TLV9064: Input resistance calculation

So,I don't need to care the input resistor here,when the voltage is in the allowed limits right? And I can even not add this resistor.

Is my understanding correct?

Hi Zzzz,

assume that the OPAmp in this example is powered by V+ = 5V. Then the input is protected against an overvoltage at "Vin" of

Vin = 10mA x 5k + V+ = 55V

If the OPAmp is unpowered (V+ = 0V), on the other hand, the allowed overvoltage at Vin is only 50V.

Two comments on this example:

1. It's recommended to keep the maximum input current flowing into the input pin of OPAmp well below 10mA, especially if such input current event happens more often than once. I would recommend a maximum current of 100µA...1mA.

2. The voltage regulator supplying the OPAmp must be able to absorb this current. And this without allowing the supply voltage to exceed the maximum supply voltage rating of OPAmp ! This is another good reason to keep the overload current as low as possible :-)

Kai

Hi Zzzz,

so you don't want to know how to calculate the 5k input protection resistor?

Kai

Hi kai,

I have understood what you mean.How should I know the maximum value of my output current?As you said, keep it at "100µA...1mA"

And I also want to know how to choose the right input protection resistor in different situations.

Thank you for your help!

Z

Hello Z,

When the op amp is operating within its limits, then adding the 5k resistor does not do much of anything.  As Clemens has stated, the input impedance of the amplifier is extremely high.

The 5k resistor can help to protect from overvoltage on the input.  If the input exceeds the V+ voltage by 500mV (from the data sheet), then you will start to have current flowing into the input.  This is in excess of the absolute maximum ratings, so you should avoid this.

However, you can try to protect the amplifier by limiting the current into the amplifier during this overvoltage with an input resistor.  So, the input resistor is really just to protect against operation where the input exceeds the absolute maximum voltage.

To size the input resistor, use the following formula: (Expected input voltage - (Supply Voltage + 500mV)) / Current Limit

The data sheet has this current limit as 10mA, but it is better to limit the current to something like 1mA or smaller.

Regards,
Daniel

Hi Daniel,

Understood ,Explained very clearly! Thanks for you kindly help ！

Z

Hi Zzzz,

I have understood what you mean.How should I know the maximum value of my output current?As you said, keep it at "100µA...1mA"

And I also want to know how to choose the right input protection resistor in different situations.

If the input you want to protect is fed by a signal from the outer "world", maybe from other equipement, then the maximum output signal of this equipment might be the maximum voltage you want to protect against. Assume the maximum voltage from the feading equipment is 10V (while the nominal output voltage is 2V or so) and your TLV9064 is still unpowered, then a 10k protection resistor would limit the input current to

10V / 10k = 1mA

Or assume that you want to limit the input voltage by the help of a TVS, then the threshold voltage of this TVS would be your maximum input voltage. Assume using a 12V TVS then a 12k input resistor would limit the input current to

12V / 12k = 1mA

Such a TVS might be useful to protect the input against ESD: The TVS limits the input voltage to about 12V and the input protection resistor finally limits the remaining ESD pulse energy by limiting the remaining current flowing into the TLV9064. Such a two stage protection regime is very often used in OPAmp circuits.

Kai