# Chat_Ghosty's PSU

Status
Not open for further replies.

#### Chat_Ghosty

Jun 30, 2012
49

I'm new to Digital Electronics.

Basically. You Turn S4 and you get 1.5, 3, 6, 9, 12 Volts.
Then Turn R6 and you get +0.1 V. (Up to 1 Volt.)
And R7 to get +0.05 Volts. (Another 1 Volt.)

So 1.5v to 14v.

Iv been told that I can pull about 10mA.

#### davenn

Moderator
Sep 5, 2009
14,040
it wont, the opamp based system wont do that

you need to use a proper voltage regulator

the LM317 is good for ~ 1 Amp (TO 220 style case) ~ 2 Amp (TO3 style case)
best would be the LM338 (TO3 case) which can do up to 5 amp.
do a google search on the LM338 and it has circuit examples in the pdf datasheet

Dave

#### Chat_Ghosty

Jun 30, 2012
49
I found pdf of the LM338.

I looked in MultiSim, and what I think is the closest is LM318 and LM339.

But I think the LM317 (2A) or LM338 will work.

#### GonzoEngineer

Dec 2, 2011
321
The LM318 is an operational amplifier, and the LM339 is a comparator.

You need to look for Voltage Regulators.

#### Chat_Ghosty

Jun 30, 2012
49
Thank you Gonzo Engineer.

I found the LM117HV in MultiSim.

Based off a pdf it can do 1.2 to 57V and 1.5A

Comes in TO3 and TO39

I think I'm going to use the TO39. I don't know how I feel about the Output being the metal Case.

Ill redesign my Circuit and report.

#### davenn

Moderator
Sep 5, 2009
14,040
follow my info in my forst post else you will end up letting the smoke out

GET a LM338 TO3 case style it WILL handle your current requirements
FORGET about multisim, there are, as I said earlier, plently of sample circuits in the pdf datasheet
the TO39 is not a high current style

Dave

#### Chat_Ghosty

Jun 30, 2012
49
I found a Nice pdf and Diagram of the LM338.

LM338 vs. LM138?

I'm having a Hard Time understanding how to set up the resisters to Change the Output Voltage.

What I'm getting from the pdf is that the LM388 Generates 1.25 v from Output and Adjust.

Vo = Vref ( 1 + (R1 / R2)) + I adj R2

Vo = 12v
R1 = 1
R2 = 1

12 v = (1.25)(1+(1/1)) + Iadj(1)

I'm confused on what I*adj means.

Last edited:

#### davenn

Moderator
Sep 5, 2009
14,040
Ummm a LM388 is a audio amplifier

try again LM338 .... or did you just typo ?

Dave

#### Chat_Ghosty

Jun 30, 2012
49
Ummm a LM388 is a audio amplifier

try again LM338 .... or did you just typo ?

Dave

As of the LM338. I looked on http://www.mouser.com to check Prices. And all I could find is the LM138K. And that about $14.50 W/ Heat sink. Is the LM138K the same as the LM338 CO3? I ask because the .pdf I read had LM138K / LM338 Last edited: #### davenn Moderator Sep 5, 2009 14,040 Mouser do list the LM338K ~$12 to 13

I cant remember how much i paid for the last ones I bought
my local ( in Australia) list them for \$14.40 so pretty much the same as in the USA

They really are worth the money tho, very easy to use and make either a switched variable PSU or use a potentiometer and make a smoothly variable PSU

Dave

#### Chat_Ghosty

Jun 30, 2012
49
Ok. Great. Just making sure we are on the Same page.

You said I could add a Potentiometer to make a Smooth Variable PSU.

I'm in need of something More Specific,

If you refer back to my diagram. There is a 5 Position Rotatory Switch that selects the Main Voltage. 1.5v, 3.0v, 6.0v, 9.0v, 12.0v.
And the fine tunes it by +0.1v and +0.05v.

I need something I can fine tune.

Say 5.25v. and Draw a Max current of 2.55.

I need something like this so when I start building projects. Then I know I have what I need.

BTW. I have OCD.

For Bonus Points. A LED basic Multimeter With a Voltage and Current Display.

#### Chat_Ghosty

Jun 30, 2012
49
After pouring me some coffee. (And not drinking any). I think the Answer hit.

Could I use a LM338 as The Main Voltage regulator, and then Two more to fine tune it?

Could I pull that off by using the Rotatory Switch with Preset Resisters then a Potentiometer? Or should I drink my Coffee??

OR

Could I some how, some way. Use the OP Amp to set the Voltage and then use the LM338 to pull my Current??

Last edited:

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
There are a number of problems with that schematic. I'd be interested to know where you got it from.

The second 9V battery, labelled V3, that provides the -9V rail for pin 11 is not needed because the LM324 is single-supply-compatible - pin 11 can be connected to the ground rail of the circuit. This change will save you a lot of hassle.

The "1V battery" labelled V1, connected to pin 3, is not a practical suggestion; it simply means that pin 3 should be provided with a 1V reference voltage. In practice it should be replaced by a voltage reference such as a TL431 (Texas Instruments, also second-sourced by various other companies with a different alphabetic prefix but the same number, 431). This device's voltage is 2.5V (nominal) so you'll need to use a voltage divider to reduce this voltage, because with this design, the voltage on pin 3 must be at least somewhat lower than the lowest output voltage you need. You can use a divider of 15K and 10K to drop 2.5V down to 1V as specified in the original diagram. The TL431 needs a current source, normally just provided from the positive supply rail through a resistor. Get a data sheet from www.ti.com or Digikey.

The circuit can only deliver around 10 mA because the op-amp output is unbuffered, and small op-amps like the LM324 aren't designed for high current output. You can add a transistor or MOSFET to increase the current output capability. The simplest option is a power Darlington NPN like the TIP120/122. Connect it between the op-amp output and the rest of the circuit, with the op-amp output to the base, emitter to the rest of the circuit, and collector to the main positive supply rail. It will need a heatsink. Power dissipation in the transistor is equal to the voltage across it (in volts) multiplied by the current through it (in amps). Maximum power dissipation will be when it's delivering 1.5V at 2.5A. If your input supply rail is 18V, voltage across the transistor will be 16.5V and current will be 2.5A so dissipation will be around 40W. To keep the transistor temperature below say 100 degrees Celsius at an ambient temperature of 25 degrees Celsius, you'll need a heatsink with a thermal resistance of about 2 degrees Celsius per watt (75/40). This is not a small heatsink. A switching power supply would be a good way to reduce power dissipation, since they are much more efficient than a linear regulator, especially when the output voltage range is wide. You really need to pay attention to layout when using them, though, and ideally use a printed circuit board. It might be a good idea to look for switching supply boards already made up - these are available from Chinese retailers and on eBay I think. The changes to add the features you need would be pretty minor, and these supplies have built-in current limiting.

The circuit has no current limiting. If the output is shorted out, the circuit is not designed specifically to limit the current it tries to deliver. With no buffer on the op-amp output, the current was supplied by the op-amp itself, which will limit it to around 20 mA (give or take), and a shorted output will make the LM324 get hot and possibly lose its magic smoke. If you add a buffer transistor, and use a power source that can deliver 2.5 amps or more, a shorted output could cause a heavy and uncontrolled current to flow, unless you provide some kind of current limiting. There are various ways to implement current limiting. I (or someone else here) will to into detail later, since I don't have that much time now.

Both of the potentiometer values are wrong. The resistances in the feedback circuit (all the resistances in the original schematic) are chosen so that when the output is at the voltage you want, the voltage on pin 2 (which is set by those resistances) will be equal to the voltage on pin 3, which is 1V from the voltage reference and voltage divider.
Because R9 is 1 kilohm, the current in R9 is 1 mA (I = V / R; V is 1V and R is 1000 ohms so I is 1 mA), and every 1 kilohm of feedback resistance will add 1V to the 1V output "base" voltage.
The five suggested values for the feedback resistors (R1~5) are close to the calculated values but the exact values are a bit strange. I don't know why the designer specified a 10977 ohm resistor; you won't be able to buy one! It would be simplest to make up the exact values by putting resistors in series and/or parallel. Also instead of having a separate resistor for each voltage, it's simpler to put the resistors in a chain, and connect the rotary switch to different places in the chain according to the desired output voltage. This means that you'll only need three different resistor values, instead of five.
If the potentiometers are supposed to add 0.1V and 0.05V then their values should be 100 ohms and 50 ohms respectively. These are at the low end of the spectrum for carbon potentiometers. You could scale up all the resistor values in the feedback circuit by a factor of ten and use 1 kilohm and 500 ohm potentiometers.
Potentiometers don't have a very accurate end-to-end resistance; typically it's around +/- 20%. If you don't mind that they don't add exactly the desired amount of voltage, then that's fine. If you do, there are ways to fix this problem that I can get into later.

Finally, you should make sure that the rotary switch is a shorting type. That is, as you turn it, the common contact connects to the next position before it disconnects from the previous position. If there is a gap during which the common contact doesn't connect to anything, the circuit's output voltage will jump up to the maximum possible voltage (about a volt less than the input supply voltage) as you rotate the rotary switch. If a load is connected at the time, it could be damaged by the high voltage glitch. Luckily, almost all rotary switches are the shorting type, but you should make a note on the schematic because it is a requirement of the circuit.

I would be happy to help with the circuit. Can you describe what you want to use it for, what you're planning to use as the input power source, and what constraints you have. Also please consider using a pre-made switching power supply (you can get these rated for 3A output, and they're not expensive) since they can probably be modified pretty easily to do what you want. Go for one that is already adjustable.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
No, simply an LM338 and switching resistors (or possibly trimpots) will probably do what you want.

#### Chat_Ghosty

Jun 30, 2012
49
Wow. KrisBlueNZ. That is a lot to Digest.

I created the schematic by redesigning a Audio Amplifier.
As of right now I don't have much money to spend on a Project.

But.. The Goal of the PSU is to Power just about any Project I might make in the Future. Even use as a Battery Charger if needed.

Ultimately I would like a PSU that I can regulate both Voltage and Current in case I need to do specific tests.

As of the resistances being funny, I used MultiSim and the Multimeter to test and find the Resister that made the Voltage Perfect.

I have thought about setting up a "Chain" of Resisters and I think I would also need Diodes?

As of a Pre-built PSU. All I have is Computer PSU. That do +12V at 6A and +5v at 18A.

Ill have to re-read what you said and let it Digest some more.

Great Job!

Last edited:

#### Chat_Ghosty

Jun 30, 2012
49
I cracked open the PSU and I found 4 Heat sinked Voltage Regulators and 2 Not.
Or at least that's what I think they are.
I'm guessing that they are, that would explain the 6 Voltages.
The not heatsink I can guess are the Negative Voltages.

I also found two 470uf Caps and a bunch of Mini Caps.

Everything else I have no idea on. I'm looking at the Post Steve gave now.

Last edited:

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
You don't need diodes with the chain of resistors. From the feedback point (pin 2 of the LM324) you start with the first resistor, 500 ohms (for 1.5V); at the far end of it, you connect the first position of the rotary switch, and the second resistor, which represents the difference between the first and second rotary switch positions (3V - 1.5V, so use. 1.5 kilohms); at the end of that you connect the second position of the rotary switch and the next resistor, which is 6V - 3V so use 3 kilohms; and so on for the next two resistors, which are also 3 kilohms because the difference in voltage between position n and position n+1 is 3V from that position onwards.

Resistor values in a circuit are normally calculated; measurement with a multimeter is not involved. Assuming R9 is 1K and the reference voltage is 1V, the feedback resistances are 1K for every 1V of output voltage above the reference voltage. That's how you should decide on the values. You can then look at how to achieve the calculated value - either by combining two or more resistors, or by choosing a close value from the E24, E48 or E192 preferred value set.

I really think you should use a switching regulator instead of a linear regulator if you want several amps and an output voltage that's adjustable over a wide range. They do produce noise on the output voltage, but this isn't an issue for most circuits. You can add an L-C filter to reduce the noise. It's not so easy to add current limiting to a switching regulator though, especially variable current limit.

So you want to use a computer power supply to power your circuit? If you want up to 12V from your circuit, you'll need a few volts more at the input, say 16~18V. You would probably need to rewind the transformer in your computer power supply for this, which is not trivial and potentially dangerous, and you would have to replace the output capacitors and possibly the output diode. I think you would be able to change the feedback circuit inside the power supply to give different output voltages directly (i.e. you wouldn't need a separate variable regulator after the power supply) but you would need to experiment. It's not simple. There are articles available on rewinding transformers in PC power supplies; try some Googling if you're keen, but follow the safety instructions carefully and don't expect everything to go smoothly!

If you want to charge batteries, you'll need more voltage flexibility. NiMH cells are nominally 1.2V not 1.5V so you'll want more voltage output options, but you'll also need a way to control the current, and ideally some timing system so you can leave it unattended. You should also have a temperature measurement system to detect end-of-charge (to prevent overcharging) and possibly to compensate for ambient temperature. This really needs a microcontroller, which requires programming, to get decent performance and avoid damaging the battery. It's not easy to do it properly.

Re your PC power supply. The four heatsinked components are probably double diodes, even though they look like three-terminal regulators. PC power supplies don't use three-terminal regulators for the high-current output rails because of the current. If you're keen and have time to spare, try to trace out the circuit diagram of the power supply, especially the output side.

#### Chat_Ghosty

Jun 30, 2012
49
Well.

I have no Idea what the difference between Switching and Linear Regulator.

But I did pop open a old PSU and I found something very nice inside.
I found two Potentiometer.
One control both 5v and 12v output.

4.8v to 5.5v
10.7v to 12.4v

No effect on Negative Voltages.
And the other does nothing.

I tried and connected the 12v and 5v Rails. And the PSU wont start. Hoping I could bridge to about 18v is out.

Ill keep looking and see what I can find.

#### Chat_Ghosty

Jun 30, 2012
49
After playing around with Two PSU. I was able to Bridge them and get a Output of 24v.

Due to the Potentiometer in one of them, I can the turn the Voltage down to 22.5v.

You said that I need 16v - 18v if I want 12v.

I..ummm... Found something interesting. I found Two SCSI CD Drives. And each has its own PSU. I pulled one out and it's really small. I tested some Voltage. And to my surprise, My Meter is reading about 18v. It keeps going up.

After adding some load. It now reads: about 12.2v. The Drive it's self says it runs on 5v @ 1A and 12v @ 0.7A. But the PSU being internal. I have no idea on what it is rated,

I figured out Two things.

1. I now have a 12v @ atleast 0.7A in a Small Form Factor W/O a Fan. (Nothing I can't add.)
2. Metal can cut you and make you bleed.

Last edited:

#### davenn

Moderator
Sep 5, 2009
14,040
There are a number of problems with that schematic. I'd be interested to know where you got it from....................

we have moved on and bypassed that circuit Kris

a 338 will do the job easily with a very small component count

as far as coarse and fine adjustment goes, there is a way to use a dual (stereo pot) and a single pot that will do that nicely. if some one else doesnt I will dig my circuit up asap....
I have to get off to work monday morn rolls around again

now if you use a multi-turn pot as shown then you can adj quite finely.

Dave

#### Attachments

• PSU1.gif
11.5 KB · Views: 1,418
Last edited:
Status
Not open for further replies.

Replies
3
Views
565
Replies
4
Views
629
Replies
8
Views
1K
Replies
2
Views
1K
Replies
10
Views
2K