I don’t have one as that’s not the schematic I worked too......

It does not matter which circuit you are working too, add a test load of say 47R, 1Watt , for a Vout set of +5V, that's ~100mA.

As 'dendad' points out use the correct resistor setting ratios/values.

If you don't have a 47R at 1W, use two 100R's in parallel.

NOTE: keep checking the temperature of the LM317 as you increase Vout above 5V and do not cook the 47R!

 

the circuit and values i chose are proven to work.....

https://makezine.com/projects/0-24-volt-2-amp-bench-top-power-supply/

 

the circuit and values i chose are proven to work.....

https://makezine.com/projects/0-24-volt-2-amp-bench-top-power-supply/

Mike, you really missed out by not taking Eric Gibbs up on his offer of free stuff, He has generously donated it all to my club now and it Is greatly appreciated.

 

the circuit and values i chose are proven to work.....

https://makezine.com/projects/0-24-volt-2-amp-bench-top-power-supply/

You may think it to be so, but it is actually, if you work out the resistor, 100R and 5K will regulate at about 65V.
So the circuit is not correct.
Work it out your self if you like.

 

Yeah i did input that and saw the 65v. Im not being ungrateful, just cant get my head around it... think i need some training!

 

Mike, you really missed out by not taking Eric Gibbs up on his offer of free stuff, He has generously donated it all to my club now and it Is greatly appreciated.

I was going to..... hadn't got around to arranging it... looks like i wont be then.

 

hi,
This is an LTS sim of your circuit.
Note: the 2k5 POT and 1k test load.
E

 

Thanks Eric, are you able to explain what that means?

 

hi,
On the Plot the left hand scale is the voltage output, each of the 6 horizontal represent the Vout for that value of Resistance I have added.
ie: Setting the POT to O ohms gives ~1.5Vout, 500R gives ~8Vout and so on in 500R steps upto 2k5, which gives a Vout of ~30.5V

The 1k Load will have a Iout [current] of Vout/1k, eg: 8Vout/1k = 8mA,, 30v/1k= 30mA.

You must ensure that the PSU has a minimal load current of approx 10mA to ensure that LM317 can regulate the Vout.

 

Thanks Eric i see now, Can i just confirm if the 1k load resistor is needed on the finished circuit or is it just for the test?

Would this Pot be suitable:
http://www.ebay.co.uk/itm/C2K5-2-5K...921945?hash=item19f1621f19:g:sPUAAOSw7NNT2koq

 

hi,
That pot has Logarithmic change in resistance value, used on Volume controls, its not suitable.
You want a Linear resistance change pot.

I would keep a 1k connected as a load while you are testing.

E

 

Is my circuit below correct to translate to PCB?

 

hi,
That pot has Logarithmic change in resistance value, used on Volume controls, its not suitable.
You want a Linear resistance change pot.



E

Any idea where i can get one at a reasonable price?

 

This company has low prices, decent P&P.
https://www.bitsbox.co.uk/index.php...rd=potentiometer&inc_subcat=0&sort=20a&page=1

 

if i understand this correctly i could use a 200 Ohm resistor and a 5k pot of the same effect?

 

 

hi,
Do not use a resistor higher than 220R, the datasheet explains that a minimum current must flow thru that resistor.
You could do it this way, ref image, this will give a max Vout of ~25V
Using a 2k2 pot
https://www.bitsbox.co.uk/index.php?main_page=product_info&cPath=94_104&products_id=816

 

ok thanks,

so the below?

 

Yes, that will give you the Vout range that you have mentioned, ie: 1.5V thru 24v.
IMO D1 and C3 are not required, add a 1K on the Vout output.

You must ensure that you do not exceed the heat dissipation rating of the LM317 over that Vout range.

eg: say you set for 5Vout at say 0.5A, that would mean that across the LM317 you would be dropping about (32v-5v)= 27V at 0.5A , that's a 13.5Watts!!!!

 

thanks eric you are a star!