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  #1  
Old 02-21-2011, 01:18 PM
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Default How to give feedback for SG3525 based smps?

Hello,

Iam designing an smps based on the attached schematic.

My requirement is,
- The input voltage for the smps is 230
- The output voltage should be 12V DC.
- I have to show a constant 12V DC for different inputs like 180v, 190v,.... 250V.

I have tested the SG3525A on a bread board. It is OK with the two complementary pulse sequences.

How can i connect the output sample to the IC to get my requirement.
I have seen some circuits in the internet. But different circuits had different techniques.

By seeing the block diagram in the datasheet, I understood that the feedback signal should adjust the reference voltage which compares with the sawtooth.

How can i implement this???

I want some detailed explanation please...

thankyou
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  #2  
Old 02-21-2011, 01:44 PM
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Your attachment pretty much shows how to do it.

There is a 4N35 optocoupler whos' IR emitter is powered via a couple of Zener diodes, current limiting resistor, and a pot to adjust the output voltage.

As the output voltage rises, the IR emitter turns on, which turns on the transistor on the output side of the 4N35, so the feedback input gets a higher voltage, causing the regulator to decrease the output duty cycle.

The optocoupler is used to provide isolation of the primary side from the secondary side.

Have a look at the attached; it's a simple flyback converter using a common 555 timer as a controller. An optocoupler is used to isolate the secondary from the primary. It's simply a variation on the same theme, but there are a lot fewer parts, so it's easier to understand.

In this case, reducing the control voltage input to the 555 timer causes both the 555 output PWM duty cycle to decrease, and an increase in frequency. The load varies from 10mA to 200mA (the red square wave). You can see how quickly the regulator responds to the load.

C5 provides a "soft start" function, so that the output from the regulator does not significantly overshoot the desired output voltage.
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File Type: png Flyback 12v to 6v.png (81.8 KB, 119 views)
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Old 02-21-2011, 02:09 PM
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Thanks Mr.SGT.
Thanks for the detailed explanation.

But still i have some doubts.
1. In your schematic, the input is 12V DC. What if it is 3V DC? How can it gives 6V output?

2. I dont know much about a zener diode. If the output is more than 5.1v, the zener should conduct.i.e., the optocoupler starts giving signal. right. Am i correct? If iam correct, when the output voltage is 6V, it should send a feedback signal. then automatically the switching frequency will be reduced. so, we cant get a 6V. I think the output will be 5.1V.

I think I was wrong
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Old 02-21-2011, 02:54 PM
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Quote:
Originally Posted by onlyvinod56 View Post
Thanks Mr.SGT.
Thanks for the detailed explanation.

But still i have some doubts.
1. In your schematic, the input is 12V DC. What if it is 3V DC? How can it gives 6V output?
You are missing the point of me posting this schematic; it was merely a simplified circuit showing a method of feedback from the secondary side using an optocoupler to change the PWM duty cycle on the primary side. If you attempted to power this circuit from 3v, it would not work, as that is below the minimum required to operate a 555 timer. If you tried to run it on less than ~7v, it wouldn't work as well either, as the MOSFET gate's Vgs would not be high enough to turn on fully. Also, the ratio of turns on the primary to the secondary transformer would need to be changed.

Quote:
2. I dont know much about a zener diode. If the output is more than 5.1v, the zener should conduct.i.e., the optocoupler starts giving signal. right. Am i correct? If iam correct, when the output voltage is 6V, it should send a feedback signal. then automatically the switching frequency will be reduced. so, we cant get a 6V. I think the output will be 5.1V.
Zeners start to conduct current before their reverse breakdown voltage is released. The IR emitter in the optocoupler has a Vf of ~1.2v when 10mA current is flowing through it, but starts conducting at a lower Vf.

If you add the 5.1v of the Zener to the 1.2v of the IR emitter, you wind up with 6.3v; however as I already mentioned, they start conducting before their breakdown or Vf is reached. And, as I previously mentioned, this is a simplified circuit.

You should try simulating this circuit and see what happens.
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Old 02-21-2011, 03:18 PM
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Quote:
Originally Posted by SgtWookie View Post
it was merely a simplified circuit showing a method of feedback from the secondary side using an optocoupler to change the PWM duty cycle on the primary side.
Ok. its a simple circuit. Iam not stressing the point.

Quote:
If you attempted to power this circuit from 3v, it would not work, as that is below the minimum required to operate a 555 timer. If you tried to run it on less than ~7v, it wouldn't work as well either, as the MOSFET gate's Vgs would not be high enough to turn on fully. Also, the ratio of turns on the primary to the secondary transformer would need to be changed.
Supplying 3V is not my intention.
For 12V input & a fixed frequency, the output will be 6V. Its perfect.
What if the input voltage is reduced a little bit....like 9V. This 9V definitely effects the output voltage. My question is, at this situation the output voltage will be less than 6V., in order to increase the output voltage, the switching frequency has to be increased.
1. How is it possible to sense the low voltage at the output?
2. How this can be given to the controller IC?

From the attacgment in my first post of this thread, I think the feedback operates only when the output voltage is greater than the required.
If the input voltage is 110V, explain how it increases the switching frequency?


Quote:
Zeners start to conduct current before their reverse breakdown voltage is released. The IR emitter in the optocoupler has a Vf of ~1.2v when 10mA current is flowing through it, but starts conducting at a lower Vf.

If you add the 5.1v of the Zener to the 1.2v of the IR emitter, you wind up with 6.3v; however as I already mentioned, they start conducting before their breakdown or Vf is reached.
ThanQ, I got the zener concept.
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Old 02-21-2011, 03:55 PM
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Quote:
Originally Posted by onlyvinod56 View Post
For 12V input & a fixed frequency, the output will be 6V. Its perfect.
What if the input voltage is reduced a little bit....like 9V. This 9V definitely effects the output voltage.
I have not run extensive simulations on the circuit, nor built it.
You are more than welcome to experiment with it.

As the input voltage decreases, greater current flow will be required in order to maintain the output voltage under load. At some point, the current required will exceed the component ratings. There are no safeguards in this circuit to prevent that occurrence; it's simplified.

Quote:
My question is, at this situation the output voltage will be less than 6V., in order to increase the output voltage, the switching frequency has to be increased.
No, the OUTPUT DUTY CYCLE needs to be increased. At some point, a real-world transformer will become saturated, causing the MOSFET to overheat - but that's beyond the scope of this thread.

Quote:
1. How is it possible to sense the low voltage at the output?
If the output of the optocoupler is not conducting, the PWM duty cycle is at maximum, causing the output voltage to increase.

Quote:
2. How this can be given to the controller IC?
My simple circuit shows how the feedback works. Don't you see the yellow trace, V(cv), on the plot?

Quote:
From the attacgment in my first post of this thread, I think the feedback operates only when the output voltage is greater than the required.
If the input voltage is 110V, explain how it increases the switching frequency?
The SG3525 operates at a fixed frequency. The regulator compares the feedback input to the reference voltage, and adjusts the duty cycle of the output in an attempt to keep them equal.
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  #7  
Old 02-21-2011, 04:19 PM
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Quote:
Originally Posted by SgtWookie View Post
No, the OUTPUT DUTY CYCLE needs to be increased. At some point, a real-world transformer will become saturated, causing the MOSFET to overheat - but that's beyond the scope of this thread.
ok


Quote:
If the output of the optocoupler is not conducting, the PWM duty cycle is at maximum, causing the output voltage to increase.
This is the answer what i want.
So, when the optocopler is off, the duty cycle increases.
Thanks Mr SGT.

I'll simulate it now. I'll be back again if i face any problems.
ThanQ
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  #8  
Old 03-10-2011, 10:22 AM
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hello, i have connected the hardware.
see the attachment.

output is connected at the pin shown by the arrow.
The puses are ok and good.

then i shorted the opto-coupler pins.(like when the opto-coupler conducts),
the output pulses are dead.

Can anybody explain how the smps o/p voltage can be adjusted with this situation????

see the video link.http://www.youtube.com/watch?v=UNyoNk6hIhE

and please explain the area shown in the attachment 2. What is the output of the error amplifier?
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File Type: jpg untitled1.JPG (89.9 KB, 93 views)
File Type: jpg untitled.JPG (63.1 KB, 83 views)

Last edited by onlyvinod56; 03-10-2011 at 10:49 AM.
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