Transformer Design

Its difficult to answe r your question because the active, positon or flyback is different, and you willl get differnet turns and gap.

Use the formula n1/n2 = V1/V2

n1 = primary turns

n2 = secondary turns

V1 = primary Voltage

V2 = secondary Voltage

Given the Voltages you specified, the turns ratio is 1:40

For current calculations, the inverse is true. That is: n1/n2 = I2/I1

For 60 Hz transformers, a rule of thumb is to use 3 to 4 turns per volt. Fewer turns per Volt are needed at highr frequencies.

The current in each winding dictates the size (cross-sectional area) of the wire. Temperature (heat dissipation) also plays a part. For 150 milliamps, you can use AWG #37 enameled copper wire for the secondary.

The wire size for the primary would be determined by calculating the primary current which would be the turns ratio (40) times I2, e.g. 40* 0.150A = 6 Amps.

AWG #22 will safely carry 7 Amps.

Thanks for the lesson. Is very nice see people who has the knowledge.

Simple and fast.

If 2000V is DC voltage you can use voltage multiplyer (by factor 4 for example).

An equivalent scheme of HF transformer is much more complicate.

You can relate to this sir in addition only to MR. JOe:

Np/Ns=Vp/Vs=Is/Ip

Are you sure it's 500KHz?

You are now in RF land. Very different type of transformer.

Wangito.

v

I think design a transformer is a complex matter. especially at high frequency condition. If we only count its turns, they have different calculate ways.I hope my answer will not misguild students learnning at the subject. just take account of the way of transformer working, they have several conditions such as positive, negative active or push-pull, bridge etc. if we calculate it working in way of sigle transistor(or others like mosfet,igbt etc) we could get this formula;
n(min) = V(min) x T(ontime)/[deltaB x Se].
where n= the minimum turns we shall get;
V== mimimum voltage applied on the primary wind;
Ton= The maximum lead on time of the active device( transistor,mosfet or igbtetc.)(us)
delta B= core flux( generally speaking we take 0.2T for core);
Se== effective section area(mm);
From this, we get priliminary winding, if the trasformer is workingat negetive active, we have to take account of gap for storage energy.
If we work at condition of pull-push. the formala could be;
v=4xBxSxfxexp(-4);where s(cm)
and if its at sine condition, we can use coefficent of
4.44 instead of 4.
This is only a spot of explain.
That s why we ask the top post for complement complete condition.
hope thsi will be help.