<p>6/6/2018 02ec59c60f75a64aa4ddd363ea02c715.doc 1/5</p><p>Feedback Stability</p><p>Recall that for the inverting amp: R 2</p><p> i 2 R 1 v v - in - oc i 1 ideal vout v + +</p><p> we have the feed-forward equation:</p><p> oc vout= - A op v- and the feed-back equation:</p><p> oc R2 vin+ R 1 v out v- = R1+ R 2</p><p>Taking derivatives of these equations, we find that:</p><p> oc � A v vout ( op - ) = = -Aop 抖v- v - 6/6/2018 02ec59c60f75a64aa4ddd363ea02c715.doc 2/5 and: �v骣 R v R v oc R - =琪 2in 1 out = 1 抖 oc oc vout v out 桫 R1+ R 2 R 1 + R 2</p><p>These derivatives are very important in determining the stability of the feedback amplifier. To see this, consider what happens when, for some reason, v- changes some small value Dv- from its nominal value of v- =0.</p><p> oc The output voltage will then likewise change by a value Dvout :</p><p> oc oc 骣vout Dvout籇琪 v- = - A op D v1 桫 v-</p><p> oc Note if Dv- is positive, then Dvout will be negative—an increase oc in v- leads to a decrease in vout .</p><p>This describes the feed-forward portion of the “loop.”</p><p>The feed-back equation states that a small change in the output oc voltage (i.e., Dvout ), will likewise result in a small change in v- :</p><p>骣 v- oc骣 R1 oc Dv- 籇琪 oc vout =琪 D v out 桫�vout 桫 R1 R 2</p><p>Note in this case, a decreasing output voltage will result in a decreasing inverting terminal voltage v-.</p><p>Thus, if the inverting terminal voltage tries to increase from its correct value of v1 =0, the 6/6/2018 02ec59c60f75a64aa4ddd363ea02c715.doc 3/5 control loop will react by decreasing the voltage v1 —essentially counteracting the initial change!</p><p>Note that the loop product:</p><p> v oc v骣 R out - = -A 琪 1 抖 oc op v- vout 桫 R1+ R 2 is a negative value; we refer to this case as negative feedback.</p><p>Negative feedback keeps the inverting voltage in place (i.e., v- =0)—it “enforces” the concept of the virtual ground!</p><p>Contrast this behavior with that of the following circuit: R 2</p><p> i 2 R 1 v v + + in oc i 1 ideal vout v - -</p><p>Q: Isn’t this precisely the same circuit as before? 6/6/2018 02ec59c60f75a64aa4ddd363ea02c715.doc 4/5</p><p>A: NO! </p><p>Note that the feedback resistor is now connected to the non- inverting terminal, and the inverting terminal is now grounded.</p><p>The feed-forward equations for this circuit are thus:</p><p> oc vout= A op v+ And so: oc A v vout ( op + ) = = Aop 抖v+ v + while the feed-back equations are:</p><p> oc R2 vin+ R 1 v out v+ = R1+ R 2 and:</p><p>�v骣 R v R v oc R + =琪 2in 1 out = 1 抖 oc oc vout v out 桫 R1+ R 2 R 1 + R 2</p><p>Note in this case, both derivatives are positive.</p><p>This means that an increase in v+ will lead to an oc increase in vout . The problem is that the feedback </p><p> will react by increasing v+ even more—the error is not corrected, it is instead reinforced!</p><p> oc The result is that the output voltage will be sent to vout = or oc vout = - (i.e., the amplifier will saturate). 6/6/2018 02ec59c60f75a64aa4ddd363ea02c715.doc 5/5</p><p>Note that the loop product for this case is positive:</p><p> v oc v骣 R out + = A 琪 1 抖 oc op v+ vout 桫 R1+ R 2</p><p>Thus, we refer to this case as positive feedback.</p><p>Positive feedback typically leads to amplifier instability!</p><p>As a result, we find that the feed-back portion of an op-amp circuit almost always is connected to its inverting (-) terminal!</p>