Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

I need an adjustable power supply

Status
Not open for further replies.

Tipsy

New Member
For my first project as detailed in another thread, I need a power supply that can simultaneously and within the same circuit, provide both 12-14V and 0-5V

I have an old mains adapter that converts 240Vac to 18Vdc @ 400mA, would this be a suitable starting point to save me buying/building one from scratch?

This is the output end after I removed the battery holder casing. It was originally used to recharge a cordless drill's battery. It appears the manufacturers bought in a ready-made adapter, complete with the common round plug (seen plugged into a socket on the far left) and added a small diode protected circuit with LED to the positive side.

**broken link removed**

**broken link removed** **broken link removed**

Assuming I've got the ok to use this adapter, my first question is; what is the function(s) of the components on the additional circuit and would it be useful for my power supply or should it be removed?
 

MikeMl

Well-Known Member
Most Helpful Member
How much current do you need at the two voltages? LM7812, LM7805, LM317 will do it, but sizing the filter capacitor ahead of the regulators and heatsinking requirements are your unknowns...
 

k7elp60

Active Member
The adapter would be suitable as long as you do not need more than 400mA for the two power supplies. The circuit board with the resistors,LED, and the diode appears to be a indicator that lights up when the battery was being charged. You could use just the LED with a higher value of a resistor for a power on indicator for the new power supply.
I would use a LM317 as the adjustable regulator and a 7805 as the 5V regulator.
 

Mr RB

Well-Known Member
Hi again Tipsy. Since you only need low power for this (based on your other thread) here are some suggestions;

You could use a 7812 regulator, these are a very simple 3 device; 18v in, ground and 12v out are the 3 pins. Google for a datasheet which will tell you all about it.

Then for the adjustable voltage you can use a LM317, this is again a simple 3 pin voltage regulator but this one just needs one resistor and one pot, and makes a variable DC voltage power supply where the pot adjusts the DC voltage that comes out.
Datasheet is available for that one too.

So those 2 simple devices and a few caps are about all you need to build your low powered dual output PSU. In the event that you want to upgrade your PSU to use in future projects, these parts are not wasted you can use exactly the same system and just replace your 18v 400mA supply with a (example) 15v 1.5 amp supply, which will give you a useful PSU for many other projects.
 

Tipsy

New Member
MikeMl, I'm confident 0.4A will be sufficient for my first project as Mr RB advises. Even if not, the experience I gain will be worth the effort of building this IMO.

Hopefully some LM317's are winging their way to me as we speak and when they arrive I shall try and absorb their datasheet.

k7elp60, would you be kind enough to elaborate more about the additional circuit pictured above? I recall when using the device as intended, the LED would glow when charging, as you rightly state. However, when no load is present, the LED remained off. Is it simply because the load completes the circuit or does it require a large enough current to be passing through before it will light? What is the purpose of the large resistor that measures 37ohms on my cheap digimeter?

Mr RB, that sounds like a plan but for possible future use, would it not be more preferable to make both outputs variable?
 

Mr RB

Well-Known Member
...
Mr RB, that sounds like a plan but for possible future use, would it not be more preferable to make both outputs variable?

Sure, what I suggested was just an expansion on what the 2 other people had already said and using the fixed 12v regulator was the simplest option.

But if you don't mind adding that little bit of complexity then 2 LM317s will give you 2 variable power supplies.

Since you have now ordered LM317s and you are going to build a power supply with them you might want to look at some simple LM317 PSU schematics;
http://home.alphalink.com.au/~parkerp/projects/psu.gif
**broken link removed**
Telephone In-Use Relay, LM317 Regulators, Lithium Charger

And also if you have a spare LM317 it only needs one resistor to operate the extra LM317 as a current limiter;
Constant current source with LM317
LM317 / LM338 / LM350 Voltage and Current Regulator Calculators
 

Tipsy

New Member
Sure, what I suggested was just an expansion on what the 2 other people had already said and using the fixed 12v regulator was the simplest option.
As I hoped, thank you. It is helpful to keep things simple (especially for me) to begin with. I appreciate that intention.

But if you don't mind adding that little bit of complexity then 2 LM317s will give you 2 variable power supplies.
My thinking is if I can make one work, the second should be even easier - as I found it more economical to order five LM317s.

I printed off and then read their datasheet over lunch but it became apparent not all datasheets are made equal. Some seem to expand greatly and give many examples of circuits the components can be used for. Unfortunately, the datasheet I printed was pretty basic.....

but these should help fill in the blanks :)

Thank you!
 

k7elp60

Active Member
k7elp60 said:
The large resistor has a voltage drop when there was a load on the circuit. I would estimate somewhere between 4 and 5 volts. The LED has about a 2V drop, the diode about 1 volt and the small resistor the voltage would vary depending upon the current for the LED. I would think 1 to 2 volts.
The circuit was designed to do exactly as you said. When charging the LED comes on and when there is no charging the LED goes off.
The resistor color code for the large resistor is hard to tell, But it could be 5.1Ω or 51Ω when it was new.
 

Tipsy

New Member
Thank you all for your kind assistance so far. I'm pleased to say I've found the time to begin this project and have already hit a snag. Following one circuit, it states a resistor should be 1/4 Watt so I googled resistor wattage and learnt, for my 18V, I may require up to 1.5Watt.

I cannot find how to determine a resistor's Watt rating - I assume the "standard" resistors I have collected are of the lower 1/8 or 1/4W. How can I tell and more importantly be sure I obtain the correct Wattage for my circuit?

Secondly, as I require two different voltage outputs from the same source, is it ok to just make the two seperate voltage regulator circuits and connect their inputs in parallel to the source or can they share some components like the Diode and Capacitors?
 

k7elp60

Active Member
Thank you all for your kind assistance so far. I'm pleased to say I've found the time to begin this project and have already hit a snag. Following one circuit, it states a resistor should be 1/4 Watt so I googled resistor wattage and learnt, for my 18V, I may require up to 1.5Watt.

I cannot find how to determine a resistor's Watt rating - I assume the "standard" resistors I have collected are of the lower 1/8 or 1/4W. How can I tell and more importantly be sure I obtain the correct Wattage for my circuit?

Secondly, as I require two different voltage outputs from the same source, is it ok to just make the two seperate voltage regulator circuits and connect their inputs in parallel to the source or can they share some components like the Diode and Capacitors?
1. If you post your schematic and ask which resistors wattage rating is in question we can help.

2. Yes you can.
 

Tipsy

New Member
My bad, I "assumed" from the above linked schematics, the adjustable power regulator circuits were pretty much all the same.

I'm going with a 5K pot and a ~220ohm resistor ala this schematic

**broken link removed**

However, the pinouts on the LM317 appear to contradict the datasheet I have, that is the gnd (adjust) is the centre leg. Which also leads me to ask, are the input and output interchangeable as even the datasheet doesn't clarify (for me) which side is facing up meaning the input could be the left or right leg. :confused:
 
Last edited:

k7elp60

Active Member
Your diagram has two errors that I see. The 220 ohm resistor is to high a value it should be 120 Ω. If you have a constant load on the regulator of greater than 10mA the 220Ω resistor would be fine, but if you are using it and there are times when there won't be a load that high of resistor could cause some instability. The 220 is okey for the LM117.

The other error is the pin connections. Take a look at the attached PDF page 2. The input/output pins cannot be interchanged.
 

Attachments

  • LM317.pdf
    105 KB · Views: 377

Tipsy

New Member
Seems I'm making all the schoolboy errors on this project. I was looking at the wrong datasheet as the pinouts I quoted were for the LM78xx series. Thank you for posting the correct sheet, you saved me from another error.
 

jeryruse

New Member
Well a small problem arise when I attached a
rheostat to my circuit. After connecting it I
came to know that the range of adjustable
resistance is lesser than what was mentioned
in it's packet. Hence a malfunctioning occured
in my circuit, I unable to make it work properly
again. What to do now ?
:confused:
 

Tipsy

New Member
Wow, I'm so pleased with all the help I've had - it's allowed me to make what I wanted. I know it's nothing great to most of you but to me it's something to get excited about - my circuit appears to be working! Thank you.

I couldn't help myself stick a digimeter on the outputs and power it up without adding the diodes but they will be part of the finished article.

The components in use are one 1000uf cap across the inputs and one 10uf across the outputs, a 5K pot between adjust and -ve source, and a pair of 55ohm resistors (in series) between the +ve output and adjust. The schematic I worked from actually has a 1uf capacitor across the outputs - is 10uf ok?

I read in the datasheet the output voltage can rise if the load reduces. Would this explain why I got a maximum reading of 21v (the source adpater is rated at 18v)? The minimum being 0.6v which excels my expectations.

For my next test I was thinking of using a motor as a load but think it best I add the diodes first. One more thing, I know a nice fat dial on the POT will allow finer adjustment but I feel it's too sensitive. Can this be improved on or should I wait to see how the load tests go?

EDIT: I just realised the 1000uf is only rated at 16v. I assume it'll die a horrible death at high load? I only have 100uf (or lower) rated at 25V (or more). Do I need to buy some beefier ones and if so, which, as I might as well get a selection?
 
Last edited:

Dalaran

New Member
@110Ohms between Vout and Vadj you need (I'm ballparking) ~2kOhm to get your 18V. With your 5kOhm pot which is probably only a 3/4 turn you have just over a 1/4 turn to go between around 1.25V and 18V. I am using just a 12V wallwart, 100Ohms between Vout and Vadj, and a 1k and 100Ohm pots in series between Vadj and ground. This way I can still get up to ~12V and have a course and fine adjustment knob. The 1k almost maxes the voltage at its max resistance, so I get the whole 3/4 turn precision out of it.

You will see a larger voltage than what it might be rated for since wall-warts can do this when the load the are made for is not applied (I'm assuming you had no load).

Cheers.
 

Tipsy

New Member
Yes, I was measuring no load and today, after adding two diodes, I measured it under load using a small pc fan. The previous max reading of 21v has now shrunk to a mere 4.6v. As you advise, I'll experiment with resistance changes next.

Aside from that, I've been trying to understand the role of the capacitors as I mentioned earlier, I'd like to know which values I can use and the differences they may make to my circuit. I've now changed the 16v to a 25v (100uf).
 

Dalaran

New Member
4.6V is the max output? You are using a 18V wall-wart right? How many amps is it rated at? You should still be able to get 18V out of it. Did you put the diodes in the correct orientation?
 

Tipsy

New Member
Good call! The diodes were the right way round - sort of - it was the regulator pinouts I got mixed up. :eek:

Starting all over again, and swapping the 1000uf 16v with a 100uf 25v I'm now seeing a range of 1.2v to a higher 27.7v unloaded and loading it (with a small 12v light doesn't change the output voltage noticeably. If I can find it, I've got a 12v motor I can use to load it up more.

If there's any more testing you think I should conduct, please advise otherwise I'll see if I can create a schematic of my circuit as I've made it so I can discuss my next steps.

By wall wart, I guess you mean what we call a mains adapter - converts mains AC to DC. The one I volunteered for this project is rated as 18vdc @400mA output.
 

Mr RB

Well-Known Member
You're doing good there Tipsy! Don't sweat that occasional "whoops I got the legs wrong" thing. I don't think we ever stop doing that. :D

Depending on your budget you might want to put a couple of panel meters on your PSU, personally I think they are indispensable, especially the current meter, here's an example;

**broken link removed**

I suggest NOT using digital panel meters, they don't have needles so you can't see the needles moving up and down when things happen. Which is REALLY handy.
 
Status
Not open for further replies.

Latest threads

Top