Finding turns ratio of an unknown transformer(small)

[savvej]

I opened up a power adapter to see whats inside.I have attached the pics of it.I tried to reverse engineer and draw the schematic of the circuit.I could'nt relate it with any of convertors learnt at the university.
Now the circuit involves a small transformer.I basically want to simulate the circuit in pspice to study its working.
I wished to know how to find the turns ratio of an unknown transformer.
The power adapter was rated for 10.5V DC 550mA.

Also when I measured AC voltage at the output of the transformer,it was around 10V(ac).
I desoldered the tx.I have a 500ma 230V -11V tx .I tried connecting this tx o/p to the power tx and find the output.From the ratio of the voltages I can find the turns ratio.But as soon as I connected the TX,it(230-11V) became hot and when I measured the voltage across input ,it was almost zero(As if equivalent to shorting).So I added a resistor in series of 33Ohm(1W) and then again the voltage across the primary was zero.Now I know that the resistance in series should be small as compared to the impedance of the transformer.I tried to measure the dc resistance of the tx using multimeter ,it was around 0.6 Ohms.I have no clue regarding its inductance hence I don't know what resistance to put in series to as to get considerable drop across the primary.

Any other idea to find the turns ratio?
One idea which I have is to measure inductance of the coils on both sides(using some inductance meter which I don't have) and considering ideal tx,turns ratio is proportional to sqrt of ratio of inductances of the coils.Actually to simulate the circuit too I require to enter the values of the inductances of the coils rather than the turns ratio.Any thoughts on easy circuits to measure inductance?

 

[Nigel Goodwin]

Presumably that's a switchmode supply, so it's NOT a conventional mains transformer.

If you want to know the turns ratio, then unwind it and count them.

 

[savvej]

btw,the power transistor ,the one in middle left used is mje13003.
Also the zener diode on top right seems to be of 6.8V rating(though its not clear)
I will be putting my drawn schematic by tonight.

 

[savvej]

@Nigel Godwin it does look like a switch mode power supply but I dont know how the switching is taking place.I am posting the recovered schematic in some time.

 

[Nigel Goodwin]

@Nigel Godwin it does look like a switch mode power supply but I dont know how the switching is taking place.I am posting the recovered schematic in some time.

The transistor is the switching element, presumably in a self oscillating configuration.

 

[eTech]

I opened up a power adapter to see whats inside.I have attached the pics of it.I tried to reverse engineer and draw the schematic of the circuit.I could'nt relate it with any of convertors learnt at the university.
Now the circuit involves a small transformer.I basically want to simulate the circuit in pspice to study its working.
I wished to know how to find the turns ratio of an unknown transformer.
The power adapter was rated for 10.5V DC 550mA.

Also when I measured AC voltage at the output of the transformer,it was around 10V(ac).
I desoldered the tx.I have a 500ma 230V -11V tx .I tried connecting this tx o/p to the power tx and find the output.From the ratio of the voltages I can find the turns ratio.But as soon as I connected the TX,it(230-11V) became hot and when I measured the voltage across input ,it was almost zero(As if equivalent to shorting).So I added a resistor in series of 33Ohm(1W) and then again the voltage across the primary was zero.Now I know that the resistance in series should be small as compared to the impedance of the transformer.I tried to measure the dc resistance of the tx using multimeter ,it was around 0.6 Ohms.I have no clue regarding its inductance hence I don't know what resistance to put in series to as to get considerable drop across the primary.

Any other idea to find the turns ratio?
One idea which I have is to measure inductance of the coils on both sides(using some inductance meter which I don't have) and considering ideal tx,turns ratio is proportional to sqrt of ratio of inductances of the coils.Actually to simulate the circuit too I require to enter the values of the inductances of the coils rather than the turns ratio.Any thoughts on easy circuits to measure inductance?

If you can determine which wires represent which coils...you can connect one coil to a signal generator, the other coil to a scope or ac voltmeter. Then feedaim a 1 vrms, 1khz signal and monitor the output voltage.
You should be able to get the ratio that way.

 

[savvej]

I checked the zener for its breakdown voltage.I found it to be 6.6V (though visibly it seems to be 6.8V) .I have attached the schematic(which I recovered from the PCB) alongwith.Please help me figure out working of the circuit.

 

[savvej]

@Etech I don't have a signal generator.Any other idea?

 

[ronsimpson]

@Etech I don't have a signal generator.Any other idea?

What do you have?
With a scope you can look at the wave form on each winding as the supply operates. "I don't have..."
This supply is on the power line so connecting a scope is not easy!

 

[eTech]

@Etech I don't have a signal generator.Any other idea?

I used to build and test magnetics and transformers in a QC dept
Years ago. That's really the only reliable way to get the ratio without
Taking it apart. You can download simple sig gen software u can use
for this on the Internet. It uses the line out on the sound card for signal output.

 

[Mr RB]

The turns ratio won't be that far off the voltage transformation ratio, although is usually less in SMPS compared to linear PS. so slightly less than 230:11 it might be 170:11 or 200:11, somewhere around there.

 

[eTech]

The turns ratio won't be that far off the voltage transformation ratio, although is usually less in SMPS compared to linear PS. so slightly less than 230:11 it might be 170:11 or 200:11, somewhere around there.

If you use that method... Then for the 230v input/10.6v output, the turns ratio is about 21.9:1.

 

[Nigel Goodwin]

If you use that method... Then for the 230v input/10.6v output, the turns ratio is about 21.9:1.

You're rather ignoring the fact that it's a SMPSU, so the output voltage is dependent on the pulse ratio as much as the windings ratio - not to mention where has be got these voltage readings from?, and how did he measure them?.

 

[eTech]

You're rather ignoring the fact that it's a SMPSU, so the output voltage is dependent on the pulse ratio as much as the windings ratio - not to mention where has be got these voltage readings from?, and how did he measure them?.

I'm ignoring that because it wasn't what was asked.
They asked how to get the turns ratio. I explained how in an earlier post.

eT

 

[Mr RB]

Yep using a signal generator and appropriately high freq and using the 'scope will be a good method but the OP asked about circuit he could build to measure inductance which probably implies he does not have much in the way of test equipment.

From my experience with simple cheap SMPS they run close to the turns ratio of Vin:Vout to give good efficiency, but with maybe 15% higher Vout ratio to allow for regulation by reducing pulse width. That should put it somewhere in the ballpark. Sometimes the turns will be even less than that as turns cost money, both for the wire and for winding time.

 

[savvej]

Can anyone throw some light explaining the topology of the convertor like buck,forward or fly-back etc.I am unable to relate it to any of those.
I am unable to decipher the working of the circuit.As someone told previously that the transistor is in self oscillating configuration.Assuming so,replacing the oscillator part by abstraction of switches,can anyone show how is the convertor working?

 

[ChrisP58]

Wikipedia has a good overview of switch-mode power supplies.
http://en.wikipedia.org/wiki/Switched-mode_power_supply

A couple of good sites for more detail are:
http://www.smps.us/
**broken link removed**


From a theoretical point of view, the primary to secondary turns ratio in a flyback converter can be almost anything. The PWM control circuit will adjust the duty cycle to compensate.

There are however, practical limits that govern the actual choice of the turns ratio.

Very high ratios result in low duty cycles with short on times. This means that all the power through the supply has to get into the transformer in short, high current pulses. This generates higher RF energy, as well as higher stress on components. The filter capacitor heats up more from the high ripple current, and the transistor sees both higher current pulses, as well has higher voltage stress from the reflected secondary voltage. Also, modulating narrow pulses is more difficult.

Low turns ratio reduces primary stress, but the secondary stresses go up. Higher ripple current in the capacitors, and the output rectifier sees both higher pulse currents, as well as higher reverse voltages.

So there is definitely a sweet spot for turns ratio, but the practical range can still be fairly broad. That is why you can have one converter that works well over an input voltage range of 3:1 or more.

 

[Mr RB]

Can anyone throw some light explaining the topology of the convertor like buck,forward or fly-back etc.I am unable to relate it to any of those.
I am unable to decipher the working of the circuit.As someone told previously that the transistor is in self oscillating configuration.Assuming so,replacing the oscillator part by abstraction of switches,can anyone show how is the convertor working?

Not from your previously posted schematic!

It's almost certainly a forward converter although you seem to have some errors in your schematic around the primary coil (connects to the collector) and probably errors elsewhere too.

 

[savvej]

@ Mr. RB ,thanks for your reply.I rechecked the schematic with pcb and indeed there was a mistake at the collector side.I have redrawn the schematic.I am posting it alongwith.Could you now explain how the circuit is working?

 

[Nigel Goodwin]

It's a crude blocking oscillator, the feedback is via the extra winding L2. The 1M resistor provides start-up current to turn the transistor on in the first place (and is a common cause of failure in such supplies).