<p>4/5/2018 0257df836ff057d5ac3663aef774ef0c.doc 1/3</p><p>Example: The Input Offset Voltage</p><p>Consider an inverting amp constructed with an op-amp exhibiting an input offset voltage of Vos: R 2</p><p> i 2</p><p>R 1 v v 1 - i - i ideal 1 vo v Vos + 2 + - + </p><p>Applying the concept of a virtual short to the ideal op-amp, we find that:</p><p> v1 0  Vos  0</p><p>Vos</p><p>Thus, v1 v 2 ! For an op-amp with an input offset voltage, the virtual “short” equation turns out to be: 4/5/2018 0257df836ff057d5ac3663aef774ef0c.doc 2/3</p><p> v1 Vos  v 2</p><p>Recall, however, that the input offset voltage is typically very small (i.e., Vos  5mV ), so that v1 v 2 .</p><p>The current into each terminal of the op-amp is still zero, so that:</p><p> i1 i 2 where:</p><p> vi  v1 i1   R1 and:</p><p> v1  vo i2   R2 Combining, we find:</p><p> v V V v i os  os o R1 R 2</p><p>Performing a little algebra, we can solve this equation for output voltage vo :</p><p>Vos R1 V os R 2  vi R 2 vo  R1 and rearranging: 4/5/2018 0257df836ff057d5ac3663aef774ef0c.doc 3/3</p><p>     vo   vi    V os    </p><p>Hey! We could have easily found this result by applying superposition!</p><p>Note that if the input offset voltage is zero (its ideal value), this expression simply reduces to the normal inverting amplifier expression:</p><p>R2  vo     vi R1 </p><p>Thus, the term: R   2 1 Vos R1  represents an output offset voltage. Note this offset voltage is a constant with respect to vi --its value does not change, even if the input voltage is zero!.</p>