TL082 dual supply problems

Thread Starter

Bod

Joined Sep 18, 2016
317
I am using some TL082's for a project. I have them connected to a dual supply of which is ±12 (sourced from a 24V 3A supply)

If I look up their datasheet, and I click on this one (The TL082xx) it says:
Screen Shot 2019-10-20 at 17.44.40.png It clearly says the absolute maximum is ±18. 12v is heaps under that maximum.
However, when I connect the +supply of my rail splitter to +Vcc and my -supply to -Vcc, the chip heats up very quickly and gets very hot.
I did some digging and come across something that said along the lines of ±xx is the range. For this, that would mean the max is ±9 which is a total range of 18. This doesn't really line up with the datasheet however, when I give it ±6 (half of 12v), it doesn't heat up. So maybe it is a range.

This is were I left off for a bit. I came back still as confused and decided to google the whole part number (TL082-CP). This has been a mistake I have made before. I found no datasheet because the '-CP' is included the datasheet above. I did loom at the specification in the mouser page though and this is what I found:
Screen Shot 2019-10-20 at 18.48.59.png
It definitely says ±12 on there.

Why is my chip heating up so much when powered with ±12? There is no other circuitry with the chip - just on its own. It might be that the chip was faulty but I'm hesitant to use my other one because if it is a fault with my circuit then I will damage the other chip and I will have none left.

Thanks,
Bod.
 
Last edited:

crutschow

Joined Mar 14, 2008
34,281
Are you certain the plus and minus connections to the opamp are correct and not reversed?

The (-) input should go to the opamp output, not ground.
 

rsjsouza

Joined Apr 21, 2014
383
I don't know which brand you are using, but TI's datasheet specifies the same +/-18V and even shows several specs that states a Vs of +/-15V.

In thia case, can you be sure that the output impedance is compatible? The same datasheet indicates that, for an output swing of +/-13.5V and a Vs of +/-15V, the load is 10k on the output pin.

The last suspicion is a faulty or counterfeit IC. Unfortunately too common these days...
 

Thread Starter

Bod

Joined Sep 18, 2016
317
Are you certain the plus and minus connections to the opamp are correct and not reversed?

The (-) input should go to the opamp output, not ground.
So you mean (crudely) like this?
Screen Shot 2019-10-20 at 19.35.58.png
I was following a circuit like this:
subsonic-audio-filter-circuit-schematic.gif
Which does what you say but also connects it to the -VCC on the chip.

I don't know which brand you are using, but TI's datasheet specifies the same +/-18V and even shows several specs that states a Vs of +/-15V.

In thia case, can you be sure that the output impedance is compatible? The same datasheet indicates that, for an output swing of +/-13.5V and a Vs of +/-15V, the load is 10k on the output pin.

The last suspicion is a faulty or counterfeit IC. Unfortunately too common these days...
Output impedance of what?

Bod
 

Thread Starter

Bod

Joined Sep 18, 2016
317
I have just realised something. I don't think the TL082 will work for my application. rsjsouza mentioned an output swing but that's not what I want. I want no negative output values
Show the pin numbers. This way you will be forced to verify your actual circuit and hopefully find your mistake.
Decoupling caps? Maybe oscillating.
TL082-Pinout-1024x534.jpg
 

crutschow

Joined Mar 14, 2008
34,281
Unfortunately, still got very hot even with those adjustments.
Then you may have a faulty device.

If you try a new one, can you bring up the voltage slowly will monitoring the current?
It should not go much above that stated in the data sheet.
 

Thread Starter

Bod

Joined Sep 18, 2016
317
Of course, it would be a faulty chip.

The faulty chip, running at 13V was drawing only 0.1mA, however thats when I felt a noticeable change in temperature.
The new chip, I got to 24V (slightly over by mistake) and it drew ~0.003mA under no load, and stayed at room temp.

That makes the previous amplifier a little difficulty with only one chip but I can figure something else out!

Thanks,
Bod
 

Audioguru again

Joined Oct 21, 2019
6,673
Look at the datasheet for a TL082. Its normal total unloaded supply current for both opamps is 2.8mA which is one thousand times higher than you measured. Maybe yours is a cheap Chinese fake?
 

Sensacell

Joined Jun 19, 2012
3,432
Supply splitters creating a virtual ground can be troublesome - they need to drive a large capacitance - which some opamps don't like.
They can become unstable and wreak havoc.
 
Last edited:

Audioguru again

Joined Oct 21, 2019
6,673
Does the TL082 get hot even when it does not power a circuit but is simply a rail splitter? It is doing nothing so it should not even get a little warm.rail splitter.png
 

DickCappels

Joined Aug 21, 2008
10,153
They are right. I have designed the TL084 (sister to the TL082) in so many products running at ±12Vand ±10V and never saw this problem. Just get a new chip (after confirming that you have the power supply pins connected properly).
 

BobTPH

Joined Jun 5, 2013
8,812
What is the virtual ground connected to? It can only sink or source what the TL082 can, which I guess is in the range of 30ma. And, with the 1K resistor in the output (which is not the usual circuit) the ground will wander around based on how much current it is sinking or sourcing at the moment.

Bob
 

Audioguru again

Joined Oct 21, 2019
6,673
The ground of the virtual ground IC will wander due to the 1k output resistor.

I have made thousands of opamp circuits and none ever used a virtual ground. Some used a dual polarity power supply or two batteries but most were properly biased and had input and output capacitors so they worked perfectly with a single positive-only supply voltage.
 

crutschow

Joined Mar 14, 2008
34,281
And, with the 1K resistor in the output (which is not the usual circuit) the ground will wander around based on how much current it is sinking or sourcing at the moment.
The posted circuit has the 1k resistor going to supply ground, not virtual ground, so its current won't affect the virtual ground voltage.
 
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