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Author Topic: 40V burning Arduino Nano?  (Read 1291 times)
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Manchester (England England)
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It is going to have a gain that is temperature and current dependent.
Yes I an describing a digital buffer that you can use to attach live inputs to an unpowered processor to avoid latch up of the processor.
I am not describing an analogue buffer because an unpowered analogue buffer has the same problem as an unpowered processor.
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Lacey, Washington, USA
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Well, the OP wants to measure an analog voltage. So an inverting digital buffer isn't much help. You see my confusion?
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Steve Greenfield AE7HD
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So an inverting digital buffer isn't much help.
Yes read reply #5 that is what I said.
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"Latch-up" is still a bit difficult phenomenon for me to understand, even after reading long descriptions about it. Especially its practical consequences. For example:

We could use a powered oscilloscope to study almost any powered (and not powered?) circuit anywhere. The probes have high resistance (e.g. 10 Megaohms?). If we have probes attached to a circuit and cut off the power of the oscilloscope, can terrible things happen to  the circuit or to the oscilloscope?  What if the power off state of the oscilloscope lasts only a very short time, milliseconds ?

If terrible things can happen, then that information is an important news for me: "unpowered anything"  is a dangerous thing smiley

« Last Edit: August 15, 2013, 04:26:56 am by optimistx » Logged

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If we have probes attached to a circuit and cut off the power of the oscilloscope, can terrible things happen to  the circuit or to the oscilloscope?
No.

Basically the oscilloscope is an input. It is quite alright to connect an input to an unpowerd circuit. The danger happens when you connect an output to an unpowered circuit.
It is the current from that output that causes the problem.
Another way to avoid a problem is to restrict the current from that input, normally with a resistor.
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Lacey, Washington, USA
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The inputs to an oscilloscope are typically pretty well protected, and may be discrete transistors (especially older models). The problem is that there is a parasitic SCR/TRIAC structure in an IC that is not there in a discrete transistor.

In that case, the fault occurs when the IC is powered on with voltage already present at some of the pins, inputs or outputs. It isn't current drawn through (in or out) of a pin, it is current flowing from Vcc to ground through the substrate inside the IC die. It does not always destroy it if you shut it off again quickly, but it will likely get hot fast. Turning off power to the IC and removing the outside voltage source resets the intrinsic TRIAC.

It can also occur in a poorly protected IC when ESD (electrostatic discharge) hits a pin, input or output. It seems a lot less prevalent with newer ICs.

http://www.fairchildsemi.com/an/AN/AN-600.pdf

http://www.ti.com/lit/an/slya014a/slya014a.pdf

That is in addition to the problems Grumpy_Mike mentioned.
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Steve Greenfield AE7HD
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Latchup is specific to CMOS due to all the complementary wells (which can act as parasitic BJT bases IIRC).
http://en.wikipedia.org/wiki/Latchup
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Lacey, Washington, USA
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And it is important to understand that latchup causes the chip to heat from current from the power supply, NOT current supplied by an outside signal. The outside voltage/current source is merely triggering latchup.

However, that outside source can damage the IC if not current limited, not by latchup but as Grumpy Mike says, overcurrent from an outside source but into an input or output. In that case, it may be forward biasing the internal protection diodes.
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Steve Greenfield AE7HD
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