Author Topic: LTC1052, fifth I have killed whilst prototyping. Any idea why they are dying?  (Read 5057 times)

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Offline king.osloTopic starter

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Hello there,

I am using the LTC1052 for my project. Just 5 minutes ago, the fifth one I've bought died for unknown reasons. Are these ultra-sensitive to  static shocks? I do not have high voltage on any of its input pins.

Symptoms at death are: V+ potential output at inverting and non-inverting input pins (and incorrect output voltage), or internally short circuiting V+ and V- so that the device gets burning hot within 10 secs of power-on (and incorrect output voltage).

What are your thoughts? Why are these chopper stabilized opamps be dropping like flies?

Thank you for your time.

Kind regards,
Marius
« Last Edit: April 28, 2012, 04:03:01 pm by king.oslo »
 

Offline mikeselectricstuff

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Sounds like latch-up, typically caused by having voltage on input pins when power is applied. Can also be caused by very fast transients on pins or supply lines.
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Offline IanB

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Yikes! That thing is expensive. And you've killed 5 of them? Ouch.
 

Offline king.osloTopic starter

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Sounds like latch-up, typically caused by having voltage on input pins when power is applied. Can also be caused by very fast transients on pins or supply lines.

Thanks. I have had voltage at input pins when power was applied to the opamp. This could be the cause.

Is voltage @ input pins when the power is off, going to be a problem, or does latch-up only occur when voltage is @ inputs at start-up?

Thanks Mike.

Yikes! That thing is expensive. And you've killed 5 of them? Ouch.

Yes, I know. So I need to take action ;) M
 

Offline amspire

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Definitely looks like that IC is very prone to latchup. The problem you have is you need to add extra protection, but in a way that does not add more than a few picoamps leakage to the inputs.

Here is my suggestion. Add 220 ohm resistors to each supply rail, just so if there is latchup again, the IC doesn't take the full short circuit current of your supply.

Add a 1N4148 type diode from the non-inverting input to the 10v reference output.  Cathode to the 10v, so that the input cannot go above 10.6 v. The reason why this should not cause leakage currents is that normally there will be less than 5uV maximum across this diode.

Last, make sure that the HV supply is only on when the opamp circuit is fully powered. The safest way to do this if to have the  100v output shorted to 0 v while the circuit is turning on and off.

Richard.
 

Offline free_electron

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are you powering this thing single ended or symmetrically ?
a typical opamp killer is a symmetrical supply that does not start up correctly. the solution is to place two shottky diodes antiparallel with the supplies. ( fat ones that can handle some current.. ) they prevent the substrate voltage reversing more than 0.3 volts. this eliminatd the intrinsic substrate thyristor from kicking in and frying the chip

Code: [Select]

+5v o-+-----
     _|_
     /_\
      |
gnd 0-+----
     _|_
     /_\
      |
-5v 0-+-----
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Offline vxp036000

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That's a neat treak, I'll need to try it sometime.  Are op amps usually that sensitive to assymetrically applied supply voltages?  I normally use discrete transistors for most of my circuits, rather than op amps, so I don't have a lot of experience in that regard. 

are you powering this thing single ended or symmetrically ?
a typical opamp killer is a symmetrical supply that does not start up correctly. the solution is to place two shottky diodes antiparallel with the supplies. ( fat ones that can handle some current.. ) they prevent the substrate voltage reversing more than 0.3 volts. this eliminatd the intrinsic substrate thyristor from kicking in and frying the chip

Code: [Select]

+5v o-+-----
     _|_
     /_\
      |
gnd 0-+----
     _|_
     /_\
      |
-5v 0-+-----
 

Offline ejeffrey

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Is voltage @ input pins when the power is off, going to be a problem, or does latch-up only occur when voltage is @ inputs at start-up?

Start-up is the real killer.  The small current drawn by the input pin is enough to turn on the parasitic SCR, which shorts out the supply, dumping its full current through the op-amp and destroying it.  Voltage on the pins with no supply probably won't hurt anything as long as the total current is limited to a small value.

One good trick when experimenting is to put resistors in series with the supplies.  They will hopefully limit the current draw to a safe value.  As long as the opamp doesn't have to drive a big load the resistors shouldn't stop the circuit from operating.  Once you get it working remove the resistors.
 

Offline free_electron

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its not the current that kicks in the parasitic... its the voltage...

if you drive any input more negative than the bottom rail , the scr may trip. it typically needs around 2 volts to kick in..
the diode trick prevents that from happening.
the problem with symmetric supplies is that they do not necessarily start up or shut down in sync with each other...

leakage in the system may force the negative rail to temporarily go above ground.... that is what the diodes protect against. they will conduct and prevent the reails from going more than 0.3 volts the 'wrong way' ( diodes will conduct. and thus clamp the voltage. ) once the regulators kick in the system normalizes.
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Offline amspire

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In this case it is a single supply, and the problem is the LTC1052 is attached to a 10:1 divider that is connected to a 150V DC source powered by a different winding.

So it is easy for the 150V to start up to the point the divider is putting more then 0.3V overvoltage into the LTC1052 non-inverting before the LTC1052 supply starts to rise.

Also it is easy for charge to remain on the 150V supply caps which may cause a problem on turn off, or make the problem wore on the next turn-on.

This is a precision circuit, so you do not want to add much to the divider output  - 1nA leakage in a diode would ruin the accuracy.  I am pretty confident my earlier suggestion will fix the problem.

Richard.
 


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