how does a bridge rectifier work?
931090
AC in and DC out.\n\nA diode only allows current to flow in one direction and blocks it if it is wrong.\n\nthe diodes in the bridge are connected together so that the +V current flows in One direction. the bridge .....( I cant be bothered)\n\nwikipedia has the answer...\n\nhttp://en.wikipedia.org/wiki/Bridge_rectifier
140958
To answer your question first: Yes, it will generate a back EMF in the circuit each time the power to the electromagnet is turned off, but this energy is never equal to the energy which was originally applied when you turned the switch on.  Nothing in life is free :)  \n\nWhat you're describing is actually called a transformer. A transformer works by applying an alternating current (equivalent to turning your hypothetical switch on and off very quickly, many times per second) to one coil, which induces current in another coil (or several, depending on the design).  \n\nWhile a transformer can be relatively efficient, it can never be 100%, and certainly not MORE than 100%, as you describe.
219925
Look at it this way. When you configure an amplifier as a common emitter, the emitter goes to ground. When a signal enters the base, it increases the emitter current. This causes a corresponding increase in the collector current. This makes the collector resistor drop _more_ voltage, producing less voltage in the load.\n\nSo consider the extreme cases. If no signal is present on the base, the transistor is effectively "off" and all the voltage at the collector resistor is avaialble to the load.\n\nThe other extreme case is if so much signal is at the base, the transistor saturates "on" -- then the collector will be at, say, 0.2V or so and the collector resistor will drop almost all the voltage, leaving nothing for the load.\n\nSo you can see that a lot of signal produces a little voltage, and no (or a little) signal produces a lot of voltage. Phase inversion!
1050520
Think about the simpliest capacitor. It is clear that two metal plates that don't touch won't let a DC current pass, right? But the spacing is very thin so that the plate's charges can effect one another. When you first put current (even DC) to a discharged capacitor, it charges up and for a brief time it does conduct current. How long? Well it will charge to 2/3 of the total voltage in RC seconds (where R is the resistance of the circuit). So when AC current is "fast" enough, the cap is constantly in a charge/discharge state and current flows. The effective resistance (reactance) of the cap is 1/(2*pi*F*C). So the resistance goes down as the capacitance and the frequency rises.\n\nThe inductor, on the other hand is just a coil of wire. Again, it is easy to imagine that it will conduct DC. But when you first apply current (even DC) to a coil, the magnetic field will interact with the coil and produce an opposing voltage. This only occurs when the magnetic field is forming or collapsing (remember, magnetic fields that do not move do not produce electricity). So right when you change the current, the inductor will "resist" the change, but if the current doesn't change, it is just a wire. The reactance of a coil is 2*pi*F*L, so the resistance increases with frequency and inductance.\n\nNow, combining inductors and capacitors is another interesting subject, but you didn't ask about that....
913471
In a purely resistive circuit, the power factor (PF) is 1 and you *can't* adjust it.\n\nAnd, since this is the 3'rd time you've asked the question and gotten the same answer from one other person and myself, do you even know what 'Power Factor' *is*?\n\n\nDoug
722546
I use a power inverter.
16972
You must first convert the 120 volt ac electricity to 12 volt ac with a transformer.  Then the ac (alternating current) must be converted to dc (direct current) with a rectifier bridge.  Both of these functions are normally handled in one unit such as the small plug in power sources for recharging cameras and cell phones and such.  You don't say how big a fan you want to power but the current rating in amperes of the converter must be great enough to handle the requirements of the fan.
229335
You can parallel the diodes to increase the current capacity of the recifier or add diodes in series to increase the voltage capacity of the rectifier.  If your looking to use diodes to drop  120v AC to 12V DC there are much easier was to do it.  The most common way would be to use a transformer to drop 120V AC to 12 - 16V AC and then use the recifier bridge to DC current.  Then place a large value capacitor in parallel with the DC output to filter the remaining pulses in the DC current.  Hope this helps!  If you have anymore questions drop me an e-mail.
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.
977954
Check out the chip below.  It is for a simple boost switching power supply.  A low input voltage is run through an inductor.  When the current builds up a bit, the current is cut off.  The collapsing magnetic field generates a higher voltage
207430
I suspect the protection devices are sharing the load or branch circuit which they are connected.\n\nThere is probably a ground fault from one of the devices or loads connected to the circuit. Disconnect each load to verify that the circuit is ok, then proceed by adding back in the loads. \n\nAs you check out each load, verify that there is no obvious wiring short to ground. If you can, use a Ohm meter to verify the load is not shorted before you connect the load back into the circuit.
478814
Both are similar devices in that they only conduct current in one direction and if you apply a voltage in the non-conducting direction and gradually increase it, for a high enough voltage it will suddenly start to conduct.   But zener diodes are designed to break down at an accurately-known voltage, specified by the manufacturer, so they're designed to be used as voltage regulators.   You can buy them in a wide range of breakdown voltage.   A rectifier diode is designed simply to convert AC to DC.
341482
Ok break it down:\n1) A power supply converts AC power (110 in the states) to some other required voltage. For this (lab) project I'd presume a DC regulated voltage from, say 1.5 to 15V something like that.\n\n2) Digital and programmable means you'd control the output voltage with some sort of digital signal like up/down buttons, a keypad, or a PC hookup.\n\n3) You are supposed to use an 8051 to accomplish this.\n\nSo there are several ways to do this. One way would be to have the 8051 accept a command to set the voltage and produce a PWM signal. Let's say the PWM signal is a 5V output switched from 0 to 100%. You can use an RC filter to produce 0V to 5V from this signal. If your instructor is nice, you could argue that this meets the criteria -- especially if you use it to drive a pass transistor to develop some serious current (but your voltage is now 0V to 4.3V because of the BE drop).\n\nBut I'd instead use an op amp with gain X3 (plus a little extra) so that I got from 0 to 15.7V and then use the pass transistor. Another idea would be to use the voltage to drive a simple current source which would program a LM317 voltage regulator.\n\nWeb searches on things like LM317, PWM, and "pass transistor" are your friends here.
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.
580159
The AC Switch Block allows a conventional 5V signal to control a much larger AC line. The switch can handle up to 16 Amps of current provided it is kept cool enough. At 110V, this is more than 1500W. We suggest that this limit be approached very carefully and recommend staying far beneath this.\n\nOnce an AC line is connected to the AC terminals, several places on the board will become live. This makes it absolutely essential that the board be mounted in a way that will prevent short circuits and accidental human contact. This is true even for rough prototypes or trial hook ups. Treat AC line voltages with much respect.\n\nHow Do I Use This?\nWith your AC device unplugged, first determine which cable from your AC device's power cord should be switched. It's best to switch the "live" or "hot" wire. On any AC outlet, there are usually 3 slots (sometimes 2). The right slot, which is slightly smaller than the left slot, is the "live" slot - check which prong plugs into this slot when you plug the AC device in to confirm which wire is live.\n\nOnce you've determined which wire is live, snip it in two somewhere along its length, strip about 1/8" insulation away from both the open ends, and tin the exposed wires. Make sure to leave enough slack on the cord for the two tinned ends to be inserted to the two AC connectors on the AC Switch Block.\n\nA small 5V signal applied to the D0 input will control the switch that turns the AC device on and off. Run a wire from the 0V connector on the AC Switch to the ground of the device providing the 5V signal. Double check that all your connections are correct, and plug in the AC device to the wall - now, a 5V signal on the D0 input turns the device on and a 0V signal turns it off.\n\nApplications\n-Automate Your Household. Many household appliances such as lights, fans, blenders, drills, vacuums, etc. run on AC power. Now you can use the behavior from a Poly Block an Intro Module, or any other device that put out 5V, to control these kinds of devices.
130611
Well I'm not sure what you mean exactly. Op amps typically will work fine at DC. The idea behind an op amp is simple. The output will do whatever it can do to make the inputs equal. For simple analysis you can pretend that the inputs are completely disconnected from anything internal to the op amp.\n\nSo if you provide a 2:1 voltage divider, for example from the output to the - input. Then when you put 3V on the + input, the op amp wil try to put 3V on the - input. How? By producing 6V on the output (6/2=3). Of course, if you put 10V on the input and the op amp "rails" are 12V, it can't do it. However, if you think about it, that explains why comparators work. They "hit the rail" trying to make the inputs equal when they can't.\n\nKeep in mind that the more current you pull from the output the further away from the rails you can go (except some op amps have special circuits to allow them to go all the way to one rail -- usually ground -- so they can work with single-ended supplies). Also there are a very few "rail to rail" output op amps.
508767
It sounds to me like you have an Interruptible Power Supply (UPS) where you have a primary AC power supply and a battery backup. It seems from your question that you are looking to automatically switch to the battery backup if the AC fails. Of course you must rectify the DC from the battery to AC of the proper current and voltage. Assuming this is the case, you can wire the load through an AC relay. Connect the AC power source to the relay coil in parallel to the load. Hot on one side, Neutral on the other.  Wire the primary AC hot wire to load  through the Normally Open contacts. Wire the DC rectified AC Hot wire through the Normally Closed contacts. When you have good AC power the relay coil will energize, closing the Normally Open contacts (and opening the Normally Closed contacts). This will energize the load with primary AC power. If you loose primary AC power, the relay will de energize. This will close the Normally Closed  contacts(and open the Normally Open contacts)  and port DC converted AC to the load. \n\nYou will probally have to come to a common point after the relay contacts such that power takes the path of either the AC or DC but goes through to the load via the same path after the relay.\nEnsure that you have compadible neutrals between the AC and the DC they will probally be common. Only the hot wires go throughthe relay. \nChoose a relay that will be able to handle the current and voltage you desire. Ie you need enough power to energize the relay coil and you do not want to melt the contacts by giving them too much current. You can obtain them as mentioned above frpm granger.com there real cheep. Order the relay and base. \nEnsure the load works well with both power sources.