12 V inverter circuit need clarification

in simple 12v inverter circuit, at the output terminal of the primary winding transformer if i connect capacitor what will be the output? please let me know your all views

It depends entirely on what the "simple 12v inverter" circuit does. You need to show us a schematic and explain how the capacitor is connected and what value you are thinking of.

In a self oscillating inverter it might just shift the frequency, in a low power inverter it might overload the output and in a square wave inverter it might just produce spikes, we need to know more to be able to advise properly.

Brian.

Thanks a lot for your reply. . Please find the circuit details as an attachment. I am having doubt whether the output is DC output ? and why need the capacitor in output terminal ?

Hi,

I am having doubt whether the output is DC output ?

Click to expand...

Why?
* The description clearely says "AC"
* and generally a transformer output never can be DC.

Regarding capacitor:
It will reduce voltage spikes.

Klaus

Its essentially a push-pull square wave driver so it will produce a waveform far from ideal and depending on the load, might have sharp spikes. As Klaus states, the capacitor is there to trap some of the spike energy to prevent it reaching the load.

I think there should also be a capacitor from the bottom of R1 to ground.

Brian.

Thanks for your suggestion. Pleas let me know your view about the below question.

I have four question
1. if i add zenor didoe parallel at the output terminal what will happen ?
2. If i add c4 630 v instead of 600 v what will happen?.
3. how long i can run BULB with this circuit for example if run for 1 hour continuously is there having any problem occur anywhere in the circuit ?
4. if i want to give TV load what type problem i can face. how to rectify that.

I want to be understand or study this circuit completely and in my mind raising this question all the time.

1. it will either burn out the Zener diode or the inverter, whichever is weakest. A Zener behaves like a normal diode in forward mode and has controlled breakdown in reverse mode so it will quite effectively 'short out' the AC from the transformer.

2. A capacitor with higher voltage rating is normally OK. Don't use one with lower voltage rating though and make sure it isn't a polarized one.

3. It depends on the power needed by the bulb. That design is not very efficient so an exact figure is difficult to predict but working on the assumption it is 50% efficient, then half the power is lost as heat or looking at it the other way, twice as much power has to go in as you get out. For example, a 10W lamp would need it to draw 20W from the battery. You can then make a rough guess at how long it would run by calculating the current it would draw from the battery (I = W/V) so for 20W load and a 12V battery it would draw 20/12 = ~1.7A and a 7AH battery would last 7/1.7 = ~4.1 hours. Obviously, a more powerful bulb needs more current and the life will be shorter. What isn't easy to predict is how it behaves as the battery voltage starts to drop, it might stop working earlier than the prediction.

4. It will work as with the light bulb and the same calculation applies. However, a TV presents a load that changes with picture and sound levels and that circuit has no voltage stabilization so you might find the voltage fluctuates. How the TV copes with that would depend on its internal design.

Brian.

Thanks a lot for giving valuable information. Now some how i can understanding this circuits.

Square wave output

Generally i am asking or related to this circuit how can i understand if the raise of voltage ie the high peak voltage in circuit?
I studied about the peak voltage in internet. with that information i am not able to find the whether the high peak voltage rises in the circuit or not . Could you please explain simplest way how the voltage rise in this circuit DC or AC?
because if i build this circuit i want to know the what are all the cause behind this circuit that would be help me to take next level in Electronics
I am very grateful to you.

I'll try...

The CD4047 is wired in astable mode, that means the parts around pins 1,2 and 3 set the frequency of a free-running oscillator. Also inside the CD4047 is a divide by two circuit fed from the oscillator, this halves the frequency but more importantly, it makes sure the waveform is a square wave with equal high and low periods.

The output of the divider is on pins 10 and 11. They carry the square waves except one is inverted, when pin 10 goes high, pin 11 goes low and vice versa. There is never a time when both pins are high or both are low. When pin 10 is high, Q1 conducts current through the transformer to ground, when pin 11 is high Q2 conducts current to ground. The transformer primary (Q1, Q2 side) is center tapped, the turns of wire on the transformer core continue in the same direction from end to end with the center tap being a connection to the middle. As the power is fed in to the tap, the current will flow through the turns then through either Q1 or Q2 to ground, however the direction through the primary will be reversed from one to the other. When the current flows through the primary, it creates a magnetic field in the transformer core and as Q1 or Q2 conduct, the field will reverse polarity. You therefore have an alternating magnetic flux in the transformer core.

The secondary winding is also wound around the transformer core and the changing magnetic flux induces a voltage into it. The more turns you add to the secondary the more voltage it produces so you can use it to 'step up' the original 12VDC from the battery to a much higher AC voltage.

The main reason for inefficiency is that after Q1 finishes its conduction time, the transformer core is still magnetized and some of the time Q2 conducts is wasted in neutralizing the magnetism before it can re-magnetize it in the other polarity. This a made worse by the square wave being used because the change of polarity is very sudden, a better circuit would use a sine wave where the change is more gradual but such a circuit would be far more complicated.

Brian.

I got more information about this circuit. This information more valuable to me. once again thank you for spending your valuable time for giving these details.

Many electronic products will not work properly when powered by a simple squarewave inverter but it can power a heater or incandescent light bulb.