EEVblog® Electronics Community Forum
Electronics => Beginners => Topic started by: jammet on May 23, 2014, 02:15:22 pm
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Ah, ye folks, its my first post here.
I am an Electrical Engineering student, not Electronics Engineering, so we deal with high power stuffs, AC circuits, PF correction, and etc, and I'm just incoming 4th year, so not much knowledge in these. I only have taken 6 units of electronics last semesters. (3 each sem).
How much current can I draw from a transformer? (yeah i just repeated that thing.)
Suppose that I have a 240VA, 24Vrms, 10Arms then rectification, filtering, regulation, etc, etc.
Can i draw full 10A from it?
So its like i can get 10A running at 20VDC?
So its like the circuit achieved resonance? purely resistive?
I'm still thinkin about this since at peak power you can get 240*sqrt(2) VA.
But it will only happen if pf is 1.
but yeah this will be rectified, so its a different thing.
Don't be afraid to give mathematical help, its okay with me. :D
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VA is rms.
you can pull 10 A from it, but it will probably get very warm at 100% loading.
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24V rms gets you about 33V DC with a cap input filter, at about 0.5 PF. So your 10A rms is drawn at only 5A DC. (Not exactly correct, but wave-of-the-hands close enough.)
Tim
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hmm, if that is the pf, 60 deg phase shift. kinda shitty. reactive power is just being wasted power.
can we get 0.8 and up pf from this? hmm, like adding choke on it?
I searched a lot about it, its hard to find data in Input choke PS, anyone help?
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Inputs and outputs are RMS values.
When you convert to DC using a bridge rectifier, you get a dc voltage with peaks of up to Vac * 1.414 = 24 x 1.414 = 33.95v .. the VA rating of the transformer remains the same, you don't magically have more voltage and the same current.
Then you have the voltage drop on diodes in the bridge rectifier, at 10A that's going to be about 1.2v x 2 = 2.4v ... so you're left with peak DC vpltage of about 31.5 volts.
Now you need to smooth that out.. capacitors.... a good enough formula is C = Current / 2 x Ac frequency x Vripple where Vripple is how much you're willing to let the voltage sag from the peak voltage.
Let's say you're in US (60 Hz mains frequency) and you want to have minimum 28v all the time.. then C = 10 / (2x60x3.5) = 10 / 420 = 0.02380 Farads or 23800 uF
This is all assuming you're going to use a bridge rectifier + capacitor, you can have some fun adding inductors after bridge rectifier, or using just two diodes instead of a bridge if your transformer has a center tap etc etc ... here's some examples:
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DPF is almost 1.0 -- it's harmonics, not phase shift. AC line PFC (of the AC steady state variety) won't do anything.
A series inductor on the AC side improves PFC, but worsens regulation (output voltage stability, i.e., change in Vout for change in Iout). Because transformers have some leakage inductance (especially class 2 and side-by-side bobbin wound types), this happens naturally, so your PF might realistically be 0.6 to 0.7 with a transformer.
It's also higher with a smaller filter capacitor, but this worsens ripple voltage.
Harmonic traps can also be used, with varying effectiveness. These are all called passive PFC.
A choke input filter has somewhat better PF, delivering Vdc = 0.9*Vrms as long as the load is heavy enough to maintain continuous current flow in the inductor (at lighter loads, the voltage floats up, until 1.41 times at open circuit). I want to say PF is 0.9 but I can't find a reference and I'm too lazy to write it out right now...
Tim
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ah, since
Capacitance = couloumbs/voltage
and current = couloumbs/time
so its
Capacitance = (Current * time) / Voltage
ah yeah i get it, the voltage ripple can be controlled, but about the current? in that center tap with choke input?
how do we compute for the value of the choke?
does it need to be resonant with the capacitor?
or wat?
sorry im really dumb about this thing.
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looks like FWB with choke input load is the best choice for me hehe.
:D
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Unless there is some magic I don't know about.
Power out = Power in minus internal losses.
The 24VA rating is input, you cannot expect to get more than that out, no matter how you manipulate the numbers.
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hmm, how do we compute for the choke?
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hmm, how do we compute for the choke?
you calculate how many joules of energy you need stored in the inductor and work backwards.
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This isn't the clearest or most accurate summary, I think, but it should be the right conclusion:
www.faculty.umassd.edu/xtras/catls/resources/binarydoc/1212.ppt (http://www.faculty.umassd.edu/xtras/catls/resources/binarydoc/1212.ppt)
Tim