# Transformer operating frequency clarification

I'm trying to design a small audio transformer. Specs are as follows:

Input impedance: 600 ohms
Output impedance: 60k ohms
Turns ratio: 1:10
Input voltage: 2V
Output voltage: 20V
Input current: 5mA
Output current: .5mA
Max power: 10mW (purely resistive load)
Frequency range: 40Hz-20kHz


Following this example: http://engineerexperiences.com/design-calculations.html#1_Core_Calculations

Since 40Hz-20kHz is much more bandwidth than say 45Hz-55Hz from the example, I figured to be safe try 20kHz as the operating frequency first.

Determining the turns per volt:

Assuming a 2" bobbin, that gives a core area of: $$\4 \ \mathrm{in}^2/1550 = 0.00258 \ \mathrm{m}^2\$$ (as the magnetic flux density uses units of $$\ \mathrm{m}^2 \$$, we must convert bobbin area to $$\ \mathrm{m}^2 \$$)

Thus, we have: $$\ Te = \frac{1}{4.44 * 20000 * 1.2 * 0.00258} = 0.0036\$$

So if my input voltage is 2 volts, when winding the transformer I need $$\ 2 * 0.0036 = 0.0072 \ \mathrm{turns} \$$

This result does not pass the sanity check, and coupled with the knowledge that transformers "need" changing current in order to work, I'm going to assume that the operating frequency is not the 3 db point of a low pass filter, but rather, the 3 db point of a high pass filter, and I can safely assume my audio transformer will pass all frequencies (up to some physical limit which is well past human hearing anyway) above the operating frequency?

Or, as another possibility, is the operating frequency some sort of middle point chosen in the bandwidth? (And then bandwidth is determined by some other sort of physical parameter?)

Basically, what exactly should I chose as the operating frequency for my transformer in that equation, given the specs?

• Lowest frequency & highest voltage first. – Spehro Pefhany Feb 11 at 22:33