Electronics > Beginners
Transformer question
xavier60:
--- Quote from: Benta on February 02, 2019, 10:31:19 pm ---
--- Quote from: ArthurDent on February 02, 2019, 09:34:26 pm ---Here is a handy reference to explain it all.
http://www.hammondmfg.com/pdf/5c007.pdf
--- End quote ---
That "handy reference" doesn't explain anything at all. Quite the opposite
Please explain to me why "FULL WAVE BRIDGE Resistive Load" tells me to multiply voltage and current by 0.90?
I don't even want to comment on the other configurations, that would be a total waste of time.
Apparently "Hammond Manufacturing" needs any excuse to derate their transformers, leading me to doubt their products.
--- End quote ---
The "0.90" is used to multiply by the RMS voltage to get the Average voltage, 0.9 x 0.707 = 0.636.
I have some doubts about some of the information in the Hammond guide.
Ian.M:
--- Quote from: xavier60 on February 03, 2019, 07:16:44 am ---I have some doubts about some of the information in the Hammond guide.
--- End quote ---
Which specific circuit(s)? Every time I've set up a LTspice sim for one of them, the worst case derating factors come out very close to the Hammond figures. I must admit I haven't tried them all though.
xavier60:
For the FULL WAVE BRIDGE Capacitor Input Load.
V (Avg) D.C. = 0.90 X Sec. V A.C.
With a 30VAC secondary, I wouldn't expect the average DC to be loaded back to 27V at the recommended load derating factor. That's a long way down from the unloaded DC of 42V.
I have started a linear bench supply project so I guess I'll have a better idea when it's fully tested.
For the FULL WAVE BRIDGE Choke Input Load.
"V (Avg) D.C. = 0.90 X Sec. V A.C" mathematically agrees with the expected DC voltage at some light load that causes continuous inductor current flow, assuming that the inductor is large enough.
At full rated load, the DC voltage would be something less due to expected resistive losses.
Ian.M:
I hadn't noticed that. |O A somewhat simplistic LTspice sim of the FULL WAVE BRIDGE Capacitor Input Load circuit that assumes all the transformer losses are resistive is attached. Its got a lot of .measure commands so view the spice error log for results.
With ideal diodes, a transformer with 10% regulation and 1000uF per A of load current, I get 38.6V average, and 35.8V ripple trough.
It also shows the Hammond 0.62 current derating factor may be insufficient for 24/7 duty if the transformer has good regulation, and the diodes and reservoir cap are large for the load current.
soldar:
If you put the mains voltage on the screen of your scope the first thing you will notice is that it is not a sine at all but the tops are flattened and that is due to all the rectifiers getting shots of current only at the top of the wave. The output resistance of the network makes the voltage fall.
If you put the current on the scope you will see short bursts of current and zero current most of the cycle. Power factor is very poor.
The behavior depends an many factors. Higher resistance and lower capacity mean a longer diode conductive duty cycle, a better power factor but worse regulation. OTOH, big capacitor and low impedance means more stable output regulation, lower ripple, but worse power factor, bigger transformer and diodes needed, etc.
Calculations are good but there is a lot of rule of thumb and pure educated estimation. If you are building a single unit the best way is to start building and testing behavior under different loads.
But, yeah, DC current out of the rectifier + capacitor is nowhere near what the transformer can supply with unity power factor on, say, a purely resistive load.
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