Output current problem

Godyreal

New member
Another thing that had been bothering me is whether the primary wire diameter determine the secondary output current? In a smps transformer
 

Silvio

Active member
The assumption is not right. You have to consider power input versus power output and adding around 20% more on the input to cover losses.
If for example I have an off line smps which works with an input voltage of 300vdc and the output is say 30v. If say the current in the output is 10A then the power absorbed is 300w. Turning back to the input power then at 300v the power in the input is 300w + 20% = 360w so the current in the primary is not 1A but 1.2A.
The wire in the primary then must be able to handle this current. On the other hand it is the same rule for the secondary winding. The wire must be thick enough to carry 10A.
I hope it is more clear for you now.
 

Godyreal

New member
thanks so much Sir though to be frank i still cannot relate it to how i use a 0.5mm diameter of wire for primary and 1.4mm for the secondary and i got 150volts at secondary output though i regulate it to 60volts but when i used it on my 200watts amplifier and tested the current the meter only read 1ampper and if i increase volume the smps dim the standby led and the sound become low and distorted, i uses uc3844 with 9n60 mosfet and the transformer is EE42 30turns for the primary and 4-0-4 turns for the secondary. The question is why am i having a small current when i used 1.4mm wire diameter which supposed ti have atleast 4amppers?
 

Silvio

Active member
thanks so much Sir though to be frank i still cannot relate it to how i use a 0.5mm diameter of wire for primary and 1.4mm for the secondary and i got 150volts at secondary output though i regulate it to 60volts but when i used it on my 200watts amplifier and tested the current the meter only read 1ampper and if i increase volume the smps dim the standby led and the sound become low and distorted, i uses uc3844 with 9n60 mosfet and the transformer is EE42 30turns for the primary and 4-0-4 turns for the secondary. The question is why am i having a small current when i used 1.4mm wire diameter which supposed ti have atleast 4amppers?
You relate power in the input and power in the output according to the voltage in each case.
Volts per turn Pri = 320v / 30 = 10.7v per turn.
Sec voltage then = 10.7 x 4 turns = 42.8v.
If you need a regulated voltage of 40v then you can clearly see that 2.8v is not enough. Your headroom voltage to allow the chip to regulate at load must be x 1.4 to 1.5 more.
So 40v x1.5 = 60v. You can see here that each secondary winding must supply 60v when the chip is at full pulse width. Remember that voltage tend to drop with load so when this happens the chip will widen the pulse width to allow more voltage at the secondary thus keeping the output voltage stabilised via the feedback loop.
Using the same reasoning then turns needed at the secondary are 60v / 10.7 = 5.5 turn. Due to that we cannont wind half a turn then we either wind 5 or 6 turns for each secondary winding.
Ampere turns.
Sec current is 10A so 80v x 10A = 800w. You can see here that an EE42 will never give you this power in flyback topology but it will in half or full bridge. 300 or 350w is a more reasonable power output for flyback.
350w / 80v = 4.4A
Using a current density of 8 for audio purpose then using 0.5mm copper wire.
This is found using the cross sectional area.
So 0.25 x 0.25 x 3.142= 0.196mm² x 8A per mm² = 1.57A per wire.
We now find how many wires needed for 4.4A.
So 4.4 A/ 1.57 = 2.8 wires (3 wires of 0.5mm for secondary).
For primary using the same rule then 350w / 320v = 1.09A. So 1 wire x 0.5mm is adequate for primary winding.
 
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Godyreal

New member
thank you very much Sir for the wonderful detailed explanation i think i get the idea now how flyback smps works, once again thanks so much
 

Godyreal

New member
sorry to bother u so much am trying to built your 700watts smps, as per your schematic there is a two point that you specify as TO 2 TURNS LOOP TO TRAFO from the triac BTA12 CAN U PLEASE EXPLAIN WHERE THAT WILL BE CONNECTED TO? thanks in advance
 

Silvio

Active member
sorry to bother u so much am trying to built your 700watts smps, as per your schematic there is a two point that you specify as TO 2 TURNS LOOP TO TRAFO from the triac BTA12 CAN U PLEASE EXPLAIN WHERE THAT WILL BE CONNECTED TO? thanks in advance
Take a look at the picture i marked it with arrow.
 

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Godyreal

New member
sorry i still cannot get it right, it seems from the picture after i have finished winding the Primary, secondary, auxiliary 1 and 2 then will now wind 2 turns the connect the 2 turn loop to the bta12 then the 80turns will now be connected the two zener diode that form the full wave rectification . please can u use words to explain to me. am so sorry to bother u this much
 

Silvio

Active member
The output from the current transformer which is center tapped is rectified by 2 small diodes 1n4148. The + side of the diodes are connected together. This forms the positive and the center tap becomes the negative. There is the burden resistor (R15 - 33R) which is tied across the pos and neg of the current transformer.
To wind the current trafo.
Take two lengths of enamelled copper wire (0.2mm) and wind 80 turns around a small ferrite ring core 15mm dia.
Solder the end as shown on the picture to 3 pieces of thin flexable wire 5cm long. It is best to use a different colour for CT not to mix it up with the others. Finally put a small piece of shrink tube on the soldered joints. Cut thin strips of about 4mm from electrical tape and tape around the core passing through the centre hole to tie the wires and winding.
I hope this helps20220822_194238.jpg
 

Godyreal

New member
am very sorry please then where is the primary of the current wind at? or the current sensor transformer does not have primary? how will voltage be induce in the secondary of the current transformer? its not very clear to me please
 

Silvio

Active member
am very sorry please then where is the primary of the current wind at? or the current sensor transformer does not have primary? how will voltage be induce in the secondary of the current transformer? its not very clear to me please
You pass the bridge from the mid point of 1uf capacitors (marked red) through the centre of the current traffo shown with blue arrow.
 

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