Frequency parameters

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myfaithnka

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Hi all,
any one pls help me finding the equation for frequency here in this ckt.




i have changed the turns of the feed back,ended up burning that two transistors.


Pls help.
 

myfaithnka said:
Hi all,
any one pls help me finding the equation for frequency here in this ckt.

View attachment 89706


i have changed the turns of the feed back,ended up burning that two transistors.


Pls help.
Click to expand...

Hi myfaithnka
What about if you tell more about frequency of what ? did you attempt to draw a royer oscillator ?
I think more explanations about what is that and what you need is required .
Best Wishes
Goldsmith
 
Hi goldsmith,

actually i am new to circuits like joule ringers and royer oscillators.

This is a cfl driving circuit for 9w 6500k with self start.

The schematic seems very similar to royer oscillators.

I found this circuit is running in 13 khz (primary),attempted to make one like the transformer in this.

Even if i try to make exact same turns in transformer,the oscillation frequency switches to 1628 khz.

I assume there is no much role for feed back other than triggering the bases(have increased feed back windings just o burn to sc 1061 transistors).

The whole circuit uses only a 6 v to run cfl.

I am wondering how exactly ths circuit is working,pls help.
 

If you have increased the feedback then you are applying more voltage across the bases, so there will be more base current and dissipation. Also when one end of the coil goes positive it is clamped to the Vbe of one transistor, so the other end goes negative and potentially blowing the other transistor by exceeding its reverse Vbe.
Frank
 
Hi chuckey,

thanks for your time.

I did so in a miscalculation of increasing the inductance in the feedback windings will decrease the frequency.

I am stuck here trying to sort out how to pull the circuit to a specific frequency ( say 20 khz). Can you mention any formulae here.
 

I would try altering the .02 MF caps, they would be resonating the primary, also they would be reducing the amplitude of Dv/Dt at the collector of each transistor as it switches off, so protecting the collector junction. One of the transistors stays "on" all the time that the base winding is delivering enough voltage. So it might be worth putting a resistor across the winding to try and reduce the voltage more quickly, say 1K for starters. Another way would be to use one or more diodes between the base and emitter, cathode to base, to clamp the negative voltage to -.8V or -1.6V, so allowing more base current for the "on "transistor and protecting the "off" transistor.
Frank
 
Chuckey,

i will try them and get back to you. Thank you very much for your time.
 

Chuckey,

i am here again on another question.

Forgive me if its a blunder.

Is there any kind of impedance matching required? Between the load and the primary circuit. Yes,there is transformer...i am just curious how it is done...
 

For a switching inverter, the power transistors just switch their supply on and off to the transformers primary, so the voltage in is 2 X Vcc. This is stepped up/down by the transformer to the desired output voltage. There is a type of inverter called a fly-back converter, with these the fast change of current in the primary (Di/Dt), causes the inductance of the transformer (L) to produce a high voltage ( V = Di/Dt X L), which produces very high voltage (the transistors need to be rated for this), but the voltage is still transformed to the output voltage.
So its voltage matching rather then impedance matching, because you would want a PSU to operate at low currents (high impedance)as well as full load (low impedance).
Frank
 
Dear frank,

thanks for your time and valuable guidance,

forgive my ignorance,i have some doubts.

1) the actual input voltage of the transformer primary is so Vcc x 2,am I right ?

2) the approximate output voltage of transformer should be so Vsec= (dI/dt)pri x Lsec, was i mistaken ?

3)the transistor rating should be matched to the Vsec ?

4) PSU, I assume it is power supply unit. I am using a 6 V battery here( 4.5 Ah ). how is voltage matching done here?


thank you,

nithin
 

1. yes
2. I am not a power supply designer, so I think that the fly back stuff is not applicable. From consulting Wikipedia, fly back converters do not use a transformer, they just switch the current through a single inductor. With a well designed transformer, the tight coupling of the primary to secondary, results in the energy being transferred from the primary to secondary to the load,so the expression V = Di/Dt X L refers to the LEAKAGE inductance, which one hopes is very small.
3. V primary + a bit more for any ringing.
4. Your CFL, what does it require? florescent tubes are current driven, thats why they have a ballast unit. They also have to have a high voltage to "strike" them on starting. At a guess, 150V + to start, and 90V to run. but the current is limited to Ptube/90 A. So I would think that the transformers in commercial CFL converters are actually made with a high leakage inductance, which limits their current but makes for high voltage over shoots at the collector of the driving transistors and provides the higher voltage for striking the tube.
Hope I have not confused you too much
Frank
 
From consulting Wikipedia, fly back converters do not use a transformer
Click to expand...
they just switch the current through a single inductor
Click to expand...
Hi Frank
I afraid but fly back converter uses a transformer .
While switch is on the core of transformer is storing the energy . and while switch is off it will deliver it's stored energy into the load network .
For more clarification : when switch is on the transformer will have the out put ! but because the diode in secondary side is in reverse biasing , so there won't be any current through the out put capacitor from the secondary side for this moment of time . but capacitor is supplying the load at this time . it is why fly back converters are dealing with higher ripples in compare with Forward converters .
So when switch is turned off , then because of the inertia ( linear density ) the primary side will try to change polarity of the voltage across itself , so voltage across the secondary will be shifted in phase by 180 degree too . thus the out put diode will be turned on and the stored energy will be gave through the out put load and capacitor . so it is a transformer .
( Most of the books are dealing with these issues too and they confess it is a transformer . )
Best Regards
Goldsmith
 


Dear frank,

i needed a bit of research on this topic to understand what you told exactly. sorry for the delay from my part, I hope i am not bugging you.

The CFL i got here is like this

**broken link removed**

2 pin,6500k,starter and condenser inside it.

I think this not exactly the cfl light but just the cfl tube.

I am not confused,please tell more about them.

nithin

- - - Updated - - -


Dear frank,
With a lot of support from you, at last i have setup a board myself using transformer of 10 primary(with centre tap),2 feed back and 400 secondary,E 25 core size.ctc 880 bjt as switch. The cfl glows.the following is the wave form,



i changed the resistance from 150 to 120 ohms.

oscillations are seen at 13 khz now. the first wave is collectors voltage, second is base voltage.( voltage per division is 2).
I am a bit confused by the wave form.

In some parts,base voltage increases where collector voltage comes down.

I don't know how that happens,another thing is the tiny spikes seen in the waveform,this spikes are not ideal i think,i dont know how they comes to the circuit.

please help.

nithin.
 

If you look at the collector wave form, there is a short "vertical" period, where the collector voltage rises very rapidly. By transformer action, the other end of the transformer must go negative at the same speed, causing the negative spike on the other collector. It is caused by the off transistor coming on really fast.
Frank
 
Dear frank,

I am a bit happy that the circuit we just made has a capacity to illuminate for two hours,while the original board available in market works only for a hour and a quarter.

I am stumbled up on another problem,after exactly 2 hours and five minutes the cfl tube is turned off,but the oscillations are still going inside,so it eats up the 6v, 4.5 Ah battery power like joule ringers.

I made a modification in the circuit,to avoid this. but it seems not working.my idea was to sense the voltage and when it reaches almost 4,cut off by grounding the two bases.but something is wrong in this,the arrangement never seem working. please see the picture.



D2 and D3 comes from the bases,of the transistors. when battery is low Zener diode stops conducting,so the first transistor is turned off,making the other one turned on.Where am I got wrong here ?

please help,

Nithin.
 

You need a collector load for Q1, else when it turns off Q2 has the Vcc driving the base , which will blow it up. Try 1K. And short out R1. when Q1 conducts you have to make sure that its collector to earth voltage is less then the Vbe of Q2, the 100 ohms is not helpful! Just noticed, when Q2 is on its collector voltage should be, say .2V, add this to the volt drop across the diodes (.8?) and there might be enough Vbe for your oscillator transistors to carry on oscillating , but at a higher frequency. - just a point to look out for.
Frank
 
Dear frank,

sorry for being this late.

I think this idea wont work as i expected,as you said,it keeps on oscillating but with reduced amperage.

Pls suggest me a method to solve this problem,I got batteries refusing to charge under normal 12 v.

pls help.

nithin.
 

dear frank,

those diode are not available in india,with out a bulk order.

I tried to turn on two sc1061 transistors in place of diodes and found working,i started with OA79 signal diodes,then to bc547 ending up on sc 1061, they seem working but i am not sure why bc547 failed.

the cost of sc1061 is high, i am looking for a low cost solution.

what is your opinion.That was an junk idea,i am looking forward a simple idea from you. what is the simplest way to make them grounded.

nithin.
 

Hi frank,

finally i got this idea.

I cut out the two biasing resisters from vcc.

then added them both to the emitter of a CTC 1061.the collector of the same connected to vcc.

i used a 10 ohm resister to base of the transistor, when put to vcc the circuit starts magically .

i used a zener diode to bias the transistor,the problem is that it needs almost full voltage at base to bias.

so i guess, it calls for a duel stages that another transistor drives this transistor. but i dont know how to do it.

It seems that it requires a thrice stage if i guess correctly. some thing like increasing of gain by a 10 times etc.

pls help.
 

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