Circuit Explanation: 12V CFL and LED Usage?

[Manoj Sahu]

I stumbled upon this inverter circuit while searching for the same on internet. Can anyone please explain working of this circuit to me.

 

[Svein]

The CD4047 is connected as an oscillator (frequency determined by19kΩ / 0.22μF). The outputs on pins 10 and 11 are in opposite phase (when one is high, the other is low) and in 50% duty cycle. Therefore the two MOSFETs are turned on alternately, driving the transformer with a square wave. This waveform is amplified through the transformer, creating a 230V square wave.

 

[Manoj Sahu]

The CD4047 is connected as an oscillator (frequency determined by19kΩ / 0.22μF). The outputs on pins 10 and 11 are in opposite phase (when one is high, the other is low) and in 50% duty cycle. Therefore the two MOSFETs are turned on alternately, driving the transformer with a square wave. This waveform is amplified through the transformer, creating a 230V square wave.

Thanks. How do I calculate the current and frequency ?

 

[Svein]

The free-running frequency (taken from the data sheet) is typically 1/(2.2*R*C) = 1/(2.2*19E3* 0.22E-6)Hz ≈ 110Hz. The output frequency is half of that.

 

[CWatters]

The current depends on the load. Note that the output transformer is a 12:230 step up transformer so the current flowing in the FETs is much much higher than the load current. You may well need heatsinks on the FETs and suitably sized conductors/wires and 12v battery.

 

[Manoj Sahu]

The current depends on the load. Note that the output transformer is a 12:230 step up transformer so the current flowing in the FETs is much much higher than the load current. You may well need heatsinks on the FETs and suitably sized conductors/wires and 12v battery.

If I use two 20w cfl and one 3w led light, will the circuit be able to draw that kind of load? How much power the circuit is capable of drawing?

 

[Manoj Sahu]

The current depends on the load. Note that the output transformer is a 12:230 step up transformer so the current flowing in the FETs is much much higher than the load current. You may well need heatsinks on the FETs and suitably sized conductors/wires and 12v battery.

As you said that FET's will draw much more current, will it not damage the transformer winding (assuming transformer is rated 12-0-12/230V, 5amp).?

 

[Manoj Sahu]

The free-running frequency (taken from the data sheet) is typically 1/(2.2*R*C) = 1/(2.2*19E3* 0.22E-6)Hz ≈ 110Hz. The output frequency is half of that.

If I have to get frequency =50.xx Hz assuming that CD4047 is used in Astable mode where Ta=4.40RC, do I have to increase the resistance (+2.5 ohm)/capacitance.

 

[Manoj Sahu]

The CD4047 is connected as an oscillator (frequency determined by19kΩ / 0.22μF). The outputs on pins 10 and 11 are in opposite phase (when one is high, the other is low) and in 50% duty cycle. Therefore the two MOSFETs are turned on alternately, driving the transformer with a square wave. This waveform is amplified through the transformer, creating a 230V square wave.

How do you make an IC to work on 50% duty cycle ?

 

[CWatters]

If I use two 20w cfl and one 3w led light, will the circuit be able to draw that kind of load? How much power the circuit is capable of drawing?

Thats a very light load. Shouldn't be a problem for that circuit.

 

[davenn]

I stumbled upon this inverter circuit while searching for the same on internet.

and just a little bit of advice
Next time post a pic with the correct orientation so everyone doesn't have to kink their neck to read it

 

[Manoj Sahu]

and just a little bit of advice
Next time post a pic with the correct orientation so everyone doesn't have to kink their neck to read it [emoji2][emoji2]

Oh yeah, that I forgot. Thanks for the bit of advice. [emoji4]

 

[Svein]

How do you make an IC to work on 50% duty cycle ?

Run the output through a "toggle" flip/flop.

 

[Manoj Sahu]

Run the output through a "toggle" flip/flop.
View attachment 195028

Thank you. I have one doubt though. How should I calculate output power ?

 

[CWatters]

You said the load is one 20W cfl and a 3W LED so the load power is 20 + 3 = 23W.

The load current is 23W/230V = 0.1A

The circuit diagram says it can deliver 5A so the load is only 0.1/5 * 100 = 2% of the maximum.

 

[CWatters]

Have you considered using a 12V cfl lamp and a 12V LED instead of 230V ?