OPAMP with Ballast resistor

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020170

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expression for the output errors in an op amp

Recently I analyzed some circuits like below.


Question :

1. How this circuit operate?

2. Why someone added to this circuit with resistor, 200 ohm ?

3. How can I analyze this curcuit? I anaysis with feedback theory, I can't get beta from feedback theory.
 

This is a unity gain buffer. You should be able to analyze it with ideal opamp approximation. The resistor value will not show in the gain formula, because the impdance of the opamp inputs is infinite (ideal opamp) and there's no current flowing through the ressitor. I've seen people add a ressitor the way you're showing before. But, unfortunately, I don't know why they do it.
 

i think the resistor is for ead issue
 

020170,
A resistor is often added to make the resistance "seen" by the inverting input equal to the resistance "seen" by the non-inverting input. This eliminates output errors due to op-amp input bias current. Some designers put in a "placeholder" resistor of low value, just in case the input source resistance changes during the course of a design, allowing the insertion of the correct value resistor without the need to change the PCB layout.
Regards,
Kral
 

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    020170

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It's interesting. you said that "the resistance "seen" by the inverting input equal to the resistance "seen" by the non-inverting input".

Could you explain in detail? And I wonder How could you analyze the circuit.

Thanks for your answer
 

020170 said:
It's interesting. you said that "the resistance "seen" by the inverting input equal to the resistance "seen" by the non-inverting input".
Click to expand...
The output resistance of the voltage source connected to the input is implied.

020170 said:
And I wonder How could you analyze the circuit.
Click to expand...
It depends on what you want to find out.
I met an old buddy of mine at a conference tonight. He works at Microsoft now, and internally they use a proverb: "analysis paralysis". Not that it necessarily applies here.

Added after 4 minutes:

mists said:
mists said:
i think the resistor is for ead issue
Click to expand...
sorry my mistake, it should be ESD
Click to expand...
Could you tell some more about it? If the positive input (inp) is exposed to ESD, the resistor doesn't protect it. If the output is exposed to ESD, the resistor doesn't protect the output, it only protects the negative input (inn).
 

here i assume the signal to inp is from chip internal, so it need not esd protect.
but as you see, the opamp out is pad_out, and usually the opamp output is mosfet source or drain, so the output has esd protection itself.
but if inn connect pad_out directly, and the inn usually connect to mos gate, it has serious esd issue. so we add a resistor for esd protection.
 

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    020170

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Great.

So far I don't know How to analyze the circuit, But your answer is very smart and very reasonable.

Thanks for your answer!
 

It could also be in there to allow an easy modification to the circuit to produce gain.
 

020170,
Let Ib be the input bias current. Let Rf = the feedback resistor. Let Ri be the source resistance of the input voltage source. In your example, bias current from the Inverting input flows through the feedback resistor to the low impedance output of the op-amp. The output resistance of the op-amp can be considered to be zero for your example. The bias current from the Non inverting input flows through Ri. The differential voltage Vd developed across the op amp input is equal to Ib(Rf-Ri). If Rf = Ri, then zero voltage is developed and the output error voltage due to input bias current is zero. The resistance "seen" by the input terminal is an expression that is commonly used to denote the total resistance through which current flows from a given terminal.
Regards,
Kral
 

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