T&L Publications

Hi.
For an audio transformer specified as 1300:8 ohm

----> http://www.ebay.com/itm/5Pcs-Audio-Tran ... fresh=true

What turns ratio should be expected ? 1300 to 8 ? --> 162:1 ? Or am I pissing out of the can ?

Am after at least 100 V when fed by 2V - 1KHz at the low impedance winding.

that's right but it sounds like you want something with a 1:50 ratio instead. How many turns on the 1 side and what gauge wire depends on the power you want to get out of it. you should be able to find online calculators. 1Khz sine wave is probably not too hard to work with, its virtually a test tone

Thank you haklesup.

The transformers markets become confusing as the ones for 50/60 Hz are expressed in volts; the ones for 1KHz are expressed in their impedance, some are listed by the inductance of windings, and using a power transformer for audio use (as I have done successfully a few times) gets awkward. Same to use an audio transformer to design a specific power supply.
As the link above, trying to achieve an over 100V tone, their data is in their winding impedances. No turn ratios, no Volts. I can de-calculate and get Henries, but still gives me little to chew.

Today I called a manufacturer for a specific transformer made by them and told they make them to customer specifications only and cannot supply such as is proprietary to a customer.

For transformers spec'ed in impedence -
turns ratio equals square root of (primary impedence divided by secondary impedence)

so 1300 / 8 = 162.5
162.5^ .5 = 12.75
2V * 12.75 = 25.5V
You need turns ratio 4 times higher, or impedence ratio 16 times higher.

21K Ohms : 8 Ohms ought to do (if current load is light)
((with 8 Ohm side as input of coarse))

Cheers,

Excellent ! Thanks, Dale.

So the turns ratio for that particular 1300 Ohms:8 Ohms transformer is actually 12.75:1. ; would need to feed 8V to obtain 100V at primary. OK. time to look for the proper one. That formula cleared the fog !

I was told there would be no math...

No math in electronics ? ? ? ?
That is called a big pile of blown fuses, fried resistors, and transistors with the magic white smoke let out; but not electronics

Hi,

Yes the turns ratio is the square root of the impedance ratio. This is because for any turns ratio both the current and the voltage are affected not just one or the other like we usually think of in order to get the proper output voltage for example.

Another way to look at it is to reflect all the impedance to one side, either primary or secondary. IF both impedances are given then you can use either one. For example, since the input is spec'd at 1300, then we can view the transformer as a perfect transformer with no resistances, but with a 1300 ohm resistor connected in series with one of the input terminals. If we were to then short circuit the new input (so as to place a 1300 ohm resistor across the primary of the perfect transformer) and measure the output impedance, we would measure 8 ohms. Similarly, if we were to place an 8 ohm resistor across the secondary and measure the input impedance, we'd measure 1300 ohms.
To calculate the input impedance knowing that 8 ohms, we multiply by the square of the turns ratio. Since the turns ratio in this case is about 12.748, if we square that we get about 162.5, and multiplying that by 8 we get 1300. Of course to get that 8 ohms we would just divide 1300 by the square of the turns ratio instead.