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convertion of acvoltage to high dcvoltage

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panchumarthy

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i want to build a ciruit to get 150v(dc) :?:
i know for getting 150v(DC),we need 166v(AC) RMS value or 236v(AC) peakTOpeak value becauce for getiing
V(DC)=[2*V(AC voltage peak to peak)]/PI.
will the diodes in the bridge rectifier circuit can withstand the value of 167V(167v rms means 236 peak to peak).
please suggest me any other possibility to get 150v dc value. :?
do anyone can suggest other possibility :?:
 
panchumarthy said:
i want to build a ciruit to get 150v(dc) :?:
i know for getting 150v(DC),we need 166v(AC) RMS value or 236v(AC) peakTOpeak value becauce for getiing
V(DC)=[2*V(AC voltage peak to peak)]/PI.
will the diodes in the bridge rectifier circuit can withstand the value of 167V(167v rms means 236 peak to peak).
please suggest me any other possibility to get 150v dc value. :?
do anyone can suggest other possibility :?:

That's incorrect, rectifying an AC voltage gives a DC voltage of the peak AC, minus the voltage drop across the rectifiers. So about 110V AC will give 150V DC - presuming you are smoothing the output with a large electrolytic!.
 
What diodes are you thinking of using? Diodes usually have a maximum forward current rating, and a peak inverse voltage (PIV) rating. Look for these on the datasheet. If the PIV isn't adequate, you can series up the diodes to give you the required rating. Depending on the load that you intend to draw through the rectifier you might have to run a few rectifiers in parallel or use different diodes.

If the load isn't too great, there *should* be moulded case rectifier bridges pre-made that should be able to handle those kinds of voltages.
 
thans for your replies, i want to use this for robot which is already builded in my department,so i need to smooth the rectifier ouput with capacitors, am i right , if i get any rectifier with this capability ,this is OK for me,do anyone has the idea about this type pf rectiefier.
 
reltime

your help is appreciated.
it is in mA range,just say 350mA, i have found one diode with PIV upto 1000v and max current have in amperes(IN5408) .
Can i use the normal bridge rectifier ciruit with the capacitor at rectifier output (for smoothing) in realtime applications or do i have to add any other components,if u have any realtime circuit please send me.
 
Re: reltime

panchumarthy said:
your help is appreciated.
it is in mA range,just say 350mA, i have found one diode with PIV upto 1000v and max current have in amperes(IN5408) .
Can i use the normal bridge rectifier ciruit with the capacitor at rectifier output (for smoothing) in realtime applications or do i have to add any other components,if u have any realtime circuit please send me.

The IN5408 is a 3A rectifier, four of them will give you a 6A (maximum) bridge rectifier - but it's good to be conservative!. All you really need more than that is the smoothing electrolytic, but it's common practice to place small capacitors directly across the recifiers to help protect then from high voltage spikes, also a surge limiting resistor from the transformer to the bridge might be required, depending on the size of capacitor you use.
 
do i have to put voltage regulator after my filter circuit (becase load will vary) ,what type of regultor do i have to use.
 
As previously stated, the nominal AC voltage is usually expressed as the RMS voltage - more or less equivalent to a DC voltage of the same magnitude. In one cycle the AC voltage rises from zero up to a peak then back down to zero. The peak voltage is about 1.4 times the RMS voltage, for a sine wave. The fact that the AC voltage varies over time is why it takes a higher peak voltage to get the same power delivered as a constant DC source.

When an AC (sine wave) voltage is rectified the result is pulsating DC - essentially half of the original sine wave. Depending on the rectifier configuration, the pulsating DC voltage might look like the original sine wave with the bottom chopped off - or like the top halves of sine waves side by side. What's important is that the voltage is still starting a zero, increasing to a peak then dropping back to zero.

If you add a capacitor filter the capacitor will charge to the peak voltage with no load and the result is more or less a smooth DC voltage at the peak value of the sine wave. As soon as you add a load the capacitor can no longer charge to the very peak and the result will be a DC voltage with vary amounts of ripple - the value will be less than the peak voltage. Another way to put it - as the load increases the output voltage will decrease and the ripple will increase.

Many of the basic electronics books - particularly those with power supply sections - explain this much better than I have here. They also offer formulas so that you can determine what AC voltage must be applied to the rectifier and filter so that you get the desired DC voltage out. I don't have a lot of experience with power supplies but it appears that for the AC input voltage, designers often use an AC (RMS) value equal to the desired DC output value. This allows for the fact that with a load, the filter capacitor cannot fully charge to the peak value.

I've taken a few shortcuts here but I think this is sufficient. Good luck.
 
Voltage regulation

yaeh ,you are right , no need to consider dc mean value because, after smoothing(with capacitor) the wave which is the output of the rectifier,we will get only ripples which are upper part of the waveform,so obviously we will get peak voltage as the dc (constant) voltage.am i right :?:
i have already done the circuit and my load will vary from 5v(dc) to 150v(dc) ,waht type of components i have to add to my circuit, in meantime i found one solution for voltage regulation by adding ZENER diode,but i am not sure how can i add this to my circuit ,can add this after the filter circuit and do i have add any resistances in the end. :?:
 
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