Odd output from Elenco XP-15K Power Supply

[alsetalokin4017]

Before taking it all apart, let's first make sure it's put together right in the first place!

I'm sorry but there are more than just the one solder joint that look questionable to me. If this were on my bench the first thing I'd do is to go back over _every_ solder joint with some good leaded solder, flux and a properly heated and tinned iron. Then inspect carefully for solder bridges (I think I see one or two) and correct them.

[ArthurDent]

I believe alsetalokin4017 is correct that there is a problem with the soldering and you should reread the section of the assembly manual on soldering (shown below) and check each solder joint. Reheat with solder with flux and make sure that the solder freely flows to both the component lead and the pad on the circuit board. As previously mentioned, also check that there are no solder bridges shorting runs together. If you're lucky, nothing will have been damaged and resoldering will cure the problem.

[tomherrick]

I suspect your suspicions are correct. I had one heck of a time soldering this project. I picked up a 25W Weller soldering iron and it took forever to heat the wire and board. I used a 250W gun on the pot leads and those worked well, but it's a tad clunky for the small stuff. Plus, it's been over 50 years since I did any electronic soldering...

[tomherrick]

Well, whaddya know???  'Twas the soldering - or lack of.  The output is, however, a tad on the high side. The range is supposed to be from 0 to 15VDC. Perhaps a resistor is a little toasty?

[rf+tech]

Congratulations 

As for the few extra volts, you deserve them 

Limiting the output to 15 Volts would require adding a 10 kOhm resistor in parallel with the pot, to reduce the effective resistance to 1.65 kOhms.

That's up to you, whether to pursue the extra effort.

RF+ Tech

[ArthurDent]

Great that you got your power supply kit working! Think of this minor problem as a positive learning experience.  Future kits you assemble will seem easier now that you've had success with your first one.

One change that has been made from when you (and I) soldered all those years ago is the use of lead free solder. If you read the construction section of your power supply manual on soldering they mention how the new lead free solder requires higher temperatures and doesn't flow as well. This obviously made your soldering job more difficult. There is a possibility that your small 25W iron doesn't get quite hot enough for lead free solder and that's why the higher temperature of your soldering gun melted the solder better even though it is physically way too big. Even if the iron gets to the required temperature you probably have to leave the tip on the spot you're soldering a few seconds longer to allow the solder to flow properly.

Here is a link to a short video that compares soldering with lead free vs lead solder and has good information. Just remember that there are videos on everything you can think of and some can be very helpful so Google is your friend. Members on this forum are always ready and willing to help as well. One thing that might help is to get some old scrap circuit board to practice soldering on. sometimes you will find a thin layer of oxide or crud on a circuit board and lightly cleaning the board with a Scotch Brite pad will help the solder adhere.

 

[tomherrick]

I think somebody else needs to own that 25W iron... I'll find another higher wattage unit that is more nimble around a circuit board than the heavy-watt gun. Oh, and I'll watch that video at lunch today. Thanks.

I do want to limit the output to 15VDC, however, as the reason for acquiring the power supply is to test and set up marine electronics off the boat in the shop. 15VDC is the top end of the range for the Garmin chartplotter we acquired, and I'd expect that's about it for most nominal 12V equipment.

I tested all the resistors: R1-2.1k, R2-147, R4-2.2. R3, the 2K pot, when turned all the way to the right at the 15V setting reads 2.38k. When turned to the left at the 0V setting it reads 0.00 ohms. I'm not getting something here. Why would a value of 0 ohms totally block the voltage output and 2.38k ohms allow almost 20VDC?

rf+tech mentioned: "Limiting the output to 15 Volts would require adding a 10 kOhm resistor in parallel with the pot, to reduce the effective resistance to 1.65 kOhms." Looking at the schematic I'm not sure where such a resistor would go to be parallel with the pot.

[rf+tech]

The extra resistor would be connected across the two ends of the potentiometer. As supplied, a 2 kOhm pot has too much resistance to limit the output to 15 Volts.

The total value of the paralleled resistor/potentiometer is derived from the LM317 datasheet formula:

V_out = 1.25V x (1 + (R_pot/R1)) + I_adj x R_pot

The last term compensates for the current flow from the adjust terminal, which is small enough to ignore.

So the simplified formula reduces to:

V_out = 1.25V x (1 + (R_pot/R1))

Substituting 147 Ohms for R1 and solving for R_pot, we arrive at 1610 Ohms, which limits V_out to 15 Volts.

By adding a resistance in parallel with the potentiometer, we can "trim" the maximum value.

Take the reciprocal of 1610 Ohms, subtract the reciprocal of 2380 Ohms, then find the reciprocal of the result.

That value represents the amount of parallel resistance to reduce the total to the calculated value of 1610 Ohms.

As for the result of that calculation, I'll leave that as an exercise for the student. 

Do post your result.

RF+ Tech

Edited for better clarity

[tomherrick]

Which would be the two ends of the pot? There are three contacts, two appear to be from P1 and P3 by various routes, and one is an output leading to P5.

[rstofer]

The center contact is usually the wiper.

Connect your Ohmmeter across the two outer pins and note that the value doesn't change when you turn the shaft.  Disconnected from the PCB, of course.

https://components101.com/potentiometer

[tomherrick]

It appears that on this pot the "wiper" is in position 1 as shown in the linked diagram. When I put a DMM lead on 1 and 2 or 1 and 3 I got the same kind of variable readings.

[ArthurDent]

The pot ends are the outside terminals and the center terminal is the wiper as shown in the attached diagram. If you measure the resistance between the outside terminals when the pot isn't connected to the board, the resistance will be always be 2K no matter how the shaft or knob is turned. However, the way this pot is wired in the circuit it is as a two terminal variable resistor so when it is on the board the resistance between the middle and bottom terminal (on the schematic) will vary from zero to 2K as the shaft and knob is turned. The middle and top terminal are wired together and the resistance between those two terminals will always be zero. 

If you haven't cut the protruding pins off the pot, wire the 10K resistor to the outside pins. 

[rf+tech]

With the wiper externally connected to one end of the pot via a copper track on the circuit board, as shown in the Elenco schematic, it is easy to come to an incorrect pin identification.

Visualize the wiper rotating with the shaft, through about 300° or rotation. One should see that the simplest construction technique would have the wiper connected to the middle terminal.

For clarity, please refer to this Wikipedia entry:

https://en.wikipedia.org/wiki/Potentiometer

On the right side of the page, an interior drawing and photo show how this type of potentiometer is constructed.

For the application in this power supply, connecting the extra resistor across the two outer-most pins is the foolproof method.

Attempting to connect the resistor between the wiper and one end has a 50% chance of not working as expected - because the resistor would be shorted out by the external connection between the wiper one end.

And by the way, if the wiper were to be externally connected to the opposite end, zero Volts would be full CW and maximum output would be at full CCW.

@ArthurDent:

With the pot measuring 2.38 k and 147 Ω, the parallel resistor value ends up less than 10 k. It's a bit far from a standard value, so the OP will likely ask, and learn more by the exercise of solving for the optimum value. Please do check me on the math.

RF+ Tech

[ArthurDent]

One thing that hasn't been mentioned is that your intended use of this power supply is for testing marine electronics. I randomly checked the specs on a Hummingbird ChartPlotter and that draws 615Ma. I suspect that other pieces of equipment might require a power supply with higher current capacity. The power supply you've put together is rated at 200Ma (.2A) at 15 volts so it could not be used for testing this particular chartplotter.

What would be a good idea is to check the manuals on the pieces of equipment you want to test power on your bench to find what amount of current they need and get a power supply with an output current capacity that is greater than the highest current drawn.  There are some supplies that output 13.8VDC at several amps that are designed to power mobile equipment.

[tomherrick]

OK, well, thanks to all for your time and assistance. I've spent too much time on this already. Apparently misread the spec as I thought it was 2A at 15VDC.

[ArthurDent]

tomherrick - You should still look at this as a positive experience as you did learn from it (and no one got injured!). We all have these little oversights and a while ago I was working on a voltage reference that I had powered on for months allowing it to burn-in when I went to make a change and forgot to remove the power and destroyed a $60 IC.

A good place to find a suitable supply would be Craigslist or Facebook market place and look for a supply that hams might use for powering mobile equipment. Most of these have 13.8VDC output and some even have a cigarette lighter socket as well as binding posts to power equipment. They generally range from a few amps up to 20 amps or more and could start at $25 and up to $100 for really big ones.

[tomherrick]

The thing that really bugs me is that I did determine the amperage of the equipment this was supposed to power. So I looked for a 12VDC 2A power supply. I probably misread or there was probably-not a typo. Simple goof on my part. Thanks again to everyone who tried to help me work out this issue. Damn nice forum you have here.

T