Power supply voltage is creeping

I built a LM317/337 bipolar power supply for circuit experimenting a while back. I fried the 317 (positive) voltage regulators not long ago by accidentally connecting something wrong, and replaced it.

Now I've got the negative side creeping down. By that I mean that I have it set to 10V, and over the course of a couple hours the voltage slowly rises (in the negative direction). The first time it happened (Tuesday) I couldn't understand why my circuit was behaving the way it was, then discovered the negative side was around 18V. Yesterday it started at 10V, and I kept turning the knob down as it crept to keep it at 10V, and by the time I turned it off a couple hours later, the knob was at the 5V mark.

Think the negative voltage regulator is going bad? I've got a spare, but is there something that might have caused this?
 

tomchr

Member
Paid Member
2009-02-11 12:58 am
Calgary
www.neurochrome.com
Measure the voltage difference between OUT and ADJ on the LM317/LM337. It should be rock solid at 1.25 V (-1.25 V for the LM337). If that's not the case, the regulator is either dead or not wired correctly. This assumes the input voltage to the regulator is within spec (i.e. min 2.5 V higher than the output voltage for the LM317; 2.5 V lower for the LM337).

~Tom
 
Measure the voltage difference between OUT and ADJ on the LM317/LM337. It should be rock solid at 1.25 V (-1.25 V for the LM337). If that's not the case, the regulator is either dead or not wired correctly. This assumes the input voltage to the regulator is within spec (i.e. min 2.5 V higher than the output voltage for the LM317; 2.5 V lower for the LM337).

Voltage between the OUT and ADJ is about -1.27V. For comparison I also checked the positive side, and it was about 1.25V (give or take a few mV on both sides). Input voltage is WELL above output.

Unfortunately while checking the input voltage, I think I must have accidentally bridged the input and output pins with the probe, and released the magic smoke from the output capacitor. I guess this is a game of "see how many times I can burn up components on this before I just go buy a power supply."

Is it getting hot? This will change the current at the adjustment pin. If your resistors/pot are too high in value then this may change the voltage.

I don't know, I haven't checked. Nothing I could notice just working next to it.

Are you using a pair of resistors to set the ADJ point?

A 240 resistor and a 5k pot.

Just to be clear - I've been using it without this issue for a while. It just recently started.
 
Last edited:
What did you do to the circuit or whatever it feeds or is fed from, just before the problem started? You seem to be a bit accident-prone with probes! Did it originally work OK, or did you not notice the problem even though it may have been present? The 240 is a resistor, and not a thermistor?

Provided the voltages are low enough to be safe, to check the temperature of the chip just put your finger on it. If you can hold your finger on it comfortably then it is probably not overheating. Also worth noting whether it heats up quickly (within a few seconds) when power applied or more slowly.

What output current are you drawing from it?
 
Hi,
if you want a 10Vdc regulated supply, use a 12Vac or 15Vac transformer.
If you want dual polarity of +-10Vdc regulated, use a 12+12Vac or 15+15Vac transformer.

Your 56Vac centre tapped is equivalent to 28+28Vac and is suitable for a dual polarity supply of +-25Vdc to +-30Vdc NOT +-10Vdc

The 56Vac rating is only applicable when the transformer is fed with it's rated supply voltage and when a resistor across the secondary draws exactly the rated AC current. If you feed the transformer with a different voltage you will get a different secondary voltage.
If you connect up a rectifier and smoothing caps with virtually no load, then the caps will charge up much higher than the secondary Vac rating would indicate.

Expect an unloaded PSU to deliver at least 5% more voltage and possibly +40% if using a very low VA transformer.

I suspect you have driven the regulators too often beyond their maximum voltage rating and eventually they have cried "enough".
 
Hi,
if you want a 10Vdc regulated supply, use a 12Vac or 15Vac transformer.
If you want dual polarity of +-10Vdc regulated, use a 12+12Vac or 15+15Vac transformer.

Your 56Vac centre tapped is equivalent to 28+28Vac and is suitable for a dual polarity supply of +-25Vdc to +-30Vdc NOT +-10Vdc

I know. The transformer is OK, I wanted a supply that was adjustable up to 20V. It's for experimenting, not for a specific circuit.

If you connect up a rectifier and smoothing caps with virtually no load, then the caps will charge up much higher than the secondary Vac rating would indicate.

Expect an unloaded PSU to deliver at least 5% more voltage and possibly +40% if using a very low VA transformer.

I suspect you have driven the regulators too often beyond their maximum voltage rating and eventually they have cried "enough".

The regulator is rated with a maximum input-output differential of 40V. The datasheet gives no absolute maximum input voltage, only the differential. I've been using the supply at 10-15V, but I expected the input voltage to be much lower than 45V. I can check the loaded input voltage once I repair it again, but what would you expect it to drop to?
 

tomchr

Member
Paid Member
2009-02-11 12:58 am
Calgary
www.neurochrome.com
I suspect you have driven the regulators too often beyond their maximum voltage rating and eventually they have cried "enough".

That would be my guess also.

To verify, disconnect the regulators and measure the DC voltage across the reservoir cap just after the bridge rectifier. Then look at the LM317/337 data sheet (www.national.com) and check the maximum input voltage spec. I think it's 36 V, but I'm not sure. There is a high-voltage version of the LM317 but not of the 337.

Unless your regulators are on a big heat sink (as in several inches by several inches in size) don't plan on drawing the full rated load current at minimum output voltage as this will overheat the regulators.

~Tom
 
Again, the datasheet gives no maximum input voltage rating. It only gives a maximum input-output differential, which is 40V. Since the output is at least 1.25V, they are essentially saying it can take up to 38.75V.

Now, maybe you guys can help me understand how to calculate the input voltage. I understand it (a filtered, rectified voltage) depends on the load resistance, but how do I figure that out when the only load is opamps? Do I use the opamp's rated supply current and the voltage I'm supplying it? Meaning: a datasheet gives a typical supply current of 8mA and I'm using it at 10V, so the effective "load" resistance for the supply is 1250 ohms?

Unfortunately when I was trying to replace the burned out components, a couple parts of the copper traces peeled away from the board. I'm going to have to etch a new board and rebuild the damned thing.
 
Unfortunately while checking the input voltage, I think I must have accidentally bridged the input and output pins with the probe, and released the magic smoke from the output capacitor. I guess this is a game of "see how many times I can burn up components on this before I just go buy a power supply."

You should be able to short the reg in & out without damage to it or anything else. Your output cap should be rated for the maximum input side voltage. I suspect that was not the case.

You can also help protect the reg and caps by putting a reverse biased diode across the outputs, which prevents reverse voltage damage. Put another from reg output to input, preventing damage due to higher output than input voltage, which can happen at turn-off.

I just realized the input voltage I measured (before the magic smoke) doesn't make sense. It was +/- 45V, but I have a 56V CT transformer, so there's no way either side should be above 28V right? Even less after the rectifier and filtering.
That could be right. 56 V CT is 28+28. 28 VAC gives 40 VDC when rectified and filtered. At no/small load, the transformer can easily give 5 or 10% higher voltage than rated. And your mains might be a little hot. Maybe the transformer is designed for 110 VAC input, but your wall socket is giving 120 VAC (or 125 VAC?).

45 VDC input is just plain too high. Not only will it potentially damage the reg due to excessive voltage, but you will disspate much more heat in the reg than necessary. The power dissipation is equal to the output current * voltage drop. If you want to output say 15 VDC and the input is 45 VDC, you are dropping 30 VDC in the reg. The thermal resistance, Junction to case, is 4 degC/W. The maximum temp is 125 degC, so assuming 25 degC ambient air temp, you can go 100 degC over ambient. With an infinite and perfect heatsink (0 degC/W), you could then dissipate (100 / (4 + 0)) = 25 W. You don't have that heatsink, yours is probably say 12 to 25 degC/W, typical of wimpy to decent PC board TO-220 heatsinks. Let's say it's 12 degC/W. Then you can handle (100 / (4 + 12)) = 6.25 W of power. At 30 V, that's only around 200 mA of current. If you have a wimpy heatsink, it's more like 100 mA. That's far from the reg's 1.5 A rated current isn't it? Don't forget to consider the current that the opamp is driving into its load, not just its own supply current.

I'd say you are probably driving the poor little reg into thermal overload. And you have probably done that so much now that it is permanently damaged. Replace the reg with a new one, and strongly consider a new lower voltage transformer. If you want to keep that transformer, consider adding a series resistor between it and the reg. This will drop some voltage, reducing the voltage drop in the regulator. In your case, maybe 10 to 20 ohms might work well.
 
You should be able to short the reg in & out without damage to it or anything else. Your output cap should be rated for the maximum input side voltage. I suspect that was not the case.

The output cap was 35V I believe.

You can also help protect the reg and caps by putting a reverse biased diode across the outputs, which prevents reverse voltage damage. Put another from reg output to input, preventing damage due to higher output than input voltage, which can happen at turn-off.

Yeah, I have those.

That could be right. 56 V CT is 28+28. 28 VAC gives 40 VDC when rectified and filtered. At no/small load, the transformer can easily give 5 or 10% higher voltage than rated. And your mains might be a little hot. Maybe the transformer is designed for 110 VAC input, but your wall socket is giving 120 VAC (or 125 VAC?).

45 VDC input is just plain too high. Not only will it potentially damage the reg due to excessive voltage, but you will disspate much more heat in the reg than necessary. The power dissipation is equal to the output current * voltage drop. If you want to output say 15 VDC and the input is 45 VDC, you are dropping 30 VDC in the reg. The thermal resistance, Junction to case, is 4 degC/W. The maximum temp is 125 degC, so assuming 25 degC ambient air temp, you can go 100 degC over ambient. With an infinite and perfect heatsink (0 degC/W), you could then dissipate (100 / (4 + 0)) = 25 W. You don't have that heatsink, yours is probably say 12 to 25 degC/W, typical of wimpy to decent PC board TO-220 heatsinks. Let's say it's 12 degC/W. Then you can handle (100 / (4 + 12)) = 6.25 W of power. At 30 V, that's only around 200 mA of current. If you have a wimpy heatsink, it's more like 100 mA. That's far from the reg's 1.5 A rated current isn't it? Don't forget to consider the current that the opamp is driving into its load, not just its own supply current.

I'd say you are probably driving the poor little reg into thermal overload. And you have probably done that so much now that it is permanently damaged. Replace the reg with a new one, and strongly consider a new lower voltage transformer. If you want to keep that transformer, consider adding a series resistor between it and the reg. This will drop some voltage, reducing the voltage drop in the regulator. In your case, maybe 10 to 20 ohms might work well.

Thanks, your post was a lot of help. After thinking about it a bit I can see I miscalculated during design. I don't want to have to buy a new transformer, but I'm sure that's a better long-term solution than the resistors, which would have to have fairly high power ratings anyway, right? And as long as I have to get a new transformer and rebuild the thing anyway, I think I might sort of redesign it to provide only up to say 15V, so I can drop less voltage across the regulator and thus handle more current. I was trying to give myself options, but realistically I don't see needing more than 15V for the circuits I'm thinking of.

Don't forget to consider the current that the opamp is driving into its load, not just its own supply current.

I don't quite understand this. How can an opamp drive current it's not pulling from the supply?

Thanks folks!