1
\$\begingroup\$

I want to create an isolation, impedance matching, balancing transformer for a DI box. I want it to work on Audio frequency and use a toroidal ferrite core.

The input is an electric guitar, so it needs 1M Ohm input impedance. The output is a line-in so it needs about 2k Ohm impedance.

I attempted to calculate the required turns using this, but I am baffled as it seems that only the ratio matters.

If it indeed is so, how many turns should I use? What determines the frequency range of the transformer?

If I am wrong(probably I am), then please explain how exactly.

\$\endgroup\$
2
\$\begingroup\$

The turns ratio will give you the impedance ratio and the voltage ratio. The choice of turns depends on the core type and size and the expected voltage that you need to drive through the transformer so that it does not saturate.

Typically one tried to have an equal size coil for primary and secondary for an isolation transformer. There will be some clever calculation to work out the optimal power transfer for a given coil size as more turns will eventually increase the DC resistance too high.

Your turns ratio will be quite high so the primary will be very fine wire. Winding so much on a toroid may be uneconomical in production volumes, pot cores and E-I stacks are still common.

Here is a link to two reference manuals Wolpert's and Whitlocks that can offer you some further guidelines. You will need to find out the parameters of the core you plan to use and see if you can cope with the required turns in the toroid.

|improve this answer|||||
\$\endgroup\$
  • 1
    \$\begingroup\$ Great links, seems I am totally getting this wrong. No way I could wind something like that at home... I'll probably resort to an Active solution. \$\endgroup\$ – akaltar Jul 7 '15 at 23:37
  • 1
    \$\begingroup\$ Based on the wolpert manual I calculated that I would need about 8000H primary inductance. And checked on wikipedia that this is 1/8 inductance of a 3000MW transformer's main winding...Great! \$\endgroup\$ – akaltar Jul 8 '15 at 0:34
4
\$\begingroup\$

1 M ohm input impedance is massive for a transformer. Consider just the primary on its own - forget about the load on the secondary. The primary is just an inductor and it must have greater than 1M ohm impedance at (say) 80 Hz.

XL = 1,000,000 = 2\$\pi f L\$ therefore L = nearly 2,000 henries.

If the ferrite has an \$A_L\$ value of 10uH per turn squared you'll need over 14,000 turns.

BTW 80 Hz corresponds to "low E" on the fat string on a guitar.

|improve this answer|||||
\$\endgroup\$
  • 1
    \$\begingroup\$ I came to the same conclusion, from KalleMP's answer. I will probably resort to an active solution. \$\endgroup\$ – akaltar Jul 8 '15 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.