Easy way to calculate losses for re-purposing a transformer?

[OldWorldBlues]

Hi everyone! This is my first thread :)
I've been working a bit with AC and radio, and would like to make a simple(-ish) circuit in which a 1-volt peak audio signal is stepped up to around 5 volts with a transformer, and is fed into a crystal oscillator to make a crude-but-effective AM transmitter. For the transformer, I "liberated" the windings from an old 120v -> 24v transformer. I have heard that transformers suffer losses when operated in frequencies over their rated frequency. How would I calculate those losses? Thanks for any help you guys can give.

 

[berkeman]

Welcome to the PF.

Fun project. A few thoughts:

• An AC Mains transformer isn't a great way to step up audio, although it might sort of work. Do you have a signal generator and and oscilloscope (or audio-frequency DVM) that you can use to test the transfer characteristics?
• Please be careful making a DIY "AM Transmitter", since these can easily interfere with normal radio transmissions (and some of those can be important, like Fire and Police radio traffic). A better choice for a DIY radio project is to build a receiver...
• You can look at using Opamps for audio waveform amplification, since those are more commonly used compared to audio transformers

 

[jrmichler]

If you transmit at low enough power (in the U.S.), you don't need a license: https://www.fcc.gov/media/radio/low-power-radio-general-information.

Lots of good information on eddy current losses if you Google transformer eddy current loss.

Good luck. When I was in high school I tried to make a transformer to make a bug zapper. I hand wound 10,000 turns on the secondary, then shorted the primary assembling the E's and I's. Our high school physics book had the equations, plus the physics teacher was more than happy to consult on the project.

 

[Baluncore]

How will you apply 5V audio to a crystal oscillator? A crystal oscillator is designed to have a high Q, typically 100 thousand or one million, so it is very stable. It will typically take an accurate crystal oscillator 0.1 seconds to start or stop oscillating. That will limit the bandwidth of the oscillator to about 1 / 0.1sec = 10 Hz.

To make an AM audio transmitter you will need an AM modulator and a stable RF carrier oscillator. The modulator will multiply the carrier amplitude by the wider band audio input.

The first problem re-purposing an AC power transformers will be greater gain at the low power frequencies because the laminations are too thick and the middle of the laminations are inaccessible to audio. The skin effect works with magnetic fields as well as electric currents. Audio transformers have very thin shim laminations or composite iron powder cores so as to reduce the core volume and the length of wire needed on windings.

 

[jim hardy]

Find in a junkshop or your attic an old transistor radio. They used an audio output transformer to drive the speaker.
With today's audio amp IC's you should be able to get something going. Thirty years ago I drove an audio interstage transformer with an LM386 , worked great.

 

[OldWorldBlues]

How will you apply 5V audio to a crystal oscillator? A crystal oscillator is designed to have a high Q, typically 100 thousand or one million, so it is very stable. It will typically take an accurate crystal oscillator 0.1 seconds to start or stop oscillating. That will limit the bandwidth of the oscillator to about 1 / 0.1sec = 10 Hz.

To make an AM audio transmitter you will need an AM modulator and a stable RF carrier oscillator. The modulator will multiply the carrier amplitude by the wider band audio input.

Ja, I've looked into LC oscillators, but they're a bit confusing when it comes to actually making them. I'm going off of some projects I saw on the web of people using 1-20 MHz clock crystals as modulators.

 

[berkeman]

If you transmit at low enough power (in the U.S.), you don't need a license:

True, but most newbies don't know how to measure or limit that Tx power...

(I sure didn't on my first transmitter that I built...)

 

[Baluncore]

Ja, I've looked into LC oscillators, but they're a bit confusing when it comes to actually making them. I'm going off of some projects I saw on the web of people using 1-20 MHz clock crystals as modulators.

It might help if you provided a link so we could understand exactly what you are doing and how you were using the transformer.

 

[sophiecentaur]

True, but most newbies don't know how to measure or limit that Tx power...

(I sure didn't on my first transmitter that I built...)

They are stuck with using commercial, type-approved equipment. (And a good thing too!)

 

[OldWorldBlues]

It might help if you provided a link so we could understand exactly what you are doing and how you were using the transformer.

Here's a page with more-or-less what I'm trying to do.

 

[berkeman]

Ah, so they are modulating the power supply to a canned oscillator circuit. Many issues exist, but that could work. Please don't couple this to an antenna directly, as you will likely spew harmonic energy across your neighborhood. And 9V seems strange for the supply voltage, but I didn't look at the component they are using for the oscillator...

 

[OldWorldBlues]

Ah, so they are modulating the power supply to a canned oscillator circuit. Many issues exist, but that could work. Please don't couple this to an antenna directly, as you will likely spew harmonic energy across your neighborhood. And 9V seems strange for the supply voltage, but I didn't look at the component they are using for the oscillator...

I agree on the antenna - 'not sure how to calculate wavelength, but I heard that the more you halve your antenna, the weaker your signal.

 

[berkeman]

I agree on the antenna - 'not sure how to calculate wavelength, but I heard that the more you halve your antenna, the weaker your signal.

Thanks. For the low-power FM radio transmitter kits that I've built, they usually use a very small capacitor (like around 10pF) to couple a small transmit signal to the antenna. That can help to keep your filtered* Tx signal down to a legal small-range-Tx level.

* The schematic you posted seems to show a square-wave canned crystal oscillator as the signal source that is AM modulated by modulating the power supply to it. That modulated square wave has many harmonics, in addition to the 1MHz fundamental. Transmitting a non-filtered waveform with so many harmonics is a big no-no in the radio community...

 

[Tom.G]

Rant
Are we supporting technical learning here or are we arguing first that something won't work then when shown it works that it shouldn't be done. I agree regulatory aspects often need to be pointed out, but the last I heard we are not the enforcement arm of a regulatory agency.
Let's help him get the project accomplished in an appropriate way... and perhaps include some of the "why" along the way.
end rant

 

[berkeman]

Rant
Are we supporting technical learning here or are we arguing first that something won't work then when shown it works that it shouldn't be done. I agree regulatory aspects often need to be pointed out, but the last I heard we are not the enforcement arm of a regulatory agency.
Let's help him get the project accomplished in an appropriate way... and perhaps include some of the "why" along the way.
end rant

That's a fair rant. But I've also spent close to a hundred hours with a portable spectrum analyzer and direction finding equipment trying to figure out what transmitter is stepping on emergency radio bands (police, fire, HAM), and sometimes the sources have been uninformed folks who didn't realize that what they were doing as experiments could step on emergency services.

The FCC rules are in place for a reason. I have been in emergency situations (as an emergency medical responder) where radio communication was critical, and it was non-operative for some dumb reason (not usually a DIY transmitter, but occasionally).

 

[Baluncore]

I believe the important thing here is to encourage an experimenter with the reward of a quick positive result. Then to improve the quality of the signal, follow a technical path that introduces RF engineering concepts as needed.

Modulating the xtal osc Vcc is a really nasty way to produce AM. I would be inclined to run the oscillator on a fixed 3V or 5V supply, then use a single transistor buffer as an output stage. The power supply to the output stage could be 5V plus the AC audio signal from the series connected transformer secondary. That would modulate the carrier oscillation without the Q of the oscillator being a problem.

Having done my time monitoring RF for the .au equivalent of the FCC, I should try to stop all experimentation that might radiate RF. But RF engineers are needed now in greater numbers than ever before, so I think encouragement here is well worth the risk. An AM signal will be easy for the national authorities to find if it does cause a problem, or if it is operated for any significant period of time.

Keep the antenna as small as possible and inside the building during the experiment.

 

[OldWorldBlues]

Thanks. For the low-power FM radio transmitter kits that I've built, they usually use a very small capacitor (like around 10pF) to couple a small transmit signal to the antenna. That can help to keep your filtered* Tx signal down to a legal small-range-Tx level.

* The schematic you posted seems to show a square-wave canned crystal oscillator as the signal source that is AM modulated by modulating the power supply to it. That modulated square wave has many harmonics, in addition to the 1MHz fundamental. Transmitting a non-filtered waveform with so many harmonics is a big no-no in the radio community...

Good point- I'll look into making/obtaining a sinusoidal oscillator that I can calculate peak voltages & wavelengths for. What exactly is a harmonic?

 

[berkeman]

What exactly is a harmonic?

https://en.wikipedia.org/wiki/Harmonic

A square wave is composed of a sine wave at the fundamental frequency (Fo = 1MHz in your case), plus odd harmonics of smaller and smaller amplitudes (3rd harmonic, 5th harmonic, etc...)

http://recordingology.com/wp-content/uploads/2013/05/Distortion_SquareWave_5Harmonics.gif

 

[OldWorldBlues]

https://en.wikipedia.org/wiki/Harmonic

A square wave is composed of a sine wave at the fundamental frequency (Fo = 1MHz in your case), plus odd harmonics of smaller and smaller amplitudes (3rd harmonic, 5th harmonic, etc...)

http://recordingology.com/wp-content/uploads/2013/05/Distortion_SquareWave_5Harmonics.gif
View attachment 223988

Ohhhh so one would use a capacitor to smooth out harmonics in the same way one uses a capacitor to smooth out, say, a rectifier output.

 

[berkeman]

Ohhhh so one would use a capacitor to smooth out harmonics in the same way one uses a capacitor to smooth out, say, a rectifier output.

Sort of. You need at least two impedances to form a filter divider. You can use an LC combination or an RC combination to form a lowpass filter...

https://upload.wikimedia.org/wikipe...svg/250px-1st_Order_Lowpass_Filter_RC.svg.png

 

[OldWorldBlues]

Sort of. You need at least two impedances to form a filter divider. You can use an LC combination or an RC combination to form a lowpass filter...

https://upload.wikimedia.org/wikipe...svg/250px-1st_Order_Lowpass_Filter_RC.svg.png
View attachment 223989

Hmmm. So I don't know what the resistor is for, but it looks like the capacitor charges like a shock absorber, to level out voltage swings? And maybe the resistor is supposed to cancel out the capacitor's discharge when the AC wave flows backwards?

 

[berkeman]

I see you have another thread asking about voltage dividers, so just think about that RC filter as a voltage divider. The impedance of the capacitor gets smaller and smaller at higher frequencies, so only the low frequency voltages make it through to Vout. The high frequency voltages are "shorted out" by the low impedance of the capacitor in that voltage divider.

 

[OldWorldBlues]

I see you have another thread asking about voltage dividers, so just think about that RC filter as a voltage divider. The impedance of the capacitor gets smaller and smaller at higher frequencies, so only the low frequency voltages make it through to Vout. The high frequency voltages are "shorted out" by the low impedance of the capacitor in that voltage divider.

That seems to make sense. What exactly is impedence? (Sorry, I have a hard time wrapping my head around the books)

 

[berkeman]

Impedance is an extension of the concept of resistance. Ideal capacitors have infinite "resistance" since they are open circuits, right? And ideal inductors have zero resistance, since they are just coils of wire. (Real inductors have a small real resistance of the wire, and real capacitors do have small "leakage" currents that flow so their "resistance" is not strictly infinite).

But we use the concept of "impedance" to calculate the effective "resistance to AC current flow" for inductors and capacitors. For now, just think of the impedance of capacitors starting very high at low frequencies and getting smaller and smaller for higher and higher frequencies. The impedance of inductors starts very low at low AC frequencies, and gets bigger and bigger as you increase the AC frequency of the current through the inductor or the voltage across it.

Hope that helps. We use the symbol "X" for reactance, which is the main part of the impedance for capacitors and inductors. This wikipedia article explains it a bit more:

https://en.wikipedia.org/wiki/Electrical_reactance

 

[OldWorldBlues]

Impedance is an extension of the concept of resistance. Ideal capacitors have infinite "resistance" since they are open circuits, right? And ideal inductors have zero resistance, since they are just coils of wire. (Real inductors have a small real resistance of the wire, and real capacitors do have small "leakage" currents that flow so their "resistance" is not strictly infinite).

But we use the concept of "impedance" to calculate the effective "resistance to AC current flow" for inductors and capacitors. For now, just think of the impedance of capacitors starting very high at low frequencies and getting smaller and smaller for higher and higher frequencies. The impedance of inductors starts very low at low AC frequencies, and gets bigger and bigger as you increase the AC frequency of the current through the inductor or the voltage across it.

Hope that helps. We use the symbol "X" for reactance, which is the main part of the impedance for capacitors and inductors. This wikipedia article explains it a bit more:

https://en.wikipedia.org/wiki/Electrical_reactance

Ok, I can see now from that article that a capacitor's reactance decreases as frequency increases, because there is less charge accumulated that a current would have to "move" in between polarity shifts. But why would that cancel out frequency ripples? (Sorry if I'm bothering you XD)

 

[OldWorldBlues]

Sort of. You need at least two impedances to form a filter divider. You can use an LC combination or an RC combination to form a lowpass filter...

https://upload.wikimedia.org/wikipe...svg/250px-1st_Order_Lowpass_Filter_RC.svg.png
View attachment 223989

Also, how does Vout have voltage? Wouldn't the capacitor just discgarge backwards and not into Vout?

 

[berkeman]

But why would that cancel out frequency ripples?

It's the voltage divider effect, but generalized from just a resistor divider to an impedance divider.

In the diagram below, there are two impedances Z1 and Z2. The voltage divider equation is:
V_o = V_i \frac{Z_2}{Z_1 + Z_2}
So if the impedance Z2 is much smaller than Z1, you get a very small voltage out for Vo. Do you see that?

So if Z2 is a capacitor, and Z1 is a resistor (to make the RC lowpass filter that I posted earler in this thread), then as the AC frequency of Vin gets higher and higher, less and less of that AC signal shows up at Vout. Makes sense?

https://upload.wikimedia.org/wikipe...r.svg/220px-Impedance_voltage_divider.svg.png

 

[OldWorldBlues]

It's the voltage divider effect, but generalized from just a resistor divider to an impedance divider.

In the diagram below, there are two impedances Z1 and Z2. The voltage divider equation is:
V_o = V_i \frac{Z_2}{Z_1 + Z_2}
So if the impedance Z2 is much smaller than Z1, you get a very small voltage out for Vo. Do you see that?

So if Z2 is a capacitor, and Z1 is a resistor (to make the RC lowpass filter that I posted earler in this thread), then as the AC frequency of Vin gets higher and higher, less and less of that AC signal shows up at Vout. Makes sense?

https://upload.wikimedia.org/wikipe...r.svg/220px-Impedance_voltage_divider.svg.png
View attachment 223993

Ohh, I think I get it now. Because in a DC voltage divider, the dropped voltage across R1 minus that of R2 is the voltage (or maybe the voltage is just R1 to ground and I'm stupid, either way) then as the capacitor R2's "resistance" decreases, there will be less of a gradient between R1's drop and R2's drop, thus less voltage. Is that sort of right?

 

[berkeman]

Yes. Basically the voltage drop across each component of the voltage divider is proportional to the value of the individual impedance compared to the overall impedance. So as the voltage drop across the capacitor goes down at higher frequencies, the voltage drop across the resistor goes up. At very high frequencies, almost all of Vin drops across the lowpass filter's resistor, and there is almost no AC voltage at Vout.