1
\$\begingroup\$

Preamble

I need an small isolation transformer for 0-6V 400Hz AC (this is my external signal, amplitude is the value):

enter image description here

I should get <10mA load from the primary signal source, and in the same time I need minimal size of the transformer. For <10mA at 6V 400 Hz the primary inductance should be >238 mH (coil impedance is \$2\pi f L\$).

Typical transformers that I found with such an inductance is pretty big, but I found also Bourns LM-LP-1001L that have very confusing values in datasheet. Help me please understand it.

Amble

The datasheet says that DC resistance of primary coil is 90Ω, and impedance at 1kHz is 600Ω, so I calculated the coil's inductance 81 mH. But datasheet says inductance is 2.8H! What is wrong here? And if 2.8H is the correct value, how do so small transformer can have so big inductance?

\$\endgroup\$
  • \$\begingroup\$ "impedance" of 600 ohms is not the impedance of the coil but the matching impedance, i.e. the source and load impedances. Primary inductance must be much higher, the LF rolloff reaches 3dB at the frequency where the primary inductance equals the source impedance. As for how it can have such a high inductance : the DC resistance tells you there are many turns of rather thin wire... \$\endgroup\$ – Brian Drummond Jun 12 '16 at 16:19
1
\$\begingroup\$

Primary inductance is 2.8 H as specified - this will have been measured/proven with the secondary open circuit because if the secondary is not open circuit then it cannot be easily measured.

At 400 Hz the impedance will be \$2\pi\times 400\times 2.8\$ = 7037 ohms reactive and draw 0.85 mA from a 6V RMS applied signal.

The 600 ohm impedance is when the secondary also has a 600 ohm load.

The 150+150 you refer to is when the primary and secondary both have centre-taps. If both primary and secondary only use one half of their respective windings, the nominal impedance becomes 150 ohms - this is because impedance is proportional to turns squared on a transformer and half the turns quarters the impedance. Given the transformer is designed for telephone applications (hence BS6306 being mentioned), the full winding is designed to be optimum for a 600 ohm load and therefore 150 ohms when half windings are used.

2.8 H does seem large but if the AL of the core is 10 uH per turn then 1000 turns would produce 10 H because inductance is proportional to turns squared. Clearly, to get 2.8 H only 529 turns are needed hence 10 uH x 529\$^2\$ = 2.8 H.

\$\endgroup\$
  • \$\begingroup\$ Thanks a lot! And small question also. For separate coils (e.g. LM-NP-1005L) datasheet says "2.8H (0.7+0.7)", so if I understand correctly, 2.8H is when coils are connected, and 0.7H is single coil, because of the square law. But on other hand, for serial connections of inductors we should just add inductances and 0.7+0.7=1.4. Confusing. \$\endgroup\$ – user54579 Jun 12 '16 at 17:49
  • 1
    \$\begingroup\$ If inductors share the same core (all flux shared) then it isn't a simple "add" anymore. \$\endgroup\$ – Andy aka Jun 12 '16 at 17:52

Your Answer

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