50 watt LED driver circuit

[ronsimpson]

The 60hz from the power line is not across the power line. The transformer is for the 100khz signal not the 60hz signal.

 

[Inquisitive]

https://www.falstad.com/circuit/e-fullrect.html

 

[Bhupesh]

The 60hz from the power line is not across the power line. The transformer is for the 100khz signal not the 60hz signal.

View attachment 99109

Yeah this is what i am unable to understand. The IC samples at 66KHz but it would sample the rectified voltage which doesnt have a negative peak anywhere. So where did the negative voltage come from?

 

[Bhupesh]

Is the negative induced in the secondary?
And i guess the IC is doing the job of switching at 66KHz the rectified output of the bridge. So where does the negative come from?
please help.

 

[ronsimpson]

"Negative" is relative. It depends on where you measure from. When you say "across the transformer" then we measure from pin to pin on the transformer not from pin to ground.

So lets only look at the black part of the picture.
When you measure from the IC's ground pin: The IC sees Vin=160 volts and Vout+Vin=240V+ring. (see in red)
One end of the transformer is connected to Vin which is 160V. From the transformer's point of view;There is -160 and +80 volts. (see in green)

 

[Bhupesh]

But how will this be rectified in the secondary if the complete waveform is positive? The schottky will be forward biased only...No swtiching will take place if there is no negative with respect to the ground..

 

[Bhupesh]

Is the capacitor, connected from primary to ground on the secondary side, doing something to shift the waveforms?
Please help...

 

[Bhupesh]

Is there any simulation software in which I could measure the waveforms at different sections? I've tried multisim and Proteus but without any results.
A simulation would provide me a great understanding about the complete circuit.
Please help.

 

[ronsimpson]

This is what is seen from pin to pin on the secondary of the transformer.
Zero voltage is the blue line. "0"
The part above the blue line is "+" and will pass through the diode and charge up a output capacitor. Note this voltage is the same at high/normal/low line.
The part that is below the line is "-" and is blocked by the output diode.
What is below the line is from the line voltage.
I tried to make the wave form at low line voltage at the top.
I tried to make the wave form at high line voltage in green at the bottom.
The red lines are input current and output current.

When the line voltage is high or low it only effect the "-" part which the output diode blocks.

 

[alec_t]

I've tried multisim and Proteus but without any results.

They must give some results. I don't see why either shouldn't be capable of indicating/measuring waveforms. Have you checked their 'Help' files?
Several members here use the free LTspice program.

 

[Bhupesh]

And what about the clamping circuit in parallel to the transformer primary? How does that work?