0
\$\begingroup\$

From what I understand, the input impedance of an op amp is very high (ideally infinite) and almost no current flows into the op amp. If that is the case why do we put a resistor in series with the input to protect it like below?

enter image description here

If the input voltage is too high (like from an ESD event or over input voltage,) won't the output just get clamped to the supply voltage anyway? By putting a resistor in series with the input, doesn't that change the input voltage seen by the op amp and introduce an error since there will be a voltage drop on the resistor? There seems to be no explanation anywhere on the Internet.

\$\endgroup\$
3
  • \$\begingroup\$ High input impedance doesn't imply infinity common mode input voltage range. \$\endgroup\$
    – Unknown123
    Feb 24 at 5:16
  • \$\begingroup\$ If the input voltage is too high the input transistors map be destroyed. There will usually be protection diodes in the IC itself, but presumably this author has found cases where they are insufficient. \$\endgroup\$ Feb 24 at 13:27
  • \$\begingroup\$ Those resistors may not be there for protection. One of them is helping set gain and the other one is helping minimize errors due to input offset of bias current. In any event, the ability of any protection diodes to clamp to the rail is limited. It is not far-fetched that a series resistor will increase survivability after ESD events. \$\endgroup\$
    – mkeith
    Feb 24 at 20:00
3
\$\begingroup\$

The input of an op amp typically looks similar to this:

op amp input stage with ESD diodes shown

Under normal operation the following hold true:

$$V_{\text{EE}} < V_{\text{P}} < V_{\text{CC}}$$

$$V_{\text{EE}} < V_{\text{N}} < V_{\text{CC}}$$

$$V_{\text{P}} \approx V_{\text{N}}$$

Under these conditions all protection diodes are reverse biased and therefore don't draw much current, as can be seen in the I-V curve for a diode:

Diode I-V curve

Similarly, the impedance looking into the bases of the input transistors is high so the input impedance is high.

However, if the voltage on the input(s) rises above \$V_{\text{CC}}\$ or falls below \$V_{\text{EE}}\$ then the protection diode(s) will be forward biased and begin to conduct current. If the overvoltage is high enough then the diode will begin to conduct significant current (refer to the diode I-V curve above) unless that current is limited by some other means (e.g. a resistor in series with the overvoltage source or current limiting by the overvoltage source). That will result in a significant drop in the effective input impedance since impedance is

$$Z = \frac{V}{I}$$

and increasing \$I\$ without a similar increase in \$V\$ will result in a lower \$Z\$. Moreover, the protection diodes in an op amp IC are small and cannot dissipate significant power without being destroyed.

By putting resistor in series with the input, doesn't that change the input voltage seen by the op amp and introduce error since there will be a voltage drop on the resistor?

Depending on how the op amp is used, putting resistors in series with the input can be the source of some error/noise. However, because the impedance of the op amp's pins are high in normal operation this is usually an insignificant error/noise -- the high impedance means that the current through the series resistors (and therefore the unwanted voltage drop across them) is small. The only time the voltage across the resistors becomes significant is during an ESD / overvoltage event in which the series resistors are limiting the current through the op amp's input protection diodes -- and in such an event it's more important to protect the op amp from damage than worry about an error voltage at the op amp inputs.

\$\endgroup\$
1
\$\begingroup\$

In normal operation there is almost no input current, so there is also no voltage drop across the input protection resistors. However, during an ESD event the protection diodes turn on, and a current may flow through the protection resistors. The protection resistors limit the input current during the ESD event so that the protection diodes are not destroyed.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.