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How to generate AC frequency from DC waveform

Harald Kapp

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I hadn't thought of using a capacitor to block the DC,

Did you read my first answer (no. 3 in this thread)?

The value of the capacitor should be chosen such that the impedance (Z=1/(2*pi*f*C)at the operating frequency (40 kHz) is considerably less than the impedance of the transformer. In the application without the H-bridge, an electrolytic C can be used. With the H-bridge the polarity reverses each cycle thus a ceramic or foil type should be used, no electrolytic.

The impedance of the transformer at 40 kHz cannot be measured by an Ohmmeter. The Ohmmeter will give yolu only the DC resistance. But the transformer is an inductor and as such will have a higher impedance at 40 kHz. You can measure the impedance by adding aseries ressitor between 555 an transformer and measuring voltage and current at the transformer. You will have to take into account the phase shift between V and I for correct calculation of the impedance, therefore an oscilloscope would be a very useful tool.
 

NomadAU

Jul 21, 2012
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Did you read my first answer (no. 3 in this thread)?

Yes, I did read it, but overlooked that (important) snippet Harald, but thanks again :)

I've put together a circuit using a small? (4.7nF) cap to connect the 555 o/p to the transformer. What I see on the scope is not really what I was expecting - the following images should help you see what I mean.

This image is the output from the 555 with no transformer connected. OutputFrom555.jpg


This shows what happens when I connect the transformer (via the cap). XformerPrimaryAttached.jpg


And the final image shows the waveform I get across the transformer secondary windings. XformerSecondary.jpg

I found these results surprising in a number of ways.
1 - the 555 o/p clean square wave became very raggedy just by connecting it to the transformer primary. Also, the frequency seems to have been changed - the scope indicates a frequency of 961KHz...but I think it must be misreading.
2 - the voltage across the secondary windings was less than double the input voltage - I was expecting it to be signifcantly higher than that
3 - the signal on the output winding is also very ragged looking - I was hoping for a smoother sine wave.

Maybe I have got the wrong capacitance (I don't know how to relate a capacitance value to an inductance value), or maybe the transformer is really not appropriate.
 

Harald Kapp

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The frequencies are the same. Don't let the scope's displayed number fool you. look at the distance between peaks, it is the same in all diagrams.

The peaks you see are possibly resonances from the LC-circuit (transformer, coupling capacitor). Depending on the impedan ce of the 555's output (btw do you use a bipolar version or a CMOS version of the 555?) this resonance is not well damped and thus you see the spikes. If you reduce the frequency of the 555, you should see more of a rectangular waveform but with the spikes at the edges.

You will not see a sinusoidal waveform on the secondary since your input is a rectangular waveform. If you want a sine output, you need a sine input to the transformer or you need to filter the non-sinusoidal output by a low-pass filter.

ASs Laplace said, the transformer may be wholly inadequate for this task.
 

NomadAU

Jul 21, 2012
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The frequencies are the same. Don't let the scope's displayed number fool you. look at the distance between peaks, it is the same in all diagrams.

You're right, thanks for pointing that out.

Depending on the impedance of the 555's output (btw do you use a bipolar version or a CMOS version of the 555?)

I'm using the bipolar version.


ASs Laplace said, the transformer may be wholly inadequate for this task.

I think this may well be the case. I thought I could use a low power transformer from Jaycar to do the job, but it looks like it isn't good enough to just look at winding ratios - there are other factors (such as impedence and resonance) that need to be considered. I'll look a bit more and see if I can find another more suitable transformer.
 

Harald Kapp

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How about a flyback transformer? Those are built for very high output voltages and can be salvaged from old CRT-TVs probably for a song.
Or use a standard mains transformer (e.g. 130 V : 9 V) in reverse.
 

NomadAU

Jul 21, 2012
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I've taken a look at some mains transformers that might give a reasonable output but one thing about their specs puzzles me. They all provide a Magnetising Current value (around 30-40mA) which I presume means the amount of current that needs to flow through the primary winding for the transformer to work? If I've got that correct, then how will that occur if I've coupled the primary to the 555 using a capacitor which blocks DC current flow?
 

Harald Kapp

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add 1) the magnetizing current is not DC. It is the current required to create a magnetic field in the transformer. And you need an alternating current to create an alternating field. A transformer cannot make use of DC.
add 2) you connect the 555 to the secondary. You want to transform upwards, ont downwards.
The specs of a mains transformer are given for the use case where mains is attached to the primary. Since you want to step up, you connect the signal source to the secondary. The magnetizing current will be different from the one given in the datasheet. You can approximate Np*Ip=Ns*Is where
Np = number of primary windings
Ns = number of secondary windings
Ip = primary current
Is = secondary current.
Therefore the magnetizing current will scale up with the transformer ratio:
Ism = Np/Ns*Ism
where the index "m" means "magnetizing".

I'm not a transformer expert, but a magnetizing current of 30 mA means ~4W at 130 V which to me looks quite a lot. I suppose you've looked at rather big transformers? Look for much smaller ones. Your output power cannot be more than the input power.
A rough calculation:
Assume the 555 gives 7 Vpeak-peak. If the signal were sinusoidal this is equivalent to 2.5 V rms (let's neglect the non-sinusiodal waveform for this calculation).
Assume furthermore that the current from the 555 into the transformer is 100 mA. So you have 2.5 V * 0.1 A = 0.25 W.
Assume an efficiency of the transformer of 90 %. This means that on the primary side (which is the secondary in this application because the transformer is used in reverse) you have only 0.8 * 0.25 W = 0.2 W available.
Assume an original transformer ration of 130 V : 9 V. Using the transformer in reverse this is 9 V : 130 V. Your 7 V input therefore gives ~ 100 V.
The available power of 0.2 W at 100 V translates into 2 mA max. output current.

From these figures you see that a very small transformer (0.25 W) is sufficient. Such a transformer will have an accordingly low magnetizing current.
 

NomadAU

Jul 21, 2012
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Thanks again Harald for taking the time to provide such clear feedback - it is very helpful in improving my understanding of this area.

As for the magnetising current comment, I don't think the transformer I was looking at is particularly large - here's the link if you are interested.

http://http://www.jaycar.com.au/productView.asp?ID=MM2006

I'm going to purchase one today to try it out - if used in reverse, it should provide somewhere around 20 times the voltage step-up. If I limit the input to around 7v, then this should give the desired 140v output, or thereabouts.
 
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