Is there a simple formula for converting AC voltage to DC voltage?

Question

I'm writing a test procedure in which a 440VAC is rectified to 170 VDC to supply some power amplifiers. But I need to know what the rectified DC voltages will be if the VAC varies from 440 +/- 20 volts.

I could just use a simple ratio, but that wouldn't be accurate enough, since any two rectifiers will vary some. It's not a matter of using an active regulator since this test procedure has to work universally in systems within the naval fleet which apparently don't have these regulators.

Answer 1

You can use a linear "simple" function, because rectifiers have linear behaviour.
What may not be linear is your transformer: It could be already within a slight saturation at 440 + 20 Volts, and therefore the DC will be somewhat less than the linear increment. But that is almost impossible to determine without knowing the core material of the transformer, etc.....

Answer 2

This Site Might Help You.

RE:
Is there a simple formula for converting AC voltage to DC voltage?
I'm writing a test procedure in which a 440VAC is rectified to 170 VDC to supply some power amplifiers. But I need to know what the rectified DC voltages will be if the VAC varies from 440 +/- 20 volts.

Answer 3

There is no simple formula that will solve your problem. It depends on the design of your rectifier. Variations in the conduction characteristics of two rectifiers are generally negligible.

When you rectify an alternating-current voltage, the peak value of the resulting direct-current will be the square root of two (1.414…) times the RMS value of the alternating current. There will be considerable ripple, especially if only half-wave rectification is used. To reduce the ripple, full-wave rectification and a filter capacitor are used.

You can achieve full-wave rectification with either a bridge rectifier using four diodes, or with a center-tapped transformer using two diodes. A full-wave bridge rectifier cannot be used if one side of the load and one side of the power source are both tied to a common “ground.”

A center-tapped transformer has the center-tap connected to one side of the load, which may be grounded or not, and a diode connected to each end of the center-tapped winding. If positive DC output is desired, with respect to the other load terminal, the anode of one diode is connected to one of the transformer winding ends, the anode of the second diode is connected to the other transformer winding end, and the cathodes of both diodes are connected together to provide a full-wave rectified output with respect to the transformer center tap. Reverse both diodes if a negative output is desired.

Selection of the filter capacitor depends on the expected load. The capacitor will charge to the peak value of the full-wave rectified voltage twice each line cycle (120 times per second for 60 Hz line voltage, or 800 times per second for 400 Hz line voltage) and more or less linearly discharge (to a first approximation) during this same interval. The resulting “saw-tooth” ripple waveform will never have an average DC value as large as the peak value, and it can be considerably less if the capacitance of the filter capacitor is too small. On the other hand, if the load is small and the capacitance is large, only a small current will be withdrawn from the capacitor between peaks of the AC line voltage and you may consider the DC output voltage to be the same as the peak voltage of the AC line voltage.

In any case, there is no way you can get a reliable 170 volts DC from a 440 volt AC source without putting something else besides a rectifier in between. There are many options depending on many things, but the current that must be supplied is a major factor. Also consider whether or not the load is constant, or will it be removed and then applied again while power is on. Such a mode of operation can result in wide variations in the output voltage, and subsequent damage to attached equipment, if substantial current is drawn when the load applied.

A “brute force” approach is to use a transformer (usually called a control transformer) to reduce the 440 VAC to 120 VAC. On the secondary (120 VAC) side, install a bridge rectifier with appropriate current handling capacity for the load you intend to connect. Add a filter capacitor of appropriate voltage capability (at least 200 volts working voltage DC) and sufficient capacitance.

To determine the capacitance, first specify how much the output voltage can drop below 170 volts and still have acceptable operation. Next, determine the current drawn by the load. Finally, use this equation to determine an approximate capacitance value (in farads):

C = I t / V where I is the current in amperes drawn by the load, t is the time interval between peaks of the AC line voltage (about 8.33 milliseconds for 60 Hz), and V is the allowable voltage drop from the peak value.

So, if you can tolerate a one volt drop and your load current is one ampere, the capacitance you need will be 8.33 millifarads or 8,330 microfarads. More is better, so I would look for a 10,000 microfarad electrolytic rated for at least 200 volts.

Now what happens if the line voltage is not exactly 440 volts? The peak output will vary proportionately, but the “sag” will depend on the nature of the load. If the load is purely resistive, the load voltage will also vary proportionately.

Answer 4

For the best answers, search on this site https://shorturl.im/aw3nI

First you need a step down transformer. A step down transformer has less turns of copper in secondary then in primary. You can calculate with this formula # of primary turns/Voltage of Primary= # of secondary turns/Voltage of secondary When you know 3 factors, you can easily find out the forth one by easy mathematical calculation From a step down transformer the 12 V AC output, needs to be converted into 12 V DC by a simple reverse bias or diode A diode does a HALF-WAY RECTIFICATION which means it allows one way of current, cut downs the reverse that is AC Current in which change of polarity takes place or a diode rectifies AC to DC or unidirectional current

Answer 5

Look, if you rectify 440V AC you have to multiple it with 1.41 and you'll get 620.4 volts of DC. If you need 170 V DC you have to transform 440 V on 120 V AC approximately. And of course, you must to use strong rectifier (diodes) if you need strong current in your device. Don't forget and electrolytic capacitors to get smooth current. Best Regards and veradisca! Neven.

Answer 6

Formula For Volts

Answer 7

I would have used a step-down transformer (220vac primary, 12vac secondary). Then run the secondary output through a full wave bridge rectifier. Filter it with large capacitors. Then run that output through a 3-terminal 12vdc regulator. That will give you about 1 amp at 12vdc.

Answer 8

You need a reducer transformer, you can get 220 VAC, then use a rectifier and a regulator to get 170 VDC

Answer 9

AC=DC/0.636

Answer 10

It depends on how you do your conversion. If you rectify and then use an active regulator, there will be no change.