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Pure sin wave inverter

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thanx tahmid ,,,
plz tell me how feedback works in sg3525?how dutycycle is changed automatically in sg3525,,??
 

There is an internal error amplifier in the SG3525. Pins 1 and 2 are the inputs to the error amplifier. They are the inverting and non-inverting pins, respectively.
Think of it as a comparator.
When voltage on pin 1 (INV) is greater than pin 2 (NINV), duty cycle is reduced. When voltage on pin 2 (NINV) is greater than pin 1 (INV), duty cycle is increased.

In the circuit I showed above, pin 2 (NINV) is connected to pin 16 (VREF). So it is at a potential of approximately 5V. so, when pin 1 is at greater than 5V, duty cycle is increased. When it is at lower than 5V, duty cycle is decreased. The pot is adjusted such that at the required output voltage, the voltage on the wiper is approximately 5V, so that by attempting to keep INV=NINV, SG3525 regulates the output by adjusting the duty cycle.

Hope this helps.
Tahmid.
 
thanx again tahmid
i hve used 3.2 volt at pin2 , and when my potential at pin1 is grater than pin2(say 3.5 volts), the output goes to 0, and when my potential at pin1 is lower than pin2(say3.0 volts), the output remains constant at 7.6 volts measuerd by the multimeter , i have used a 12volts to supply the sg 3525 bu at output i get 7.6volts only....so where is the problem ... my feedback is not wrking why?
 

You have probably attempted to witness the feedback yourself by manually connecting the said voltages to the relevant pins. Problem with this is that, the change in duty cycle is so quick that when a duty cycle decreasing repeatedly occurs, it very quickly falls to 0 before you can measure it. Similarly with increasing duty cycle. Very soon it goes up to maximum duty cycle.

Construct the circuit with feedback connected from the output. Then you can observe feedback at work.

Hope this helps.
Tahmid.
 
tahmid , suppose at the output, the voltage is 500 volts at no load, and then by using a voltage divider i give around 5 volts to pin1, now consider two cases
case(1) when i put a small load my terminal voltage goes to 430volts,without any feedback.
case(2) when i put a 100watt bulb ,my terminal voltage goes to 275 volts ,without any feedback.
now as i applied the feedback by considering 500volts at the output and my transformer is designed such that it will give 350 volts at the , so what will happen in this case??
 

If in case 2, when the voltage dropped to 275V without feedback, the duty cycle reached the maximum possible value, then feedback can't "work" or do anything since the voltage can not be increased as duty cycle can not be increased any further.

Hope this helps.
Tahmid.
 

na na tahmid , u hve wrongly percieved my question... suppose my transformer design is such that when input voltage is 10.5 then the output voltage should be 330v
so if my input voltage is 12.5 my output voltage will be around 392volts
now when i connect a bridge rectifier at the output voltage will be 1.414*V(output) i.e 1.414*392 =554.288v. now my question is , should i apply feedback to this voltage using voltage divider or should i connect a small load then i apply the feedback?, what should i do so that , my output voltage goes to 330vdc at the output?
 

You should obviously use feedback. But there must always be a minimum load. If you don't have the minimum required load, the feedback loop may become unstable and not function as desired.

output voltage will be 1.414*V(output) i.e 1.414*392 =554.288v
Click to expand...

No, output voltage will be V(output), not V(output) * 1.414V.

V(output) * 1.414 is only for sine waves. Here you have square wave. At maximum duty cycle Vout = Vpeak = V(output).

Hope this helps.
Tahmid.
 
but tahmid , when i put a capacitor at the output, the no load voltage increases ,as measured by the multimeter it goes to above 500volts, now as u said that put a small load at the output , now suppose i put a 250k resistor 2 watts in series with a LED, then there is no significant change at the output..say it is around 400volts ,now as u said that ,after applying the small load ,apply the feedback , now after applying the feedback ,output voltage should be 400v constant, but our aim is to get 330v at the output ,,so how it can be fulfilled?
 

When you put a capacitor at the output, it charges to the peak voltage and thus you get that reading. To prevent this from happening and to obtain an "average" voltage output, an inductor must be used that must be placed between the bridge rectifier output and the capacitor. After this, you must put a load. When I say small, I mean in the order of 10W to 20W or more. 2 250k resistors in series with a LED is not really much load and won't be enough as "minimum load" unless you have quite a large inductance. Once you have the proper inductance in place, then connect the feedback circuitry. Since your transformer is designed to give a large enough output voltage, feedback should be able to get the output stable at 330V.

Hope this helps.
Tahmid.
 
thanx tahmid ,,,,u r saying that i should apply the feedback when my output voltage is 330volts at the terminal ,,,(1) am i right?

u hve said that use 2mh to 3mh inductor at the output ,
2) my question is how can i calculate this value,
3) i dont have lcr meter ,,so is there any way to design the inductor without lcr meter?
 

You should connect the feedback circuit such that it sets the output voltage at 330V. Since you mentioned that you use a 3.2V reference, your calculation should be that the output of the voltage divider will be 3.2V when input voltage to the voltage divider is 330V.

2) There are quite a few formulae involved. I had come across them in an application note by ST Microelectronics on a DC-AC inverter.

3) If you know the parameters of the core, specifically the AL value (inductance per turns squared), then you can wind an inductor by calculation without the need to use an LCR meter. However, if you don't know that, you will definitely need an LCR meter. Since high accuracy or precision in measurement is not required, you can even make one yourself. There are quite a few available online. The IronBark LC meter was a good design. You can use that.

Hope this helps.
Tahmid.
 
thamid ,still one small confusion... u have said that i should connect the feedback circuit such that it sets the output voltage at 330volt , but my question is ,, should i first pull down my outut voltage to 330v and then by using voltage divider ,i connect the feedback, or should i connect the feedback first then adjust the voltage? if so , then how can i adjust the voltage if my feedback is once connected, for example - say at the output i m getting 392volts and i hve connected the feedback, now this 392v will be constant , then how can i adjust this 392volts to get output 330volts , because i hve connected the feedback, n feedback will try to keep the constant 392volts at the output..
 

No. The output voltage will be compared against a reference voltage, which the feedback circuit will try to establish.

You said your reference is 3.2V and 330V is your desired output voltage. To gain a 3.2V output from a 330V input, the required voltage divider will have resistances in the ratio 326.8:3.2 = 102.125:1. I'll round that off to 102 for simplicity's sake. So you can have a 102k resistor and a 1k resistor for the voltage divider. Then you connect this. Now if you have the feedback circuit, the reference is now 3.2V (which is for an output of approximately 330V). So, reference voltage / target voltage is 330V. So, the feedback circuit will attempt to keep output constant at 330V. So voltage won't rise to 392V, but stay fixed at 330V since the SG3525 will limit PWM duty cycle such that the output stays constant at about 330V.

I hope I could make you understand this. If you are still in doubt, feel free to ask questions.

Hope this helps.
Tahmid.
 
tahmid ,,now i will tell u, what i hve understood till now from u, please correct me if i m wrong anywhere
1) set the refrence voltage first say 3.2 volts
2) now design a voltage divider such that when the input of the voltage divider is 330volts ,output will be 3.2volts,
3) now as my transformer is giving output of 392volts which means when this 392volts goes to my designed voltage divider circuit , then the output of the voltage divider circuit will be greater than 3.2volts (say 3.6 volts), now 3.2volts is compared with 3.6volts and the duty cycle will be reduced such that the output voltage will be equal to 330volts so that the output of the voltage divider is 3.2volts
4) if i apply a load , then output voltage will decrease (say 275volts), now the voltage coming out of the voltage divider will be less than 3.2volts(say2.9volts ),now again this 2.9volts is compared with 3.2volts ,and the dutycycle is automatically increased such that, the output voltage goes to 330volts.


in some thread i hve read that we can use a pot in the voltage divider at pin 1 ,by adjusting this pot ,we adjust the output voltage ,,how it is possible? as per as i learned from u is ,,i shoud connect the pot to pin2 so that by adjusting the refrence voltage ,we can adjust the output voltage, am i right?
 

Your understanding is correct. There is a mistake in one place.

Your transformer is designed such that the output increases to 392V at 12V (or thereabouts) input.

When SG3525 operates, the duty cycle increases slowly from 0 upwards. When it goes to that duty cycle where voltage just crossed 330V, duty cycle no longer increases. So, voltage no longer increases. So, the output voltage does not go all the way to 392V. But from 0 to 330V upwards and then regulated around 330V.

You can adjust the reference voltage, or you can adjust the input voltage. It depends on your circuit configuration. In the circuit I posted (post #81), it uses a pot on INV pin. There is a fixed reference and the adjustment of the pot basically adjusts for the voltage division ratio. Either way works. Both methods do the same thing: compare output voltage against reference voltage.

Hope this helps.
Tahmid.
 
tahmid in your blog https://tahmidmc.blogspot.in/2012/12/low-side-mosfet-drive-circuits-and_23.html
1) among fig 4, 5, 6 , which circuit should i select for sg3525
and how to calculate the value of r28, r39 in fig 4
and which bjt should i use at Q9,Q11,Q12?
2) and when i change the operating frequency of sg3525 ,my output voltage also changes ,why?
3) what is the estimated power that ETD44 core can provide at 50khz frequency?
 

1) Use Fig. 5. You can see its implementation in the circuit presented on this post: https://www.edaboard.com/threads/272940/#post1182819
2) By how much does the output voltage change? How much did you change the frequency?
3) My guess is that it should be greater than 300W. According to TDK document e141, the estimated power (forward topology) is 523W at 100kHz.

Hope this helps.
Tahmid.
 

i hve connected a 100w bulb at the output and then measured the voltage
at 35khz voltage was 286v
at 70khz voltage was 232v

- - - Updated - - -

tahmid what frequency should i use for etd44 core? as at higher frequency output is also high, but is there any disadvantage of using higher frequency?

in input of my transformer i hve only three turns,, i hve closely wind the three turns and it covered only half the area of the bobbin .so my question is should i closely wound the thee turns or widely wind the three turns so that whole bobbin get covered?
 

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