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How to make Regulated Power Supply

I will borrow books from the library l8r to understand and gain more knowledge. But i believe your circuit would work fine.You seem like an expert/connoisseur.
Smoothing capacitor for 10% ripple, C =(5 × Io)/ ( Vs × f)
Are you from EEE also?
 
Pump control

Yes i try to help where i can, as over the years others have helped me, thats right 0 volts neutral connection, and posative volts in, but its AC and either way round is ok, but if the line in is marked as neutral and live in, then observe the input status. And the trasformer bracket, and the chassis case is grounded with a bolt washer and nut connected to the wire ground link from the supply input. Good luck with the project, i will keep an eye out for updates. Dave.
 
Pump control

Whats EEE, i dont know that. Expert no not me, keen hobbyist best sums it up, i have long term illness to contend with, but stil enjoy electronics on a hobby level, when my health allows me. My forum activaty is part of it all, and lets me have some electronic contacts even when not well enough to build any circuits, anyway good luck with it all. Dave.
 
"chassis case is grounded with a bolt washer and nut connected to the wire ground link from the supply input." I will have to use the earth wire for this grounding?
EEE means Electrical and Electronic Engineering. Sorry to hear that. You will get better :) you can take your hobby to the next lvl.
 
this circuit can be designed in 3 stages..
firstly,the ac voltage from the mains supply must be reduced to 24 V ac voltage by using a transformer
next is converting the 24v ac to unregulated dc by using a bridge rectifier...
then the obtained voltage must be smoothened by connecting a capacitor in parallel
next it is reguleted by using a simple zener diode or by some regulator ics...
the resulting will be a regulated dc supply...:)
 
Thanks Srikanth but if my output is not 24V what can i do to increase it ?use op-amps
But one qn if you dont supply enough current required for the motor will it run?
the motor needs ard 2A. if i supply 1A it wont run right.
 
op-amps cannot be used...coz they also require dc power supply( +Vcc and -Vcc)to operate...if u want to increase the current drawn from the transformer,u hav to buy the transformer with high KVA (kilo-volt ampere) rating...
 
ok. I got 24.5V from the tansformer but secondary current is only 1.1A dunmo why. transformer power rating is 50VA.i will use op-amp at the end if needed. i get lets say 20V i can use a om-amp and increase. A capacitor of 10000uF s very big? does a capacitor increase voltage ?
 
how can u provide power supply to an op-amp?...transformer of 50 VA cannot provide your required current of 2A becoz its not ideal...u still have to increase..
 
its is written as 2A in secondary but does not come out..50/24=2.08A
that is what was written on the transformer. u can provide dc voltage of 24 to an op-amp i guess.
 
opamp also requires a negative power supply...if u can provide +24 volts..u must also provide -24....and moreover how can u use your input to the opamp as its power supply...i think its not possible...the above calculation which u made is just theoritical...most of the transformers we use are practical...so eventhough the specified VA rating is 50...it may just provide some voltage around 30 or 40...
 
oh. its said as 2.08A on the transformer. but i think most of it is lost. yeah op=amp need negative 24v dc as well. i will to connect the bridge recitifier etc and see whether it works.
 
Lol. Even u made it? I guess you just started using this forum and you are already helping me. You are a Electrical Engineering student? Which college you are from?

does capacitor increase voltage?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
+/-24V seems a little high for post op-amps. +/-18 is a more common maximum.

When you rectify AC you create a DC rail that can be as high as the peak voltage of the transformer. This is 1.4 times the RMS value. It also means that the current available is roughly limited to 0.71 of the nominal transformer rating. (1/1.414 = 0.707).

In practice it may not be so simple, but you get the idea.
 
Yeah i am not very sure abt op-amps. after rectification vpeak= sqrt2* V rms and current reduces ?
i need the motor to work and an output of 24V .thats the main objective.
 
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