It must be because you used a 220V and I used a 120V supply. Although looking at your turns ratio, I would have to guess it is a full range supply, 80-260V or something like that.Originally posted by marsjell:
<STRONG>Hi presslab,
again sorry for bothering you but...
I've just unwound my transformer, and the ratio's are way of compared to your transformer I have 2 x 21 windings on the prim. site, and 2x 3turns, 1x 4 and 1x 5 turns on the sec. side.
I think this means that I can't take the windings on the sec side the way you did can I?
I'm using an old 220V PSU BTW.
please help?
BTW2 : this won't affect the circuit which drives the transformer (prim side) will it?
[ 10-18-2001: Message edited by: marsjell ]</STRONG>
Warning, class is now in session!![]()
The way the windings work is like this. First, let's look at the real secondary voltage. I'll use 3.3V as an example. There is a schottky rectifier diode before the final output, so the secondary voltage will be 3.3V + 0.4V (diode drop) = 3.7V. So let's look at our turns ratio now. 3:4 is the primary vs 3.3V secondary ratio (on my xformer).
Max duty cycle is 50%, or 0.5
Vin = Vout * pri / sec / duty
Vin = 3.7 * 3 / 4 / 0.5
Vin = 5.55V
This is our minimum input voltage for the transformer indicated in my drawing.
So, assmuing the 3 turn winding is the 3.3V winding, a look at your transformer reveals:
Vin = Vout * pri / sec / duty
Vin = 3.7V * 21 / 3 / 0.5
Vin = 51.8V
Now convert DC to AC(rms).
51.8VDC = (51.8V + 1.4V) * 0.707 = 37.6VAC
That's pretty low voltage. My minimum on the old transformer was around 70VAC for a full range supply.
So let's figure out your duty cycle at 220V.
220VAC = 220 / 0.707 + 1.4
220VAC = 312.6VDC
Vin = Vout * pri / sec / duty
312.6 = 3.7 * 21 / 3 / duty
duty = 3.7 * 21 / 3 / 312.6
duty = 8.3%
That's kind of a low duty cycle, but it should work at 220V at 8.3%
So you read all this, and are asking, "NOW WHAT??"Well okay, give this a shot.
Wind 2 turns on the primary.This will give you a turns ratio of 2:3. Let's figure out minimum input voltage.
Vin = Vout * pri / sec / duty
Vin = 3.7 * 2 / 3 / 0.5
Vin = 4.93V
The inductance of the primary is low, which can be bad (core saturation). This leads to high primary current, and inefficient operation (running hot).
The next proposed winding will take a lot more bobbin space, so it probably won't fit on your bobbin. It will have considerably less currents in the primary though. The calculation for minimum Vin again with the new primary (5 turns) and secondary (6 turns). This means you multiply the number of turns on EACH secondary winding by 2. The 3 becomes 6, the 4 becomes 8, etc etc.
Vin = Vout * pri / sec / duty
Vin = 3.7V * 5 / 6 / 0.5
Vin = 6.17V
Not too bad, eh? Try the 2 turn primary first. If it gets hot and works funny, then see if you can make the 5 turn primary one work.
Presslab



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Well okay, give this a shot.
This will give you a turns ratio of 2:3. Let's figure out minimum input voltage.

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