What is output voltage of fullwave rectifier if input is 230v AC?

hi,
i am designing a super led night lamp using high power leds. ...

i have attached that ckt ...


in that ckt i need to build a fullwave rectifier,so made it using doides IN4007 input voltage is 230v AC ,i am not using any Transformer at the input, getting ouput voltage of 300v DC .........

my problem is i dont know at what voltage this leds glow,
.is rectifier ouput voltage(300 )correct output voltage giving ? if it is correct how can i reduce that voltage?..........



attached ckt....

can any body help me out............

230Vas is rms, if you rectify it (no smoothing capacitor) you'll have dc voltage of 0.9*230 => 207V ..

See:
http://www.hammondmfg.com/pdf/5c007.pdf

:wink:
IanP

But keep in mind that you have 1uF cap and it represents (at 50Hz) series resistance of roughly 3kohm ..

At a first approximation, 25 LEDs at 2V/LED = 50V. So with the LEds connected this will be the voltage at the output of the bridge. When the mains is at a peak, you get 230 X 1.4 ~ 331V . At the bridge input you will have your 50V + 1.6V (diode drop), so as mentioned you have a voltage drop of 279 and the main impedance is the cap (3K) so the current flowing through the circuit will be 93 mA (279/3 K).
Frank

I want to supplement chuckeys excellent explanation in a single regard. If the LEDs are white, as the schematic suggests, then the forward voltage can be rather expected around 3.5 V. So the LED current achieved with an 1 uF capacitor will be slightly smaller. Basically, you have to check the allowed current for your LED type (also considering self heating of the enclosure) and determine an appropriate capacitor. With white LEDs and the planned current level, I assume that nightlight will be sufficient for a large hallway!

You can do away with the plain diodes altogether. You also don't need the 10uF capacitor.

You can use the same operating method as with similar circuits using one LED and one diode, anti-parallel. Or two LED's, anti-parallel.

Consider creating two strings of LED's. Hook them up anti-parallel. One string will pass current one way to charge the capacitor. The other string will pass current in the other direction to discharge it.

Important: You should add a plain diode in each string, which has a high enough reverse V rating to prevent reverse flow through the LED's in a worst case scenario.

The LED's will flicker at 60 Hz. If it's unacceptable then disregard this post.

By the way conventional circuits show the limiting resistor having a higher value than 100 ohms.

haroon_tech said:

my problem is i dont know at what voltage this leds glow,
.is rectifier ouput voltage(300 )correct output voltage giving ? if it is correct how can i reduce that voltage?..........

Click to expand...

The high brightness LEDs that you usually have to use voltage drop of approx. 3.5V
Therefore, in the LEDs you will Vled = 24 * 3.5 = 84V

Due to the large difference between the line voltage and the output circuit behavior is similar to a current source, then don't worry about the output voltage without load.


As the average current that delivers this full-wave capacitive source is:




So Iout = 65 * (1-84/325) * 1 = 48 mA


If your LEDs have rated this then you are OK, if not, you must modify the value of the capacitor series and probably the series resistance of 100 ohms (more watts) if you need more current.

The LED's will flicker at 60 Hz. If it's unacceptable then disregard this post.

Click to expand...

Even more unwanted, operating the LEDs with a pulsating current reduces their efficiency and usable intensity.

By the way conventional circuits show the limiting resistor having a higher value than 100 ohms.

Click to expand...

These resistors are only provided to limit inrush and voltage transient related currents. They are already dissipating about 1/4 W each. The main current limiting function is provided by the 1 u capacitor, as explained in other posts.

The 230VAC x 1.414=345VDC with a small 10ufd/400VDC as tank..For a string of 25LEDs with forward drop of 1.6VDC we get 40VDC
This gives a 300VDC to be used for the constant drive currents .Kindly choose an appropriate resistance .
I would suggest a IR540 mosfet with res. at gate to control the channel currents needed.