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What determines the voltage/current ratings for smps transformers?

ronsimpson

Well-Known Member
Most Helpful Member
#2
The data sheet is not good.
It was designed for 60khz switching. Vin= 85 to 265V, Vout 12V @ 500mA 6W
The breakdown voltage from any primary to any secondary is 4000 volts for 1 minutes.
 
Thread starter #3
The data sheet is not good.
It was designed for 60khz switching. Vin= 85 to 265V, Vout 12V @ 500mA 6W
The breakdown voltage from any primary to any secondary is 4000 volts for 1 minutes.
Yeah I know it is designed for switching and thats how I plan to use it, what I am confused about is how to determine the max current/voltage rating for particular windings based off the datasheet.
 
#4
Fusion, are you having trouble reading?
Vin= 85 to 265V, Vout 12V @ 500mA 6W
The breakdown voltage from any primary to any secondary is 4000 volts for 1 minutes.
Answered your post.
 
#6
Vout 12V @ 500mA 6W
That's pretty clear, 6VA output limit. Pulse current can be much higher RMS must equal the VA limit listed. It's power limited not current limited.

What 'single winding' are you referring to? Pins 7 and 8 are listed with no ratings with the exclusion of a DC resistance, if the winding is the same wire thickness as the 6-9 coil then you can determine the turns ratio using the resistance ratio that's listed, if they use the same insulator their voltage can be considered the same.

There is no information to go on for 7-8, it's simply not there. You'd have to empirically measure it.
 
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Thread starter #7
That's pretty clear, 6VA output limit. Pulse current can be much higher RMS must equal the VA limit listed. It's power limited not current limited.

What 'single winding' are you referring to? Pins 7 and 8 are listed with no ratings with the exclusion of a DC resistance, if the winding is the same wire thickness as the 6-9 coil then you can determine the turns ratio using the resistance ratio that's listed, if they use the same insulator their voltage can be considered the same.

There is no information to go on for 7-8, it's simply not there. You'd have to empirically measure it.
The windings I am interested in are 4-5 and 7-8. I guess the information I really need is the wire gauge which doesn't appear to be on the datasheet. I guess I'll have to email the manufacturer to get that information.
 
#8
Well that would probably be faster but you can infer the primary current from the secondary(ies) current but adjusting for the turns ratio and working backwards.

The output is 12V 610ma max (10ma for the aux) The input is 120V, so it's 10 to 1. So the input current is 61ma, adjust for roughly 80% efficiency and you'll get 73ma. You can calculate required wire thickness from that.

The big unknown is the 7-8 winding, nothing about it is specified at all, if it's just a sense coil with little to no load applied then it can be ignored, but if it draws any kind of load it will invalidate the above math.
 
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ronsimpson

Well-Known Member
Most Helpful Member
#9
Turn ration is 9.555:1:1:1 So three of the windings have the same number of turns.

Secondary 1 pins 6-9 is rated for 600mA RMS 0.16 ohms.
Secondary 2 pins 7-8 is not rated 0.21 ohms

Secondary 3 (on the primary side of isolation) pins 1-2 1.00 ohms
Primary 1 pins 4-5 You can find the current by 0.6A/9.555

If you know the resistance of each winding and you know the turns you can find the current rating.
 
#10
Thanks ron, you provided far more detail and reference than I could.
 
#11
This does not explain the problem fully. With a switched mode power supply you work with peak currents, not RMS currents. I have found Wurth particularly bad at specifying this in datasheets to the point where I normally choose another manufacturer whose datasheet is better written. In an SMPS, you need to know the current at which the ferrite saturates and once you know this, with your Vout and Vin you can work out the duty cycle, hence how much power you can put into the primary, hence how much power you can get out of the secondary. RMS values really dont come into the design.

I would phone the local support office of Wurth and ask them the saturation current of the transformer and do your off line converter design based on that
 
#12
I understand what you're talking about. There seems to be no formula for peak current in primary winding. I = U*T/L is working only without core. But core such as ferrite has its hysteresis curve that determines flux density and magnetizing force loop area with its saturation points. I'am currently writing a code in javascript for trying out those magnetism formulas. It is counter intuitive, but this peak current actually depends on the load on the secondary winding. Therefore some simulations yield 1.9 KW and only a few hundred milliampers of peak current. This is without the load I guess.
 

unclejed613

Well-Known Member
#13
it would be interesting to see how your code turns out. welcome to ETO.
just one comment about this thread though..... you dug up an old one here....

but seriously post some code so we can see how it works.
 
#14
When I am very sure it works, I will share. First there are many computer SMPS transformers to analyse. All of their primary windings have 38 to 40 turns no matter if they are rated 200W or 600W. Maybe this info helps someone. Want more power, pick bigger transformer or use many in parallel.
 
#15
The more I dig in magnetism formulae the more vague things get. So I decided to make a few experiments. First I took a transformer and short-circuited its secondary winding. Result was drastic drop in primary windings inductance. There lies the answer. You need to know your output load (resistance) and connect equivalent resistor to transformers secondary winding. Then measure primarys inductance. Knowing your switching frequency and voltage assuming less than 80% duty cycle, finding the period (t) should not be very confusing. Then use this formula: (I = U*t/L) and you'll get your amperage which the transistors or fets must tolerate. It was supposed to be obvious because I recently measured effect of resistance on capacitance. It works vice versa. The more resistance the less capacity, at least from oscillation perspective. This makes contactless heart rate monitor possible.

Second experiment was about giving a high voltage surge into transformers secondary and measuring voltage of capacitor connected via diode to primary winding. Result was astounding. a 211.2 uF got 15.05 volts yielding 23.92 mJ of energy. Discharge capacitors held up 311 volts. Taking the Root-Mean-Square (U/sqrt(2)) we obtain 220 volts as their voltage drops during discharge. From (E = L*I*I/2) I calculated equivalent amperage that must have been the peak value for such energy transfer. 11.69 A. Next the period (t) from (t = I*L/U) I obtained 18.6uS. This is 53.76 kHz at 100% duty cycle. this times energy yields 1286W, but 600W is also fine. Transformers origin was 300 watts SMPS. Actually they operate below 50% duty cycle for safety reasons. It's very difficult to cool down a transformer and room temperature might rise above normal. Second nuance is that I didn't measure the magnetic field spreading out as the core saturated.

My conclusion: if you need a lot of power for short period of time it's perfectly fine to use these transformers at their absolute maximum (for charging Blumlein device for instance).

Use the resistor and ohms law (I=U/R)
 

dr pepper

Well-Known Member
Most Helpful Member
#16
Interesting concept charging a cap through a diode from a comparatively long input pulse.
Of course the caps esr will make a low pass filter, but still you'll get some useful info from such a test.
I look forward to seeing your software too, at the moment I like excellent It, from russia, pretty good software.
 

ChrisP58

Well-Known Member
#17
The more I dig in magnetism formulae the more vague things get. So I decided to make a few experiments. First I took a transformer and short-circuited its secondary winding. Result was drastic drop in primary windings inductance. There lies the answer. You need to know your output load (resistance) and connect equivalent resistor to transformers secondary winding. Then measure primarys inductance. Knowing your switching frequency and voltage assuming less than 80% duty cycle, finding the period (t) should not be very confusing. Then use this formula: (I = U*t/L) and you'll get your amperage which the transistors or fets must tolerate. It was supposed to be obvious because I recently measured effect of resistance on capacitance. It works vice versa. The more resistance the less capacity, at least from oscillation perspective. This makes contactless heart rate monitor possible.
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Measuring the primary inductance with the secondary shorted is a standard test for transformers to determine their leakage inductance. The lower, the better. For an ideal transformer, it would be zero.

In switch mode power supplies, the output of a transformer is intended to be connected, through a rectifier, to an inductor (note 1). It is that inductance, and a capacitor, that forms the integrator that smooths the pulses from the transformer into DC. The control loop of the SMPS looks at the output voltage and adjust the duty cycle to keep the voltage where it's supposed to be. Use of a resistor instead of an inductor to limit the current would be an unnecessary waste of energy.

Most power supplies that are used in desktop computers fall into two groups. Half bridge (these will have a 115/230 Volt switch) or full bridge (these generally have PFC also)

In the half bridge supplies, one end of the primary sits at ~170 Volts DC and the other end gets alternately pulled up to +340 Volts or down to 0 Volts.

In full bridge supplies, both ends of the primary are switched. First the left side to 0 Volts and the right to ~ +390 Volts. Then both are off, Then they change polarity. then both off again. Repeat. (note 2)

Note 1. This assumes that we are talking about forward topology based SMPS using true transformers, and not flyback type SMPS that use coupled inductors.

Note 2. Some systems, such as phase shift switching, use different timings.
 

dr pepper

Well-Known Member
Most Helpful Member
#18
I sometimes have used the measuring leakage inductance trick to make a start calcing the snubber for the primary, also handy for knowing if you made any kind of reasonable job winding the trans before you put it in circuit.
I have a jig somewhere that displays the Bh curve on a 'scope, I used it to sort out a load of cores I had.
 

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