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24v 2.5 amp power supply......some questions

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New Member
For my Senior Design I have to supply 24v and be able to handle 2.5 amps.
I have a 24vAC 2.5A transformer and a simple circuit with 4 diodes and a 3300uF capacitor......

When I measure the voltage at the output its about 35 volts. I need it to be 24V.

Is there a voltage regulator for 24V 3Amp???? or what would you guys recommend to get it down to 24V volts?

Any advice or suggestions is greatly appreciated.


New Member
hey....thanks for replying

I see there is also a LM350 which is for 3 amps....the LM338K would work, but now I'm kinda confused about the resistors......

What watt resistor should I got with? 25Watt? and when looking at the circuit for the regulator http://www.electro-tech-online.com/custompdfs/2003/02/lm350rev0f.pdf

there are 2 resistors there and I want to be able to put in permanent resistors, not a variable one. Any ideas on what values the resistors should be for 24V? and what wattage resistors?

Thanks again,


New Member
I just realized there is an equation with VRef and the resistors, and in the equation is Iadj.. do I use the current value that the load will pull when operating (2-2.5amps??????) I just want to get this right.


New Member
I think your 24Vac transformer will not able to give you a 24 Vdc at full load... a volatge of 24Vac rectified with a full diode brigde and a big capacitor will give you an ouput voltage of 20-22Vdc...

have you tested it at full load?


You don't need the current value - the Iadj term in the equation is very small, and can be ignored.

So, you end up with:

Vout = 1.25(1 + R2/R1)

As for wattage, the resistors draw very little current (microamps), so you can use 1/8W - 1/4W resistors.


Active Member

yes, you should check that you get your 25 volts
when its delivering 2.5 Amperes.

loosely wire up the diode bridge and your cap,
you say you get about 35 volts, thats about right.

get a 1KW item, such as a 1 bar electric fire,
and stick it on the supply, the voltage will drop
cos that will take between 2.6 to 3 Amperes if you
still have at least 24 volts then you can go ahead.

If this is for battery charging you wont need much
of a capacitor, and the open voltage is a bit high.

If it is to be made into a stabilised supply,
the easiest way is with a heavy current TO5 like
a 3055 on a decent heat sink and a zener.

Ready made stabilised supply circuits are also
available, with very good stabilising, and S/C
protection, useful if its for general use.

Best of luck with it,


New Member
thanks for all the replys.....

This power supply is used to give power to soleniods that control hydrolic pumps. Its a huge garage door basically.

So lets say that get the diodes and capacitor wired up.....put in a LM350 or a LM338K....there are the 2 resistors there......Are those used to get my 24Volts?? Do I have to choose certain ones to get the 24Volts? Do those need to be high wattage resistors because thats where the load will be comming from?

Any help on the resistors will be greatly appreciated.


I assume you are talking about the 2 resistors that are used to set the regulator output voltage. See my last post.

R1 is usually about 240 ohms. Therefore, by calculation, to get 24V, R2 is

(24/1.25 - 1) * 240 = 4368 ohms.

Wattage is not important.


Active Member
For solenoids You don`t need filter cap. and stabilizer, just hook up to bridge rectifier output.


Active Member
yes, as Sebi says, if this is just for the solenoids,
dont worry about stabilising it.
Use it as it is.
Check that with the smallest solenoid operating,
it doesn't get too hot, it will be a bit over the
25 you want, but less than the 35 you're getting.
Sometimes small solenoids will 'buzz' if there is
no smoothing at all, so you may need some capacitor
across it.

If there is other stuff on this supply as well as
the solenoids, it might not like the higher voltage
on 'no load' conditions, sometimes a small load like
an indicator lamp can trim that peak down to a more
reasonable voltage.

With no load, the voltage can gradually assume the
peak of the applied wave form, sometimes a little
higher if the wave form is rough.

Best of luck with it,
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