purpose of protecting comparator inputs

Thread Starter

MikeJacobs

Joined Dec 7, 2019
123
So generally, in the pst I have protected the inputs of a comparator like an LM139 with a clamping diode if needed.
This is assuming there is a potential risk of over voltageing the pins for whatever reason.

In general however, I see many circuits that just appear to stick a resistor in series with the inputs.
What is the purpose of this. In the idea case the input impedance is already very high.

Any current input on the pins with a large resistor is going to cause a reasonable voltage drop across this resistor causing error.
Can someone explain me what the purpose is and how to size this magical resistor?

Thanks
 

MrChips

Joined Oct 2, 2009
21,156
I have seen many LM311 comparators destroyed from over-voltage on inputs.

Without knowing the internals of the chip I would suspect it has something to do with breakdown between input pins and chip substrates, and nothing to do with the high impedance inputs.

1kΩ is usually high enough to limit the current that might destroy the chip.
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
123
I have seen many LM311 comparators destroyed from over-voltage on inputs.

Without knowing the internals of the chip I would suspect it has something to do with breakdown between input pins and chip substrates, and nothing to do with the high impedance inputs.

1kΩ is usually high enough to limit the current that might destroy the chip.
thanks for the reply.
Tell me more about this 1k if you don't mind.

I am confused how an (op amp) really needs to be current limited. Is this if the chip itself fails internally and starts to draw more current?

If you do have an over voltage condition, the resistor is surely not going to save you if the input current is micro amps. The voltage drop across the resistor will be nothing and the voltage at the pin will still be very high

still fairly confused on this.
 

crutschow

Joined Mar 14, 2008
24,988
Generally if there is over or under voltage on the inputs, the input impedance drops (due to diode protection or the substrate diode) and current starts to flow.
If you limit that current to a small value, than the input won't be damaged.
 

ebeowulf17

Joined Aug 12, 2014
3,274
Still not getting this
How does under voltage cause over current
I don't know about comparators specifically, but many integrated circuits have built in protection diodes, conceptually the same as the external ones you sometimes add.

The built in diodes prevent over voltage (or under/reverse voltage) from reaching parts of the IC which would be damaged by the voltage... but there's nothing in the IC to protect those built in diodes from too much current running through them. An external resistor reduces the amount of current that can flow though the built-in protection diode, thereby protection the internal diode. The internal diode is what's preventing the rest of the chip from being damaged.

If you measured voltages during an over voltage event in this scenario, you'd find that the voltage at the pin would be roughly one diode drop above the IC positive supply voltage, and that any remaining voltage is being dropped across the the external resistor. Make sense?
 

ebeowulf17

Joined Aug 12, 2014
3,274
Still not getting this
How does under voltage cause over current
Oh, as for under voltage, there are often internal diodes for that too, so if the input sees voltage more than one diode drop below the negative supply rail, then that diode starts conducting, again without any current limiting unless you provide that externally.
 

ronsimpson

Joined Oct 7, 2019
684
How does under voltage cause over current
"under voltage" is a voltage below ground. The part is built to work with inputs slightly below ground. If the inputs are forced below 1.5V current climbs. In the data sheet the part will not die with -50mA in this condition. (the part will not work right)

I did not see what happens if the inputs are above the supply. Many parts have 1mA or 10mA diodes to supplies. My guess is that 1mA when above supply and 50mA when below supply.

Any current input on the pins with a large resistor is going to cause a reasonable voltage drop across this resistor causing error.
Look at input bias current and input offset current. Generally min=? typical is stated Max is what you to design for. Also note that there is a number for room temperature and one for full range temp. It looks like worst case at low temperature = -300nA. That could be -25nA. Note watch out for (-) or (+).
So you need to think about two currents. The 1mA or 10mA or what ever the max input current is AND think about what -300nA to -10nA will do to your errors.
 
Last edited:

MrChips

Joined Oct 2, 2009
21,156
Look at it this way.

Suppose that your chip is powered from two power rails 0V and 5V.
If an input voltage exceeds these two voltages, i.e. goes below 0V or above 5V, nasty things happen.
Many years ago CMOS circuits would enter a short circuit mode and the chip would be fried. This is known as CMOS latch-up.
Today, they put protection diodes on the pins in order to clamp the voltage to one diode voltage drop below 0V and above 5V. These protection diodes are built into the design of the chip.

If the diode experiences a current that is too high it will blow like a fuse. The chip is no longer protected and it goes too.
Adding an external resistor in series with the input limits the short circuit current to a safe value.

1587334722597.png
 

ronsimpson

Joined Oct 7, 2019
684
Vdd-0.3 and Vss+0.3
These numbers vary widely depending on the process for making the IC. CMOS is very different than transistors. Digital different than analog. (read the data sheet)
It is common for comparators to have PNP input transistors so they can work with signals slightly below the supply. The LM139 is a good example of that.
Again, read the data sheet.
Latch up is certainly a problem. The LM139 and its brothers; pulling high input current "may cause output to be incorrect."
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
123
Look at it this way.

Suppose that your chip is powered from two power rails 0V and 5V.
If an input voltage exceeds these two voltages, i.e. goes below 0V or above 5V, nasty things happen.
Many years ago CMOS circuits would enter a short circuit mode and the chip would be fried. This is known as CMOS latch-up.
Today, they put protection diodes on the pins in order to clamp the voltage to one diode voltage drop below 0V and above 5V. These protection diodes are built into the design of the chip.

If the diode experiences a current that is too high it will blow like a fuse. The chip is no longer protected and it goes too.
Adding an external resistor in series with the input limits the short circuit current to a safe value.

View attachment 204820
Really interesting

Arnt your clamping diodes backwards?
Thanks for taking the time to write all this
really nice of you
 

MrChips

Joined Oct 2, 2009
21,156
Really interesting

Arnt your clamping diodes backwards?
Thanks for taking the time to write all this
really nice of you
The diagram is just for explanation. The diodes are installed in the IC fabrication process.
Diodes are oriented correctly. Why do you think they are backwards?
 
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