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|26th August 2008, 09:35 AM||#1|
Join Date: Aug 2008
i asked someone about the current of my 9-0-0 20va 220v transformer and he showed me a drawing and some answers here as follows
"The FILter cap should be 10000 uF and not less than 4700 uF. That's generally a good suggestion, if the unregulated voltage is required to have a lower ripple. 10000 uF means about 0.8 Vss ripple at 1 A DC load with a 9V transformer, 2200 uF means somewhat below 2.5 Vss, leaving minimal difference voltage of > 4 V at the 7805. A cheap design would possibly use 1000 uF with around 4V ripple and a capacitor AC current near to the maximum rating, resulting in a reduced lifetime.
Most DIY electronicans (and some professionals as well) would simply assemble a circuit and watch the results - that's also a way to learn. "
1000uf = 0.8vss at 1ADC?
2200uf = somewhat below 2.5 vss?
leaving minimal diff voltage of >4v at 7805?
I have no idea how he got all these figures
anyone know how to calculate these, please help
|27th August 2008, 09:51 PM||#2|
Join Date: Aug 2008
Location: High Wycombe
Well I would work the problem like this:
Assume the line frequency is 50Hz and we have a full wave bridge, giving 100Hz ripple fundamental.
Further assume that the diode conduction time is short compared to the cycle time, you can calculate without this assumption, but it gets into nasty trig or numerical methods.
Then the voltage across the cap is approximately a sawtooth at the ripple frequency:
dv/dt = C/I where C is capacitence and I is current from the basic capacitor equations.
At 100hz, dt is 10ms, so dv = C/(.01I).
For a 10,000uF cap at 1A discharge current this gives:
1E-2/0.01 = about a volt ripple. With your small transformer, the assumption about the diode conduction angle being small is likely false, as both the winding resistance and leakage inductance will work against it, so 0.8V is reasonable.
Check at the end that the calculated ripple is smallish compared to the transformers output (required if the assumptions above are to be valid).
I would note that in this sort of power supply input stage the VA rating of the transformer will have to be a fair bit higher then the desired DC power output in watts due to the short conduction time (Implies large current spikes in the transformer and subsequent large I^2R losses) .
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