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I’m trying to construct a small transformer to step up the audio of a crystal radio to line level. I can actually hear the radio faintly using a conventional earbud (not piezoelectric), but I’d like to increase the volume with a transformer.

This is more of a learning exercise than anything else. If it’s not clear: I have almost no idea what I’m doing.

The radio output is on the order of 300 mV and based on Wikipedia line level is a little below 1 V, so I should need about a 1:5 ratio. For the core, I’m attempting to use a small piece of ferrite (about 1.5” long and 1.25” wide). I’m using 28 awg magnet wire for both sides.

My first attempt was to wind 100 turns up and down the core as the primary, then in the same way 500 turns of a second bit of wire on top of that for the secondary. It doesn’t perform very well. I can hear the signal, but it’s actually weaker than it was without the transformer.

I based the design off of this:

enter image description here

Based on what I’ve described, what am I doing wrong? Why isn’t this working?

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  • \$\begingroup\$ a crystal radio is a high-impedance source, but I can't think of a good way to explain why this is significant. \$\endgroup\$ Commented Aug 2, 2022 at 22:26

3 Answers 3

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You need to match the impedance of the crystal radio, to the impedance of the phones, for maximum audio signal transfer.

Depending on its design, the impedance of the crystal radio could range over hundreds of ohms.

Likewise, the impedance of the phones could range from tens of ohms to kilohms.

For a proper match a 'crystal radio audio-matching transformer', with multiple impedance taps, would come in handy. Such transformers are available off-the-shelf.

enter image description here

For example, a crystal radio with an impedance of 300 ohms would have to be connected to the '0' and '300' terminals and 8 ohm phones to the '0' and '8' terminals.

In the absence of such a transformer, vintage transistor audio transformers may be tried out.

Another alternative would be a 240 V / 24 V power transformer that would give a 100 : 1 impedance ratio. Impedance ratio is the square of the turns ratio.

A transformer wound on a ferrite rod would not work, on account of flux leakage.

It would need to be wound, using ferrite 'pot' cores / 'E' cores, intended for audio frequency applications.

enter image description here

enter image description here

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There's two things that are killing you here:

First, an earbud is not an efficient transducer. For a crystal radio you need a transducer (i.e. piezo earphone or old-style high-impedance headphones) that is made to be very efficient at converting electrical energy to sound energy. This is usually done by sacrificing fidelity.

Modern earbuds go exactly the other way -- they throw efficiency under the bus, all in the name of sounding good.

So you need to amplify :(.

If that wasn't enough, you need to step down in impedance, not up. Crystal radios need a high-impedance load (like 10 k-ohm or higher). Your earbud is probably around 100 ohms.

To add one more element to your misery, it looks like you're using a ferrite for an antenna. This will give you far too small of an inductance, because a ferrite won't develop enough inductance without way too much wire. You would need an audio transformer that's designed for the impedances involved, and that involves a full "E" core with laminated iron, and lots of windings of very fine wire.

You might be able to get by with a new-old-stock transistor radio output transformer or interstage coupling transformer, but in step down mode, not step up. Back in the vacuum-tube era this was a catalog item you could mail-order, or in a decent sized city, there was a store a car ride away. Today it's a lot of digging on antique electronics sites. With such a transformer to give you a proper impedance match you may be able to hear the radio very quietly -- or, maybe not.

Bottom line -- you need a transducer (the earphone) that's crazy efficient, and you need to match the impedance of the transducer to an impedance that the radio can feed -- which means way higher than a current-production consumer earbud.

Or, you just need a high impedance input amplifier, listen with the earbud, and be happy that the radio part works as expected.

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  • \$\begingroup\$ Thanks. I have already been using a piezoelectric earpiece and it works nicely. I wanted to try this because of some videos that I’ve seen demonstrating that a transformer and crystal radio can drive a modern earbud. I’m having a hard time grasping how the impedance of the radio affects things. I understand that higher impedance is better for the earpiece because of voltage divider, but if you can step up the input voltage enough it shouldn’t matter whether the earpiece has high impedance, right? \$\endgroup\$
    – shwoseph
    Commented Aug 3, 2022 at 1:51
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    \$\begingroup\$ Stepping up the input means that the impedance presented to the radio is lower. This makes things even further from an impedance match, which reduces the radio's ability to transfer power to the earphones -- and since the only energy you have is what the radio harvests from the aether, you can't throw away available power. electronics.stackexchange.com/questions/13631/…. \$\endgroup\$
    – TimWescott
    Commented Aug 3, 2022 at 2:04
  • \$\begingroup\$ This video really helped me to begin to understand impedance matching: m.youtube.com/watch?v=DovunOxlY1k \$\endgroup\$
    – shwoseph
    Commented Aug 4, 2022 at 19:05
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A ferrite bar has a great deal of leakage inductance, and makes a poor transformer at audio frequencies. You need much higher inductance to get sufficient impedance to match the output of your crystal radio. You would need many more turns to avoid loading down the radio output, and you would need a magnetic core with much higher inductance per square root of turns. From another answer:

Inductance formula

You would do much better with a simple transistor amplifier, but you could try using a small power transformer or modem transformer.

More information:

Why is inductance (L) proportional with turns-square (N²)?

https://www.learnabout-electronics.org/ac_theory/transformers02.php

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    \$\begingroup\$ A ferrite doesn't necessarily create significant leakage inductance but it does leak flux quite a lot. \$\endgroup\$
    – Andy aka
    Commented Aug 3, 2022 at 7:28
  • \$\begingroup\$ You may be right - magnetics is not my strong point. But a bar, whether ferrite or iron, does not make for a good transformer, except maybe for RF. \$\endgroup\$
    – PStechPaul
    Commented Aug 3, 2022 at 18:47
  • \$\begingroup\$ It makes a poor transformer at audio because the magnetization inductance is pitifully small when you have such a large air-gap. \$\endgroup\$
    – Andy aka
    Commented Aug 3, 2022 at 20:02

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