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What is the formula for calculating DC votage from a 3 phase rectifier bridge using AC voltage across one phase?
Wiki User
∙ 10y ago
A 3-phase rectifier bridge can be used with a single phase supply, it just means that four of the diodes are not connected. The peak voltage (if a reservoir capacitor is used) is sqrt(2) times the rms supply voltage and the average voltage using inductor smoothing is 0.9 times the rms voltage.
Wiki User
∙ 10y ago
Wiki User
∙ 11y agoVdc
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How do you convert dc to ac current?
The AC current is fed into a rectifier, which is a set of four diodes that force the current at the output to be one direction. A capacitor across the rectifier output is then used to smooth out the voltage to a level higher than the desired DC output (eliminating, for example, the zero-voltage portions of the original AC sine-wave) A voltage regulator then regulates the voltage to a constant level.
Application of full wave rectifier?
Transformer is used to step down the voltage. Your normal supply voltage is 230V but diodes used in rectifier can't handle that huge voltage hence this voltage needs to be reduced. This is done by transformer. Transformer brings down 230V voltage to say 10V. If you are talking about step down transformer than you should know that step down transformer produces equal voltages although opposite in polarity but equal in magnitude(ie +V and -V) across both diodes. If you don't use it the voltage across one diode may be greater than voltage across other diode. Say non-center tapped transformer produces 10V across diode D1 & -8V across diode D2. In next half cycle, -10V will be produced across d1 & +8V across D2. So in first half cycle current due to forward biased diode D1(as voltage across it is greater) will be greater than the current produced due to diode D2 in next half cycle. So rectified current waveform won't be equal.The transformer is for isolation so the full wave rectifier can float free of the line hot & neutral wires. The power supply can then define its own ground node without "fighting" the line, this is also much safer for the user. Step-up, Step-down, or one-to-one transformer types are irrelevant to full wave rectifier but are selected by the needs of the application. There are solid state diodes that can handle several thousand volts if the application needs it and vacuum tube diodes that can go even higher.Full wave bridge rectifiers do not use center tapped transformer secondaries, but require the isolation the transformer provides or diodes will blow out in operation!
The voltage measured directly across an open switch in a circuit will be?
The voltage measured across an open in a series circuit is the equivalent of the sourse voltage.
How a can capacitor smoothen or reduce the ripple of the voltage produce by the rectifier?
Rectifiers will not give a smooth DC voltage. There are ripples in the voltage given the rectifier. So in order to smoothen the voltage we use capacitor in parallel to the rectifier output. Now lets see how the capacitor smoothen the voltage that is coming from a rectifier...... Capacitor blocks DC and allows AC...... If we take the voltage that is coming from the rectifier it has some ripples in addition to DC, these ripples can be divided in to sinusoidal wave forms ( fictitious )according to the Fourier series. So the rippled DC now divided ( fictitious ) in to a pure DC and sinusoidal AC wave forms having the frequency that is multiples of ripple frequency. Now the DC current will not pass through the capacitor as the capacitor blocks DC. But the AC will pass through it i.e the ripple wave forms that are divided ( fictitious ) in to sinusoidal AC wave forms will pass through the capacitor. So only DC current enters in to the load, which will produce a pure DC voltage drop across the load. In this manner the capacitor smoothens the voltage.
Measure the voltage across a voltmeter and a switch?
If the switch is closed (connected) the voltage across it will read 0V. If the switch is open (disconnected) the voltage across it could be anything, it just depends on what the voltage between the wires going into the switch is.
Formula for peak inverse voltage?
piv:the maximum value of reverse voltage across a diode that occurs at the peak of the input cycle when the diode is reversed-biased.
Why do rectifier diodes in power supplies have small capacitors fitted across them?
To bypass the rapid high voltage spikes that could damage the diode.
How do you convert dc to ac current?
The AC current is fed into a rectifier, which is a set of four diodes that force the current at the output to be one direction. A capacitor across the rectifier output is then used to smooth out the voltage to a level higher than the desired DC output (eliminating, for example, the zero-voltage portions of the original AC sine-wave) A voltage regulator then regulates the voltage to a constant level.
What is the nature of the output voltage of half wave and full wave rectifiers with that of the input voltage?
In a half wave rectifier voltage across load resistance is not consistent, because for positive pulse of input voltage diode work as a forward bias i,e half wave rectifier treat as closed circuit and for negative pulse of a input voltage diode work as a reverse bias so no current flow through circuit. therefore voltage output is not consistent. In full wave rectifier two diodes are used at the both side of secondary coil of transformer. due to that for positive pulse of input voltage one diode diode work as a forward bias another as a reverse bias. for negative pulse of a input voltage second diode work as a forward bias another as a reverse bias,so consistent voltage can be provided by full wave rectifier.the nature of output voltage of half wave rectifier and full wave rectifier is that it flows through with only one polarity either in positive or negative in the circuit.
What is the maximum DC voltage you could expect to obtain from a transformer with an 18V rms secondary using a bridge rectifier circuit with a filter capacitor?
The maximum DC voltage you could expect to obtain from a transformer with an 18V rms secondary using a bridge rectifier circuit with a filter capacitor is about 24V.This assumes a truly sinusoidal AC waveform, and a forward conductioin voltage of 0.7 volts across each diode.Multiply 18 by the square root of two, and subtract two times the diode voltage.The maximum is the peak value. If there is any load on the output, there will be some ripple, but the peak value will still be around 24V.To calculate the output voltage of single phase diode bridge it is reasonable to assume a filter capacitor exists across the output and realize that it will be charged to the maximum voltage available to it.
Why is a full wave rectifier less than 100 percent efficient?
Any diode is less than 100% efficient because there is a forward voltage drop across the diode. This translates to power loss, because power is voltage times current.
Why we don't get negative half cycle in rectifier outpus?
We don't get negative half cycles at a rectifier's output only in a positive supply.If the supply is for a negative voltage, then there will be no positive half cycles at the output.Read a bit more about rectification.We don't get negative half cycle in rectifier outputs because the negative part of the supply only shows up across the diode when it is in the reverse bias and blocked (acting like an open switch) there by making the output voltage zero at that instant.
What value resistor and diode would you need to drop voltage from 12 volts ac to 1.2 volts dc?
I'm not sure you understand what you're asking. A diode will have a voltage drop of ~.5-.7 volts. If you put a diode and resistor in series, the voltage across the diode will be .5 - .7 volts, and the voltage drop across the resistor will be (supply voltage - diode voltage drop). If you are trying to rectify to DC, you need at the least a half wave rectifier (two diodes), and some system to remove the ripple. The rectifier simply chops the AC waveform, so for the part of the supply sine wave that is ~.5 or less, the output of the rectifier will be zero. The top part of the sign wave will show up on the output of the rectifier, but will be slightly smaller (due to the voltage drop across the diode). You'll need to get ride of this rippling for true DC. One fairly easy way to do this is to use a zener diode. It will attempt to keep the voltage drop across it the same, so purchase a 1.2 volt zener diode. The problem with this is the zener diode will saturate if you have too heavy a load. What I've done on simple projects is to use a zener diode to bias the base to collector voltage of a transistor, with the emitter acting as the output (an NPN BJT usually). I also used a fairly large capacitor to help minimize the ripple as well, although this may not be necessary for you.
Rectifier How do you change the 2 arrows direction in the circuit to make it positive or negative In Bridge?
A: Actually there is no changing of diodes required to get negative voltage out put the two cathode to ground if you have a load the voltage across will be negative. Electrons only flow in one direction where the measurement point are located across the load makes positive or negative.
What can be inferred about voltage across the bulb?
what can be inferred about the voltage across the bulb
What is piv requirement of a diode in the center tapped full wave rectifier?
peak inverse voltage of a center tapped full wave rectifier is 2Vwhere the maximum secondary voltage be VProof :- recall the diagram of the centre-tapped full wave rectifier ,during positive cycle the whole of the secondary voltage rests on the upper half of the transformer making D1 forward biased, but consider KVL in mesh D2 which is reverse biased so no current flows through it .KVL is ,VD=VR+VTwhere VR is drop across resistorand VT be the drop on the lower half of the transformersincs both are equal to Vwe get.VD=2V
How does the voltage measured across a dry cell ompare with the voltage drop measured across three bulbs in series?
How does the voltage measured across a dry cell ompare with the voltage drop measured across three bulbs in series?
