turns in the primary and secondary windings of the transformer determines the step-up or step-down ratio of the transformer.  if you have a 3:1 inductor then you will step down the voltage by a factor of 3, for example.\n\nthird, the transformed AC signal gets converted to DC by a bridge rectifier or some other AC->DC circuit.  generally speaking, a bridge will output a DC voltage that is simply proportional to whatever AC you put into it.  (of course if you way over- or under-drive it, it won't work right.)\n\nanyway, if you have a single missing turn on either side of your transformer, that will change the step-up/down ratio.  my 3:1 example would become a 2:1 transformer and then you'd get an output voltage that was too high (1/2 of the input, instead of 1/3).  \n\nthen the rectifier will just flatten out whatever voltage you put into it.\n\nhope this helps!
775866
I don't think you can do that. you need the number of amps and the power factor for that device.
902006
You might try these folks.  I have used them for low voltage (24 VDC) with great success.\n\nTransorb Electrical Products Ltd, PO Box 1799. Grand Forks, BC, V0H ...
443133
hi,\n\ntransformers you can buy for pretty cheap at any decent electronics or  hobby stores (radio shack sucks these days, but they probably still have them).  \n\nhowever you can make them really easily.  all you need is a hunk of steel or iron (some kind of rod-shaped thing would work, as would a donut-shaped loop).  take one wire and wind it around the piece of steel N times.  take another wire and wind it around a different part of the steel M times.  congratulations, you have just built a transformer that will step up/down voltages by a factor of M/N (or N/M, if you hook it up the other way.)\n\ncheers-
227723
Regardless of what size capacitor you choose, there will be a small ripple on the output voltage, which of course gets bigger as the load increases.  You need to determine what is an acceptable ripple voltage for your load to calculate out a value.\n\nIf you want a totally smooth DC output, you'll need to first rectify and filter, then use a voltage regulator with a voltage set point that's a few volts below the voltage output from the capacitor.  Follow that by a smaller capacitor after the regulator to help catch any remaining transients and response after regulation.\n\nSomething to note:   Your capacitor's voltage rating should be much higher than the actual working voltage you will subject it to.  Eletrolytics like to blow up (literally) if you run them past their maximum rating, even for a short time.   Also remember this, the AC voltage that you measure at the output of the transformer (or the output of the rectifier, unfiltered) is the RMS voltage, not the peak voltage of the AC waveform.  Once you add the filter cap, your DC voltage will appear to be higher because the capacitor is charging to the peak voltage, which is basically the RMS voltage multiplied by 1.414   Keep this in mind as well when selecting the working voltage rating for your capacitor.
568250
Where are you going to plug in your DC heater?
1172658
Actually it's a question in physics. I will try to explain it anyway:\nassume we have two windings w1 and w2 with  n1 rounds and n2 rounds respectively. If an A.C. voltage of V1 is connected to w1 then a voltage V2 will be induced in w2, satisfying the following equation:\nV1/V2=n1/n2 and for currents we have: I1/I2=n2/n1\nand it doesnt convert ac to dc to do so we should use a rectifier such as bridge rectifier etc
107063
Yes, a full-wave rectifier can be made with only four diodes. No transformer is required.
399258
just put a limiting resistor. 60mA and 12V. Limiting resistor =R=(V-Vled)/I. assuming the extreme, 0V for Vled. R=12/6e-2=200ohm. that way the current will never exceed 60mA.