{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

09_op-amps

# Imagine hooking the output to the inverting terminal

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: without burdening the source of V in with any current! • Imagine hooking the output to the inverting terminal: • If the output is less than Vin, it shoots positive • If the output is greater than Vin, it shoots negative – result is that output quickly forces itself to be exactly V in negative feedback loop Vin + Vin Winter 2008 5 + Important note: op-amp output terminal sources/sinks current at will: not like inputs that have no current flow Winter 2008 6 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 Positive feedback pathology Op-Amp “Golden Rules” Rules” • In the configuration below, if the + input is even a smidge higher than Vin, the output goes way positive • This makes the + terminal even more positive than more positive Vin, making the situation worse • This system will immediately “rail” at the supply rail” voltage • When an op-amp is configured in any negativeany negativefeedback arrangement, it will obey the following two rules: – The inputs to the op-amp draw or source no current (true whether negative feedback or not) – could rail either direction, depending on initial offset – The op-amp output will do whatever it can (within its limitations) to make the voltage difference between the two inputs zero Vin + Winter 2008 Lecture 9 positive feedback: BAD 7 Winter 2008 8 2 Op-Amps 02/14/2008 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 Inverting amplifier example Non-inverting Amplifier R2 R2 R1 R1 Vin Vout + • Applying the rules: Vout + Vin • Now neg. terminal held at V in neg. terminal at “virtual ground” terminal ground” – so current through R 1 is If = V in/R1 (to left, into ground) – so current through R1 is If = V in/R 1 • This current cannot come from op-amp input • Current does not flow into op-amp (one of our rules) – – – – – so the current through R1 must go through R 2 – voltage drop across R 2 is then IfR2 = V in (R2/R 1) • So Vout = 0 Vin (R2/R1) = Vin (R2/R1) • Thus we amplify Vin by factor R2/R1 by so comes through R2 ( delive...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online