How does a UPS inverter works?

[Farrukh Jamal]

Hello dear group! I have just registered. I am actually stuck in the middle of my semester project. Kindly help me. Here are the details:

I have implemented an inverter circuit. Implementation was easy, of course, as i had the circuit diagram. Now i want to know how it exactly works. I mean the whole process of converting the Dc signal in to a square wave and then that in to an AC output. I also want to know the role of each component in this process. (The IC 4047B, Heatsync transistors etc...)
Here is the link to the circuit i used...

http://www.circuitstoday.com/simple-40w-inverter
Regards,
Farrukh Jamal

 

[(*steve*)]

Unfortunately we don't have access to your C drive.

 

[davenn]

As Steve said we cant acces your HDD, upload a suitably resized pic to the forum or maybe to a image site like flicker so we can see it.

if you do some googling on inverters you will find there are 2 basic types pure sinewave and modified sinewave
Both them involve producing an oscillator from an input DC voltage, the resulting square or sinewave voltage is fed to a transformer where it is stepped up to the 110VAC / 240VAC (depending on your country) for use by your equip.

Thats the basics and your unit will prob follow that model

Dave

 

[Farrukh Jamal]

Pardon me for the mistake. I have updated the link. Kindly check it now.
Basically, the one i have implemented give square wave form as the output to the inverter circuit. That means a squarewave on transformer's primary side. That means, the stepped up 240V signal will also be squarewave.. Right??

 

[Farrukh Jamal]

Sorry! pardon me..
I have updated the link, by the way.
Kindly check it and provide me with some help..

 

[duke37]

The 4047 contains an oscillator whose frequency is set by a resistor and capacitor ( F = 1/(4.4*R*C) There is an internal divide by two circiut and two outputs, the division makes an equal on/off times and the outputs are in antiphase, When one is high, the other is low so that either one or the other pair of transistors is turned on.
Since the 4047 can only provide a low current, this has to be amplified by a BC557 and further amplified by the TIP3055. When turned on the transformer connection is pulled towards ground so that there is ~12V across that half of the winding. The other half of the winding is coupled to the first and will rise to ~24V.
Before the transformer saturates the input needs to be changed over (using the correct frequency). The square wave output is taken from the other winding.
The transistor connection is known as a Darlington connection and gives very high current gain but cannot work at less than a volt when turned on, this voltage drop means a power loss and a heat sink is needed to spread the heat to keep the temperature of the TIP3055 within acceptable levels.
In the link you gave, there is a 100W invertor using IRF540 mosfets instead of the bipolar transistors, the mos fet can be turned on with negligible current and provide a very low resistance when turned on. The power loss is therefore very low and a heat sink may not be needed.

 

[Farrukh Jamal]

Thankyou so much duke. You helped a lot..