What is a full wave bridge rectifier?
89554
A bridge rectifier makes use of four diodes in a bridge arrangement to achieve full-wave rectification ... component bridges where the diode bridge is wired internally ...
437532
Possibly Rectangular Series Transformer (RST).  See Page 7 of the link.
688125
This is a good question.  Many of the electronic devices today use semi-conductor devices that work at low voltages.  However, that has only been the case since the mid-1960's on average.  Prior to this electronic devices used Vacuum Tubes.  These vacuum tubes used very high voltage DC to cause the flow from an cathode to an anode to flow.\n\nUntil the development of the High Voltage semi-conductor or Silicon Diode most electronics circuits used Vacuum Tube Diodes.  These Diodes would allow for current to flow in only one polarity the same as silicon diodes.\n\nAn intermediate high voltage diode between the vacuum tube diode and the silicon diode was the Selenium Diode.  This diode was the first solid state diode; however, they were not very efficient nor were very effective when high currents were required.  However, today selenium diodes are still in use where size is not an issue.  These diodes generally can work with voltages up to about 2000 VAC and currents as high as 100 amps.\n\nGenerally, most high voltage circuits today use silicon diodes with a high Peak Inverse Voltage (PIC).  On average most cost effective Silicon diodes have a PIV of 600 volts DC.  To increase the voltage that diodes can work you can put them in series, or connected one after then other.  This way you can daisy chain several diodes together to achieve a PIV of up to around (3 in a row) or 1800 volts before the capacitance could be a problem.  This same type of diode works at one amp, if you place them in parallel or side by side you can increase the amount of current your bridge can support.  Again I would not recommend using more then 3 semi-conductor or silicon diodes in parallel, or use for rectifying more 3 Amps of voltage.\n\nTo make your Neon Sign DC voltage converter I would recommend using selenium diodes specifically designed for the high voltage you are talking about.  These diodes and semi-conductor diodes are still available from a number of retail suppliers.  If you would like to try a Vacuum Tube diode, similar to the one used in the high voltage section of a Color TV, Ratheon is the only supplier I can recall that still makes vacuum tube diodes.
471894
a drawbridge or a circuit??
1065990
In Electronics Physics:\nthe use of an oscillator feedback loop as a power supply. Add a bridge rectifier to the output and feed the DC current Resistance limited back to the oscillator amplifier setup. Like using an auditorium feedback signal to drive the DC system.
418745
An uninterruptable power supply.  'UPS'
1164919
you need to describe what you mean by "small" better.\nis it too small vertically, horizontally, or both?\nAlso, how good is your electronics background?\nDo you know how to discharge the CRT and large capacitors so you don't get zapped with 330 to 10,000 volts while attempting to repair it?\nYou also need to define "wavy" better.\nWith the tin amount of information you have given though, it sounds like the power supply is failing, but in which part is impossible to say.
53767
Inductance is the "definite answer" that I was given 30 years ago. I don't suppose that my confirmation is any better than yours, but it's a vote anyway.
46766
DC generator would have a rectifier built in to convert the AC to DC.
471780
I prefer you buying one.  But there are always schematics out there that show you how to create your own inverters.  \n\nhttp://www.aaroncake.net/circuits/inverter.htm
399669
Do you mean for a transistor (or even a valve) circuit?\n\nVOLTS on x axis, AMPS on y axis, Plot on x-axis full voltage at zero current through the device, even if the device could not actually do that.  Calculate what full current would be through load resistor with zero voltage across the device and plot on y-axis (again even though the device might never do that, the load resistor governs what voltage remains across the device at any given current).\n\nThe chosen bias current or voltage will determine the no-signal "Q-point" somewhere along this line. Any signal altering the bias will move the volts/current  up & down the diagonal load-line you drew.\n\nHowever, to complicate the matter, the ac load line may not be on the dc load line you drew.  You calculated this on the basis of a resistor dropping the supply voltage. You could have a reactive (eg capacitative) load.\nThe Q-point is the same, but the signal will operate up and down a line through the Q-point.  The line will have a gradient according to the net impedance of the load.\n\nHope this helps.\n\nPLEASE NOTE I HAVE EDITED x- AND y- AXIS SINCE THE ORIGINAL.
1170454
Inverter: DC to AC\n\nRectifier: AC to DC.\n\nthey do opposite functions.
749686
Have a look at my project.  Several links for pictures, wiring, charge controller, and so on.\nhttp://members.rennlist.org/warren/wind.html
978617
Try this web page:\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/electronic/zenereg.html\n\nIt shows the basic circuit and has calculator to figure out component values.\n\nThere are much better solutions out there for a voltage regulators than zeners... such as the three terminal regulators like 78xx or LM317 types...
827050
Yupp... but unfortunately is a hardcopy on my desk. Perhaps you may straight to tell us what\n\n1. You are designing from HT to Transformer 11kV / 415V\n2. You are designinig from Transformer to LV ??\n3. What is the kVA of the transfomer?
700902
Ok, at the risk of being blunt, if you need to ask this, you might be better of leaving things well alone - or doing as the other person suggested and using a transormer to lower the voltage to begin with.\n\nHowever.........\n\nFirst of all, you need to plonk a diode in line really, LED's do not like AC so the diode will make things into "raggy DC", using a bridge rectifier would be better and would reduce the flickering, as would plonking a capacitor over the DC output - typically with AT LEAST a 380V rating if using a bridge and a 200V rating if using a single diode.\n\nStill with me?\n\nYou now need to work out the voltage drop over the LED's - this varies according to type - and mainly colour. About 1.8V for a typical RED LED and around 4v for a white one.\n\nYou then need to look at how many volts you have "left" after you have taken into account all of the LED's in series.\n\nEG:  You take 220V, use a single diode (110 v) and then pop a cap over to stop the flicker (around 160V)\n\nYou now have 30 red, 20 green and 10 blue LED's in series. Assume 1.8V drop on Red, 2.2V on Green and 4.1V on blue.\n54v  + 44v + 22v = 120v\nThis gives us 40v to "burn off" with a resistor.\n\nLet us assume you want to drive the LED's at 20mA\n\nR = V/I   therefore  R = 40/.02 = 2000 Ohms\n\nPower = I x I x R  (current sqaured 8 resistance) so the resistor will be dissipating 800mW - Allowing for plenty of play and wanting a warm rather than hot resistor, then 2w ideally.\n\n\nSo a 2K2 resistor - or 1K8 for a bit brighter would be your answer here (for the number of LEDS and type), rated at 2W or better.\n\nAs I said at the beginning, if ONE step of this doesn't make GOOD sense to you, PLEASE leave WELL alone.\n\n\n\nPLEASE DO NOT use this as a way to hook up the leds exactly as shown WITHOUT first checking their "forward voltage losses", I have used typical examples, different models can vary considerably.
119135
10mH is 10 milli-Henries.  A Henry is a measure of inductance, and is defined as L=di/dt.  Inductors resist changes in current flow.  You have to use calculus to get the equation for an RL (resistor-inductor) circuit, which is what you have.  You are subjecting an RL system to a step input of 10 V from a "voltage source."  See the more technical, long-winded explanation of inductance.\n\nThe second site actually shows you an animated plot of the exact ciruit you are interested in.  \n\nAs a note of interest, large iron core inductors were developed for power supply filters in Vacuum Tube radios and amplifiers.  They had huge inductances of 7H or more.  These days, everything is in microhenries or millihenries because the frequencies in today's electronic systems have gone past 1 gigahertz.  Inductors can act as frequency-dependent resistors.  \n\nThese days, one hugely popular use of an inductor is to store energy in a magnetic field for switching power supplies in conjunction with electrolytic capacitors and FET modulated drives.  Wound 60Hz transformers are quite expensive, large, and heat producing.  The days of linear power supplies with large iron core transformers have given way to tiny circuits with driving transistors, an inductor, and some capacitors.  Switching supplies running at 125kHz or more can use transformers with ferrite (powdered iron) cores and far fewer windings.  The higher frequency packs much more energy than a 60Hz electrical source.\n\nAs a point of interest, try to find an old computer power supply which should be very easy to get.  Open it up.  Look at the parts, and you will see a high frequency transformer, inductors, capacitors, and drive transistors with modulating chips.  The 120VAC from the wall is put through 4 diode rectifiers to make 169V DC with some high voltage capacitors.  This voltage is made available to be switched at relatively high freqencies into a powdered core transformer primary.  The Secondary makes high frequency AC voltage which is rectified and filtered with capacitors and/or inductors.  A voltage detector circuit, usually isolated from the high voltage DC side by an optical isolator, tells the modulating chip what the load is.  The modulator will then vary the duty cycle of the transistors at a constant frequency to move more power across the transformer.  Have fun with electronics!  I love it...