Maker Pro
Maker Pro

How to make Regulated Power Supply

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
yeah steve....but i think we get the average value((1/pi)Vmax) and not the peak value when ac is converted to dc...

It depends a lot on your load.

Also note that current won't flow from the transformer until the AC voltage exceeds the capacitor voltage (plus the Vf of one or 2 diodes). Ths means you're drawing peaks of high current from the transformer rather than a nice continuous load.

where does 1/pi come from?
 
yeah steve..I agree with you...now i understood the concept clearly...thaanks!!!
i though that the output dc we get will be the average value of ac....but now i understood...average value of the sinusoidal ac signal Vmsin(wt) is (1/pi)Vm..
 
i have a 10000uF capacitor rated 50v. that mean i cant use that capacitor?i got one 470uF which one shd i use?
for the rectifier i got 1a4001 and 6a10 which is better for stripboard
 
more the value of capacitor more constant dc will be....but as the value of capacitor increases its size increases...So u hav to make compromise between the compactness of your ckt and the steadiness in dc waveform
 
ok."current won't flow from the transformer until the AC voltage exceeds the capacitor voltage (plus the Vf of one or 2 diodes)." i thought no current go through to the load since capacitor voltage rating is 50 and sec v is 24.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
That's no current will flow through the secondary until the instantaneous voltage across the winding exceeds the instantaneous voltage across the capacitor (plus losses in the rectifier(s))

If you imagine a completely discharged capacitor, current will flow from the secondary through the rectifiers into the capacitor as soon as the voltage across the secondary is high enough to forward bias the diodes. In a bridge rectifier this is approx 1.2V. So at the voltage rises from zero to 1.3V, there is about 0.1 of a volt greater than the capacitor voltage (0) and thus current starts to flow and the capacitor will charge up to 0.1V.

As the half cycle proceeds all the way up to 34 volts (24 x 1.414) the capacitor is charged by the transformer.

As the voltage starts to drop back down to zero, no current flows into the capacitor because its voltage is higher than the voltage supplied by the transformer.

Once you apply a load to the output, it causes the capacitor to discharge. The rate of this discharge (in terms of voltage) is determined by the current drawn and the size (in uF) of the capacitor.

If we assume that the load is able to discharge the capacitor about 1 volt before the secondary voltage again exceeds the capacitor voltage (approx 33V now) then current flows for a very brief period while the secondary voltage is between 33 and 34 volts. During this time (perhaps 5% of the time) the entire amount of charge used by the load needs to be replenished. Thus the secondary current is approximately 20 times the load current for this brief period.

This high current may cause several effects, resistive losses, larger losses in diodes, transformer core saturation, etc., which mean that the capacitors cannot actually be charged back up to 34 volts. Thus, under load the voltage drops, and the ripple increases.

The voltage is not determined by the rating of the capacitor. You use a 50V capacitor in this design because 34 volts is less than 50V and it provides a comfortable margin. You could certainly not use a 24 volt capacitor (it would probably explode in fairly short order) and a 35 volt capacitor is very marginal and I wouldn't recommend it.

Indeed you may find that the unloaded voltage from the power supply is higher than calculated. This may be due to a number of factors:

1) At low currents the Vf of the diodes can be very small.
2) the transformer may be rated for 24V at normal load, and may have a higher voltage unloaded.
3) your mains may be slightly higher than it's rated voltage
 
Wow. thanks a lot for the detailed explanation.I can understand it very well. But i forgot to drill holes in the circuit where i needed to .I think i saw smoke from the diode
 
I am getting 0 dc output.when i used 1A4001 diode it burned.I changed to 6A10 diode.I am not sure why i get 0 output.maybe the 10000uF capacitor take very long to charge up?or i connected the diodes wrongly?
can u help me
 
Last edited:

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
First thing to check is that you had the diodes connected around the right way.

One diode around the wrong way will cause a short which will probably destroy at least 2 diodes.

the capacitor shouldn't take more than a couple of half-cycles (i.e. a handfull or two of milliseconds) to charge.
 
maybe diodes are wrong.i got 33v dc at the end but no current.can u show me a stripboard planning for the 4 diodes?can u check my planning?i will uplaod it
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Show me what you've done and I will try to tell you what you've done wrong.
 
Manick9
if you not sure of the writing ask me.i want to get this done .http://s1202.photobucket.com/albums/bb380/Manick9/?action=view&current=planning.jpg
 
i was actually referring to it all this time.
i got 33v dc and 15amp output not sure why so much current.so i am using 7824 voltage regulator at the end to reduce 33V to 24V and hopefully the current also.
1.The bleeder resistor shd be at the end of the regulator right?
2. the fuse i have put is on the primary side of the transformer. shd it be on the secondary side?
3.I think 10,000uF seems too big. i see black color on the back of the stripboard near the capacitor.
 
Last edited:
oh k.....if your secondary voltage and current itself are 24v and 1.1 amp....how can you get output dc as 33V and 15 amps...its against conservation of energy principle...i think u hav measured them wrong...verify again...
 
Top