I am planning on winding an isolation transformer to couple an AC signal onto a DC source. The load will be capacitive and I plan on varying frequency, so I would like to investigate using a third winding to regulate the output pk-pk. Does anyone have any good references/thoughts on this?
It won't work very well. Why not use an optoisolator from the DC output back to the control logic? That's the usual way.
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How much DC current is flowing through it? How much capacitance, what frequency range, how much current are you driving through the transformer?
Transformers are pretty versatile but the performance you get out of them is directly proportional to the amount of effort you use winding them. A capacitive load is probably going to need unusually low leakage inductance, which means many interleaved layers.
Tim
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Assuming you said what you really meant....
The load is capacitive, so there's zero average current in the DC source and the transformer. You get to worry less about saturation. I've obviously made some unwarranted assumptions about the nature of your AC signal. I think this falls under GIGO. If you wind a transformer that has flat amplitude response over the frequency and amplitude ranges of interest, you should be able to just control the p-p voltage of the input side. If your transformer is no good, the third winding should be equally no good. Could be better if you put a fixed load impedance on the third winding equal to the main load and sensed that. Resonances will bite you in the A$$.
In the general case, in an imperfect world, you might have better luck with two separate transformers. With the current lack of specificity, I can't determine whether that's trivial or weighs 400 pounds.
---------- @John L.
Thanks for the tip I will look into it to see if it meets my needs.
---------- @Tim W.
*There is very little dc current in this application (uA range).To give more details (several of you pointed out that I didn't give enough the first go around).
*load = 1uF->10uF *f range -- 100Hz -> 100kHZ *i_ac = 3A *v_ac = 5V---------- @mike
I plan on having a frequency sweep of the input ac voltage. My worry was that as I move up in frequency, I will see a slight increase in the pkpk V on the output. I wound a quick example on a ferrite torroid and saw this effect.
In terms of getting more of a flat response, I have also considered having two transformers and splitting up the range. Still deciding on that though.
first go around).
as I move up in frequency, I will see a slight increase in the pkpk V on the output. I wound a quick example on a ferrite torroid and saw this effect.
transformers and splitting up the range. Still deciding on that though.
Also, I am planning on coupling the ac signal to a dc source in the 100V-1kV range.
the first go around).
as I move up in
frequency, I will see a slight increase in the pkpk V on the output. I wound a quick example on a
ferrite torroid and saw this effect.
two transformers
and splitting up the range. Still deciding on that though.
range.
Assuming a sine wave, the back of my envelope calculation says you're gonna need closer to 30Amps peak to make 5V RMS into 10uF at 100KHZ. Wouldn't be the first time I got math wrong by an order or magnitude, but I did the calculation 3 times.
You're likely to have resonance problems. A resistor across the secondary can damp it some. If my current numbers are right, it's likely that a second sense transformer will be less hassle than making a more perfect big one.
the first go around).
that as I move up in
two transformers
range.
Hey Mike, you bring up a good point. In truth, In needed to be more careful with my calculations. In the situation that I described I believe that you would need to say:
I = C * dv/dt = C*V_pkpk*2*f = (10uF)*(5V)*2*(100kHz) = 10A -- If that is correct, our numbers are fairly close.
------ In my original post, I was interested in controlling output voltage of the transformer. Let's set up a senario where this make a little more sense:
Using a transformer to couple an AC signal onto a DC source. Then use this signal as the input to a lowpass filter with an R to be determined and a C of
10uF.
the first go around).
that as I move up in
two transformers
range.
with my calculations. In the situation that I described I believe that you would need to say:
correct, our numbers are fairly close.
transformer. Let's set up a senario where this make a little more sense:
signal as the input to a lowpass filter with an R to be determined and a C of
10uF.Saying that the DC current is in the microamp region is a LOT different from saying that the DC source has to be able to source and sink the peak AC current. Might be easier to build a big honkin' amplifier that can do 10V RMS at 100KHz and use a cap to couple to the load. Then the 1KV supply doesn't have to supply hardly any peak current.
You've probably got something pretty close in the basement gathering dust. And when you're not using it on the cap, you can plug it into your mp3 player.
The first part makes sense you have a capacitative load so there won't be continuous DC current through the trasnsformer so a normal transformer will work to couple the signal
The second part, using a third winding to modulate the output sounds a bit like a magnetic amplifier.
When winding a magnetic amplifier on an E-I core you wind it on the outer limbs and connect the coils in opposite directions (anti-series) so that when you energise the coils they'll act summetrically saturating the outer limbs and reducing the coupling and the inductance of the transformer.
If your AC signal is a current source this could work well, if it's a voltage source it'll need some series resistance added
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that sounds like saturation. try two beads or more turns (assuming voltage source - fewer turns if a current source)
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