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Re: gEDA-user: HPS announce



47,000uf ? thats huge.
lets think. 10ms period (50hz full wave)
1A (for this reg)
i = c dv/dt => c = i * dt / dv
= ,01/2
= 5000uf

I suspect you have a kind of a lot of significant figures for the accuracy of supply voltage too :)

an extra couple of volts from the transformer would help the cap size
and hurt power. Thats linear supplies for you.


john




Robert Riemer wrote:

After a few quick calculations regarding the power supply filter capacitance, I think you should spec a different transformer. With the transformer you have specified: Vfl = 16.968Vp, I used this for my calculations even though you are not using full rated current draw for the transformer. Vout = 12.0V. According to National Semiconductor the voltage accros the regulator could be 2.5V worse case. I used Vf = 0.7V for the diode. Subtract all the drops from the transformer voltage, this leaves you with the maxinum ripple voltage accros the capacitor. dV = 1.768V. Assuming a full load current the filter capacitor for the 12 volt supply should be >47000 mfd. The 5V supply will require an even larger capacitor.

I may be under estimating the transformer ratings, and over estimating the dropout voltage, but is that what a good designer should do?





On October 29, 2004 06:16 pm, Robert Riemer wrote:

Some technical tips:

1) Use fewer 500 mfd capacitors. There may be too much capacitance. If
not,
use fewer larger value capacitors. Probabaly the same price and size approx.. The value of capacitance depends on the transformer voltage,
amount
of allowable output ripple, and current drawn from the supply. If the capacitance is too large, the regulator could dissapate too much power. If it is too small the output ripple may be large, our the regulator can drop out of regulation.

2) The transformer is potentially unevenly loaded. There is a split 12v supply but not 5v.

3) The terminal labled +Ureg (unregulated) should be as variable, not unregulated.