ripple issues

[simonwai999]

hi folks
I asked someone about the filter cap
and i got these answers

"The drawing suggest 10000 uF and the text says 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. "

my question is
how come 100000uf =0.8vss ripple at 1ADC?
2200 uf means somewhat below 2.5vss
and minimal diff voltage of >4v at 7805?
how do we calculate all these figures?
thanks

[SY]

Ignore the regulator for a moment. You can do a rough calculation of the ripple from first principles (see, for example, the power supply section of the Radio Amateur's Handbook). A useful approximation is V(ripple) = It/C, where I is the load current, t is the period of one-half cycle of the mains supply, and C is the filter capacitance.

If your mains is 50Hz, t = 0.01. With 10mF filtering, V(ripple) = 1V, pretty close to your estimate.

The minimum differential voltage is a function of load current and can be found on the 7805 data sheet. Remember, the ripple has to be added to this so that the regulator doesn't drop out 100 times a second!

[simonwai999]

hi
thanks for your reply moderator

(A useful approximation is V(ripple) = It/C, where I is the load current, t is the period of one-half cycle of the mains supply, and C is the filter capacitance.

If your mains is 50Hz, t = 0.01. With 10mF filtering, V(ripple) = 1V, pretty close to your estimate.)
I understand this part
the current =1A
1 times 0.01/ (10/1000)=1

but I do not understand this part
why 50 hertz = 0.01 for one-half cycle of the mains supply?

please clear my doubts again
thanks a lot

[AndrewT]

@ 50Hz one cycle lasts 1000mS/50= 20mS.
A full wave rectifier converts all the voltage excursions to the same polarity resulting in two voltages peaks for each cycle.
The timing between these peaks is 20mS/2=10mS =0.01S
t=0.01

[AndrewT]

when mains voltage is at minimum i.e. 0.9*220Vac~200Vac, then what is the peak voltage from the transformer?
Subtract the diode voltage drop, ~1.4V
Now calculate the maximum ripple if maximum load current is drawn while mains is at minimum supply voltage.
subtract ripple from peak transformer voltage.
Now what voltage sag will occurr when maximum current is drawn? This could be between 5% and 30% of open circuit voltage depending on transformer regulation and ratio of VA to peak power demand.
Subtract the voltage sag from the last voltage figure.
Now look up the regulator datasheet and find the drop out voltage for the maximum current you require.
Subtract drop out from last voltage figure you calculated.
What is left? is it sufficient to meet your output voltage needs?

Now increase the mains voltage to maximum tolerance. ~1.1*220Vac~240Vac
What is the peak voltage from the transformer.
Subtract the minimum current diode voltage drop,~1V.
reduce the current draw to minimum. ripple and voltage sag will be at minimum.
What is the maximum supply voltage to the regulator. Can the regulator survive this highest input voltage?

Now look at typical current demand and typical voltage supply and calculate the regulator dissipation.
How big a heatsink is needed. Will this heatsink be sufficient for short term high current demand?
Will the heatsink be sufficient for prolonged high supply voltage?

[simonwai999]

hi all
thanks for all replies
i understand it now
cool
cheers