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# Transformer input and output with varying frequency

#### Thesecret20111

Apr 11, 2014
18
Hello guys!

I've been doing experiments with a relatively strong step-up transformer which I've been using to test how frequency on the primary coil affects output voltage. I've done research online and many times I've read that frequency will have very little effect on output voltage but my findings contradict this.

Here's what I found from doing tests (all at low frequency and low voltage):

At 20Hz and 1.5 volt on the primary I got on average 90.2 V on the secondary.
At 16Hz and 1.5 volt on the primary I got on average 62.2 V on the secondary.
At 12Hz and 1.5 volt on the primary I got on average 50.1 V on the secondary.

Does this make sense to anyone? Or is there something seriously wrong with my experiment?
Thanks!

#### duke37

Jan 9, 2011
5,364
As the frequency reduces, the reactance of the primary reduces so more inductive current is needed to maintain the voltage. At some stage the current will be sufficient to saturate the core and the primary winding will look more like a short circuit.

To maintain the output at low frequencies you will need a low impedance source to provide the current. Do you have this?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,728
Here is a very good read on this (and other transformer related) topic(s). See chapter "11.6 voltage & frequency" (it may be in another part of the series than I linked too - sorry, I have a printout without the assignment of pages to parts of the web series.

#### Thesecret20111

Apr 11, 2014
18
As the frequency reduces, the reactance of the primary reduces so more inductive current is needed to maintain the voltage. At some stage the current will be sufficient to saturate the core and the primary winding will look more like a short circuit.

To maintain the output at low frequencies you will need a low impedance source to provide the current. Do you have this?

The power source I'm using is a single AA battery.

#### Thesecret20111

Apr 11, 2014
18
Here is a very good read on this (and other transformer related) topic(s). See chapter "11.6 voltage & frequency" (it may be in another part of the series than I linked too - sorry, I have a printout without the assignment of pages to parts of the web series.

Thank you for the article link! Its very well written and I found it very informative. At the risk of sounding completely ignorant, I have a few questions regarding what is said in the article. Where he writes:
60Hz is 20% greater than 50Hz
how did he work this out? I thought 60Hz would be only 10% higher than 50Hz? Also where he says
Operating a 60Hz transformer at 50Hz is effectively the same as a 20% increase in mains voltage, but note that this does not mean that the secondary voltage is increased. For a 230V transformer that's the same as running at 60Hz, but at a supply voltage of 276V. The core will be seriously saturated, and the magnetising current will be increased dramatically.
wouldn't that mean that operating a transformer at a lower frequency than its rated for would in fact increase the output voltage? I know he says that it doesn't actually correlate to a increased output as the core gets saturated but thats using voltage that its rated for. The setup I've got, I've decreased the voltage so much that surely its not reaching saturation. Wouldn't this mean I would be expecting the opposite of what I found? With output voltage actually increasing with lower frequency?

Thanks so much for your help!

#### duke37

Jan 9, 2011
5,364
60/50 = 1.2 i.e. 20% more.

A single AA cell will produce DC which is a frequency of 0Hz and will not produce much current.

The faster you drive the transformer, the fewer turns you need and the transformer can be smaller. Aircraft transformers were run on 400Hz. I found out that they would not run on 50Hz (big bang), it was one way of learning.

If the transformer is running in a linear manner and it has little leakage inductance, the output voltage will just be due to the input voltage and the turns ratio.

#### Thesecret20111

Apr 11, 2014
18
60/50 = 1.2 i.e. 20% more.

A single AA cell will produce DC which is a frequency of 0Hz and will not produce much current.

The faster you drive the transformer, the fewer turns you need and the transformer can be smaller. Aircraft transformers were run on 400Hz. I found out that they would not run on 50Hz (big bang), it was one way of learning.

If the transformer is running in a linear manner and it has little leakage inductance, the output voltage will just be due to the input voltage and the turns ratio.

Ah, that makes sense.

I should have said before, the AA battery is run through a little low frequency AC generator that I've made, so it is infact reaching the frequencies I specified.

So it's more to do with how heavy duty the transformer has to be? Haha that's hilarious! 50Hz on an 400Hz transformer would basically be a short right? That's certainly is a way to learn what not to do very quickly!

I guess why I'm confused is I've been thinking back to the basic induction experiment using a bar magnet and a solenoid. When you move a bar magnet in and out of the solenoid the voltage increases with the speed at which you move it. Is this idea true to a certain extent with transformers when using very low voltages and currents to avoid transformer failure? And if so would higher frequency = higher output V or vise versa?

Thanks so much for taking the time to answer my silly questions! I really appreciate it!

#### duke37

Jan 9, 2011
5,364
Your little AC generator will not be able to produce much current and when the frequency is low, a lot of inductive current will be needed even if saturation is not reached. The input voltage will drop and so will the output voltage.

Xl=2*pi*f*L
I=V/Xl

Last edited:

#### Thesecret20111

Apr 11, 2014
18
Your little AC generator will not be able to produce much current and when the frequency is low, a lot of inductive current will be needed even if saturation is not reached. The input voltage will drop and so will the output voltage.

Xl=2*pi*f*L I=V/Xl

Interesting! That makes sense now! Would you just mind explaining what X is in the equation above? Or perhaps if these equations have names you could tell me and I'll do some reading myself.

Thank you very much for all your help!

#### duke37

Jan 9, 2011
5,364
There are two equations there, the space was deleted so I have now put them on separate lines.

X is the normal symbol for reactance. It can sometimes be used as a replacement for resistance in Ohms law but cannot be simply added to resistance. Xl is the reactance due to inductance. Xc would be the reactance due to capacitance, not used here.

L is the inductance.

I is the current. My display does not distinguish between I (capital i) and l (lower case L) which causes confusion.

#### Thesecret20111

Apr 11, 2014
18
There are two equations there, the space was deleted so I have now put them on separate lines.

X is the normal symbol for reactance. It can sometimes be used as a replacement for resistance in Ohms law but cannot be simply added to resistance. Xl is the reactance due to inductance. Xc would be the reactance due to capacitance, not used here.

L is the inductance.

I is the current. My display does not distinguish between I (capital i) and l (lower case L) which causes confusion.

That clears it all up for me! Thanks again for all your help!

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