6 volt transformer

[M0BSW]

Hi , I have a 250v in to 6 volt out transformer, and I want to use this as a small power supply,, I'm stuck on the smoothing capacitor I was thinking around  440UF  ,by using 2X 220 uf in parallel, as that's what I have handy, any ideas
Paul

[madires]

The well known rule of thumb is 2200µF per 1A. How much current does the transformer deliver?

[M0BSW]

The well known rule of thumb is 2200µF per 1A. How much current does the transformer deliver?

250MA so am I right in thinking 550uf

[madires]

250MA so am I right in thinking 550uf

It doesn't need to be that value exactly ;-) A 470µF cap will do the job just fine. Or the two 220µF caps you already got.

[Kevin.D]

you didnt say what voltage  and current you expect.  but 470uf ? is that enough for what you want  at 250 mA  ? . 470 uF at 250mA  get's you ~ 5V ripple (p-p).
so if your peak dc voltage at the caps  is ~ 8V  (6X1.41- 2 diode drops) then your left with valleys down to three volts   for your regulator .
approx ripple(p-p) at caps is I/2fC for full wave (I/fc for half wave) .where f=mains frequency ,I = current .

[mariush]

A basic formula I memorized is:

Capacitance = 0.7 x Current / 2x AC Frequency x Vripple

where Vripple is how much you're willing to have the voltage go up and down on the dc output.

You have a 6v AC transformer ... that means the peak voltage is about 6x1.41 = ~ 8.5v and if you use a bridge rectifier you may have about 1-1.2v drop on the diodes, so you end up with a peak voltage of about 7.2v.
Now if you want a minimum of 6v, it means you can afford a Vripple of 1.2v

For example, if you want 0.25A output, 6v minimum  and you're in US (so frequency is 60Hz)  then Capacitance = 0.7x0.25 / (2x60x1.2) = 0.175/144 = 0.001215 Farads = 1215uF


Another simple formula is basically  Vripple = Iout / 2fC  (for a full wave rectifier)   
... so as you increase the capacitance the Vripple decreases.   For 1215uF, we have  Vripple = 0.25/ 2x60x0.001215 = 0.25/0.1458 = 1.71 volts  (so as you can see not quite 1.2v but both formulas are simplifications and give you a reasonable value)
You can basically lower the ripple by increasing capacitance, you don't have to resume to exactly these values.

Not sure for what you want this... if you want it to obtain a 5v power supply, you should make sure the dc voltage is always higher than 5v + whatever voltage drop the linear regulator requires. A 7805 needs about 1.5v , so you'd need a minimum of 6.5v at input.. a LD1117 or *1117  needs only about 1v-1.1v etc etc

[Kevin.D]

Here's a usefull little single sheet guide to transformer / rectifier / filter relationships that was in an old components catalogue I have. 

Addendum
the equation posted by mariush .
V ripple (p-p) = 0.7I/2fC gives you a much closer approximation of V ripple than the formulea I/2fC . But dont forget the large tolerance of capacitors +/-20% means you have to oversize by at least 20% above what you require +10% for luck . 

[M0BSW]

Thanks everyone , nearly all the circuits I build run at between 5 & 8 volts, so I want to build this small power supply, as it will probably be cheaper than batteries, so nothing fancy is required. I had good advice from you all, which has helped my learning process. Thank You All 

[Zero999]

You'll probably need a low drop-out regulator to get 5V out with any decent level of regulation.

You on't be able to get the full 250mA out at 5V, 100mA is probably a safer bet.