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In this diagram, there's an op-amp with no negative feedback. What is V+ (voltage at positive end of op-amp)?enter image description here

I'm self-learning circuits, and a resource I'm using says V+ = V3 / 2R. However, I don't understand why it's V3 / 2R. I know V3 is the voltage source thus there's the V3 there, but why isn't it V3 / 2 for example?

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2 Answers 2

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The source you are using is wrong. You know because the units are wrong -- a voltage divided by a resistance is a current, not a voltage. The correct expression is yielded using voltage division: \$\frac{V_3 \cdot R}{R+R}\$, which reduces to \$\frac{V_3}{2}\$. Notice now that the units are \$\frac{Volt \cdot Ohms}{Ohms}\$, or volts, and everything is consistent.

Good catch.

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  • \$\begingroup\$ Thank you for reducing 100% of my cognitive dissonance!!! Awesome, circuits make a whole lotta more sense now! \$\endgroup\$
    – Laura K
    Commented Dec 5, 2016 at 14:17
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Laura - because you have mentioned "self-learning", here are some additional hints for you:

  • The result V3/2 is correct - as long as you are allowed to assume an infinite input resistance at the non-inverting opamp input terminal. Normally, this is the case if the external resistors are in the lower to mid kOhm range. Otherwise this input resistance Rin is to be considered in parallel to the grounded resistor R.

  • Furthermore, please realize that such an opamp application (without negative feeedback for setting the desired gain value) has no practical meaning for amplication purposes because the device does not work in its linear region. The reason: An unavoidable input imbalance causes an input DC offset voltage, which appears at the output - multiplied with the very large open-loop gain (app. 1E5). As a result, the DC output voltage saturates and is at the upper or lower supply voltage rail. This effect is also drastically reduced using negative feedback.

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    \$\begingroup\$ It has plenty of practical meaning as a comparator. \$\endgroup\$ Commented Dec 5, 2016 at 14:36
  • \$\begingroup\$ Yes - you are right. I should say "as an amplifier". Thank you. \$\endgroup\$
    – LvW
    Commented Dec 5, 2016 at 15:00
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    \$\begingroup\$ Note that the effective input resistance of an op-amp used as a comparator may be much lower than that of the same op-amp operating in it's linear region. In some cases the op-amp may even be damaged by excessive input currents. \$\endgroup\$ Commented Dec 5, 2016 at 15:48
  • \$\begingroup\$ Theoretically it should operate as a comparator (but some op-amps suffer from phase inversion (also called phase reversal), and there would be a problem with the missing input hysteresis) - I think your answer still doesn't read as if covering that scenario (or rather excluding it) - perhaps "amplication purposes" should be emphasised somehow(?). \$\endgroup\$ Commented Dec 5, 2016 at 20:17
  • \$\begingroup\$ "amplication purposes"? Don't you mean "amplification purposes"? \$\endgroup\$ Commented Dec 5, 2016 at 20:19

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