Winding ratio on a transformer for impedance matching

BkraM

Joined Jan 19, 2014
24
Hi all,

I've a question regarding impedance matching using a toroid transformer.

I have a source, 50kHz with an internal resistance of 50 Ohms.
I want to connect this to a 1000 ohm resistor and get maximal power transfer.
(no real goal to this other than some experimenting with electronics)

As far as i understand the winding ratio of the transformer (Ns/Np) should be equal to root of the impedance difference of the loads ( sqrt(Zs/Zp) ).
resulting in approx Ns/Np = 9/2, which I've wound on a ferrite toroid (2 turns primary, 9 turns secondary).

When connecting the source and the load, the resulting voltage at the primary side of the transformer drops to almost 0.
Which does seem to make sense, as I've more or less shorted the source with only two winding.

So my question is, is next to the turns ratio (9/2) the absolute number of turns also important here?
Should I wind the primary side of the transformer with such an amount of turns the the impedance at 50kHz is also 50 Ohms?
This would be 16mH -> about 292 turns primary? (and 1314 secondary)
Or am i missing something else in my approach?

Thanks,

GopherT

Joined Nov 23, 2012
8,012
Hi all,

I've a question regarding impedance matching using a toroid transformer.

I have a source, 50kHz with an internal resistance of 50 Ohms.
I want to connect this to a 1000 ohm resistor and get maximal power transfer.
(no real goal to this other than some experimenting with electronics)

As far as i understand the winding ratio of the transformer (Ns/Np) should be equal to root of the impedance difference of the loads ( sqrt(Zs/Zp) ).
resulting in approx Ns/Np = 9/2, which I've wound on a ferrite toroid (2 turns primary, 9 turns secondary).

When connecting the source and the load, the resulting voltage at the primary side of the transformer drops to almost 0.
Which does seem to make sense, as I've more or less shorted the source with only two winding.

So my question is, is next to the turns ratio (9/2) the absolute number of turns also important here?
Should I wind the primary side of the transformer with such an amount of turns the the impedance at 50kHz is also 50 Ohms?
This would be 16mH -> about 292 turns primary? (and 1314 secondary)
Or am i missing something else in my approach?

Thanks,

YES, the number of turns determines he strength of the magnetic field you are creating. The current, number of turns, length of the wrapped coil, diameter of the wire, are all in the formula that determine the magnetic flux produced (in the ferrite core).

Also, the ferrite is only sustain a certain magnetic field where it is said to be saturated. Look up that specification from your core supplier’s datasheet.

DickCappels

Joined Aug 21, 2008
8,150
Another way to look at it is that the number of turns affects the inductance of the winding. The current resulting from the voltage across the primary inductance is often referred to as magnetizing current. E/XL

In general you want the magnetizing to be small compared to the load current. Since the magnetizing current is through the primary inductance and the load current is through a resistor the two currents will be 90° out from one-another, so the total current is the square root of the sum of the squares of the two currents, so its not as serious as it might first seem.

BkraM

Joined Jan 19, 2014
24
Thanks for the replies.

Do I understand correctly that without saturation limitations of a core (theorectical), 2 and 9 turns would have be sufficient to transform the impedance from 50 to 1000 ohms at any frequency (>0)?

Thanks,

crutschow

Joined Mar 14, 2008
28,489
Do I understand correctly that without saturation limitations of a core (theorectical), 2 and 9 turns would have be sufficient to transform the impedance from 50 to 1000 ohms at any frequency (>0)?
No.
Core saturation is only one consideration.
You also want enough turns so that the magnetizing inductance is sufficient, at the lowest frequency, to provide a primary impedance much larger than 50 ohms (>10 times) with no secondary load.
Since inductive impedance is proportional to frequency, the lower the frequency, the more turns are required for a given impedance.

Last edited:
• GopherT