Power Supply voltage drop issue

[dwn]

I've built a small voltage regulator that I can use with my breadboard or to test circuits. It is the first circuit that I've actually soldered together, so please disregard the hack job. The problem with the circuit is that when I hook up a very small load to the supply, the voltage plummets. A few things I did wrong keeping track of wires and component terminals. I replaced two capacitors that I managed to overload and they swelled up like a frozen soda can. The LM317, I had the ports all mixed up. Instead of looking at the data sheet for the unit, I was referencing the circuit schematic. I had the Vin passing through the Adj pin, the Vout passing from the Vin pin, and the Adj lead (10K pot) hooked up to the Vin. I have to say, I've learned more from this little DIY than about any class I've taken.

When I connect the supply, the voltage readout is fine (about 2V less than the input); however, when a small fan is hooked up, the output drops to ~2 Volts. Could it be the voltage regulator that is no good, my soldering joints (which doesn't really make sense, since the output is fine without a load), or the way I have the circuit connected.

 

[milesyoung]

Your schematic seems a bit odd to me. The LM317 is a three-terminal device, so why are there four connections to it?

Could you clean it up a bit?

Edit:
Is that a node connection "jumping" over the LM317?

What is the input voltage and what load are you connecting?

 

[dwn]

Sorry, , my drawing was not very good--was not planning ahead. You are right, that is the resistor jumping over the LM317. The resistor connects to the adjustor pin and Vout pin. :)

The input voltage is about 18V and when I test the output without a load it reads about ~16V, which from what I have learned, is what you should expect from the LM317. I'm not sure what the exact load is, the fan in the picture was just something I had laying around. When I have the circuit hooked up to the breadboard though, the fan functioned perfectly fine. It required about 5 V to get it started on my breadboard circuit. .

 

[milesyoung]

While unloaded, does it regulate the output as you'd expect when you trim the pot?

I can't really tell much from the pictures you've taken. I'm just assuming that your schematic corresponds to the reference shown in the datasheet, and you've made sure your actual connections follow the schematic. I've abused LM317's plenty without them taking notice.

 

[dwn]

That last bit is good news. The unloaded voltage range is 7.7 to 16.4 V from the ~18V input. I thought the LM317 voltage range was 1.25 to 37 V. Would the 100 Ohm resistor that connect the Adj and Vout ports affect the output voltage?

Do you have any tips on how I can troubleshoot the problem? What generally causes such a drop?

 

[milesyoung]

The unloaded voltage range is 7.7 to 16.4 V from the ~18V input. I thought the LM317 voltage range was 1.25 to 37 V. Would the 100 Ohm resistor that connect the Adj and Vout ports affect the output voltage?

It should go down to the ~1.25V reference if your pot range is 0-10k. What pot is it?

100 ohm is in the low end, but nothing to worry about. It just means the LM317 will draw a bit more current.

What generally causes such a drop?

Typically, if you've connected it right, it means that it's overloaded.

 

[dwn]

It is a linear taper - 500VDC - 0.5W pot.

 

[milesyoung]

If you have a range of resistors lying around, you could try loading it that way going from high to low resistance, and see how it responds.

 

[berkeman]

@dwn -- can you re-post a cleaned-up version of your schematic? Please try to draw it as close as possible to the datasheet suggested schematic. Thanks!

 

[Dax331]

circuit looks fine. you might have trouble setting the capacitors which might affect the feedback of the adjust and make problems with output
the resistor seems well placed in Vout but the pot is not it should be driven to ground through adjust for regulating the current of the output
instead of the voltage. the input capacitor might not be necesaries as a output cap and a diode. adjust resistor should be 0.5w eventually when regulating
voltage not current because the pots might become unusable easily and then provide erratic R and burn circuitry and stuff

 

[berkeman]

voltage not current because the pots might become unusable easily and then provide erratic R and burn circuitry and stuff

Welcome to the PF.

If you mean too little wiper current can cause potentiometer reliability problems, that is correct. That's one reason why I'd like to see a cleaned-up schematic.

 

[dwn]

I tested the load using some resistors (100 - 100K ohm) and the output fluctuated as you might expect as the resistor values went up. I decided to add an LED at the end of the resistor to see if it would light up, but the voltage dropped. I flipped the LED around to see if it might be a polarity issue, to my surprise, the voltage did NOT drop, but the LED still didn't light up.

@Dax, I do have the potentiometer hooked up to ground (pin 1 of the pot).

@berkeman, hopefully this will be easier to read. I was trying to draw the first one from my pcb layout, since I know how confusing it can be to interpret someone else's work.

This seems like a circuit that is too easy and dummy proof. Love proving the world wrong! haha

 

[dwn]

Hey Dax, I was rereading your post and I'm not sure what you mean about the pot. I have pin 2 of the pot (Vin) connected to the Adj pin of the LM317, are you saying that is incorrect?

As far as I understand it, which is not, the input caps smooth out the voltage flow acting as a filter. When I had the circuit on the breadboard, I played around with different size caps and it appeared the voltage reading on my DMM remained at a steadier rate with hight cap values. It is difficult to tell for sure without the proper equipment, but my electronics bible (Practical Electronics For Inventors, Scherz and Monk) said as much. Not sure if its entirely necessary for such a small supply.

 

[dwn]

Could you guys recommend a good brand for electronic measurement equipment, or more specifically oscilloscopes?

 

[dwn]

I think I may have the potentiometer mixed up, as Dax suggested. Looking at the pot from above, the first pin (left to right) is attached to ground, the second pin is attached to the Adj of the regulator, and the third pin is disconnected. Should I have all 3 pins of the pot hooked up?

 

[Dax331]

3 pins of the pot certify you that its resistance will be "mostly" regular through the vector even if some overheating occurs at its path
it supposedly modifies the scale from linear to quasi logarithmic not sure actually. used for full current drops
the ~2 volt drop might be relative to the cooling fan circuitry used for sustained current. it may prove its transistors are old.
to be certain your circuit is working near properly first measure the current. probably 25 to 75mA if the Vout load is working correctly
that's all i know

 

[dwn]

Thank you for the tip. The current is pretty jumpy. I attached a 220 Ohm resistor to test the current and the highest read out was from 2.66 mA down to about 0. The DMM was jumping all over the place.

 

[davenn]

Your main problem is that you are using a 1k resistor from the output pin back to the adj pin
this is way too high and should be around 240 Ohms as per the datasheet

cheers
Dave

 

[dwn]

haha, thank you Dave, I was actually just in the process on switching out that resistor.

Update

I think that did the job Dave. I hooked up the fan and it work, max current was around 75mA.

Just to clarify, basically all the pot is doing on this circuit is lowering its resistance to direct more current its direction and if the resistance on the pot is high (10K) all of the current will be direction to the output. If I were not using a voltage regulator would it be possible to run the supply output from the voltage out terminal of the potentiometer (it just wouldn't be able to handle any irregularities)..?

 

[davenn]

That resistor ( the 240 Ω one) must be connected between the Vout and the adj pins, else the circuit won't work
The regulator provides a 1.25V reference voltage between the output pin and the adj pin. That Vref is used to set a
constant current flow through the voltage divider consisting of the 240 Ω resistor and the variable resistor.

If I were not using a voltage regulator would it be possible to run the supply from the voltage out terminal of the potentiometer (it just wouldn't be able to handle any irregularities)..?

yes you could, but there would be no voltage regulation and the voltage out of the pot would vary depending on the load resistance
Also, a normal pot cannot handle much current, maybe a 100mA or so at most.

those 2 things are the main reason for the use of regulator chips

cheers
Dave

 

[dwn]

The Vout terminal of the potentiometer or the regulator? Please excuse the density.

That all makes sense, I just wanted to make sure I had my ducks in a row.

 

[Dax331]

remember sink is way important beyond 75mA because if the circuitry becomes jumpy or unstable the current scalates immediatly to 1 amp and lm317 starts smoking

 

[davenn]

The Vout terminal of the potentiometer or the regulator? Please excuse the density.

the pot, as you stated you weren't using a regulator

If I were not using a voltage regulator would it be possible to run the supply from the voltage out terminal of the potentiometer

 

[davenn]

remember sink is way important beyond 75mA because if the circuitry becomes jumpy or unstable the current scalates immediatly to 1 amp and lm317 starts smoking

that doesn't really make much sense ?