ronsimpson wrote:
JimB. Do you think the 9V transformer will work or does he need a 12V transformer?
A 9v transformer would be fine for a 5v supply, but not good enough for a 12v supply.
9 x 1.4 = 12.6v
Allowing for ONE diode drop (we are considering a biphase rectifier here rather than a full wave bridge), that would leave 11.9 volts.
Just plain not enough for a 7812 regulator.
Musicmanager wrote:
Is this load x C / ripple freq ?
Approaching the calculation from both ends.
Assuming:
a 15-0-15v transformer,
biphase rectification with silicon diodes,
a 7812 voltage regulator.
Rectifying to 15 volts, we will have a maximum DC voltage of (15 - 0.7) x 1.414 = 20.2 volts.
This voltage can only go downwards from there due to:
IxR losses in the transformer, wiring and diodes
Ripple due to discharging the reservoir capacitor.
Now consider the 7812.
The Fairchild datasheet gives a dropout voltage of 2 volts, so we need a minimum of 14 volts at the input of the 7812.
To find the capacitor we need:
The maximum voltage is 20.2, the minimum is 14v, so we have a maximum ripple of 20.2 - 14 = 6.2 volts.
One of the accepted formulae for calulating the required capacitance is:
C = (IL x t)/Vripp
Where:
C is the capacitance in Farads
IL is the load current in Amps
t is the period of the ripple in seconds. (for a full wave rectifier with a 50hz supply t = 10ms, for a 60hz supply t = 8.33ms.
Vripp is ripple voltage.
Using our values:
C = (1 x 0.01)/6.2 = 0.0016 F = 1600uF
So , 1600uF is the very MINIMUM capacitance for this thing to work correctly.
The next highest common value is 2200uF, this could be a but marginal, allowing for wide tolerances in electrolytic capacitor values, and loss of value with age.
A 4700uF capacitor would be my choice to give a robust reliable power supply.
The ripple voltage at the reservoir capacitor will be about 2.1 volts, giving us 6.2 - 2.1 = 4.1 volts "headroom" for unknowns and "Ooooh, I never thought of that" items.
JimB