What is the relation between transformer turns and frequency of input?
I hear there is a transformer coupling in stratix 2 FPGA. I went through the data sheet and found the lower cut-off input is 400 Khz. Why is there a lower cut off for a transformer?
Also what would happen if I replace 1000:100 transformer with a 10:1 transformer? Answer this not from EMF or flux point of view, but from input frequency point of view or input current point of view.
Put simply lets consider an inductor. Transformers are just coupled inductors.
Some of the key relationships are:
$$V = L \cdot \dfrac{d}{dt} I = N \cdot \dfrac{d}{dt} \Phi = N \cdot A \cdot \dfrac{d}{dt} B$$
Where:
\$ A \$ is Area (square meter)
\$ B \$ is Flux density (Tesla)
\$ I \$ is Current (amp)
\$ L \$ is Inductance (henry)
\$ N \$ is Turns
\$ V \$ is Voltage (volt)
\$ \Phi \$ Flux (weber)
\$ \dfrac{d}{dt} \$ is rate of change with respect to time, differential.
If we take transformer then and reduce the frequency we are in effect increasing the \$ dt \$ so all other things being equal you would expect to see an increase in both the flux density and the current. The best you can hope to see from this would be for the transformer to run hotter because it will have higher core and copper losses.
All other things are not equal however and high frequency magnetics are designed to run at certain flux densities: Too high and they risk saturation, too low and the part is over-designed and costs more than it needs to.
As you reduce the frequency to maintain the optimum flux density swing you will either need to increase the the effective area of the core or add more turns. In practice this usually means an increase in both and requires a larger, more expensive core.
In short : the relationship between turns and input frequency is inductance.
The lower cutoff frequency is determined by the point where the transformer's primary inductance has the same impedance as the impedance driving it. (Specifically, this is the -3dB point in the frequency response).
So if the transformer is fed from a 50 ohm source, and its LF cutoff (-3dB point) is 400 kHz, Xl = 50 ohms = jwL = 2*Pi*400,000*L so L = 50/2/Pi/400000 = 19.9 microhenries.
Assuming by a 1000:100 transformer you mean one with a 1000 turn primary, if you replace that with a 10 turn primary (and it is wound on the original core) you have reduced the primary inductance by 10,000, and increased the low frequency point by 10,000.
