For the circuit shown below, the OPAMP is ideal and \(V_{REF}\) is fixed. Find the value of:
\[ \frac{R_F}{R_{IN}} \]Given:
- \(V_{OUT}=1V\) when \(V_{IN}=0.1V\)
- \(V_{OUT}=6V\) when \(V_{IN}=1V\)
Solution
Ideal OPAMP Rules
For an ideal op-amp:1. Input current is zero: \[ i_+=i_-=0 \] 2. With negative feedback: \[ V_+=V_- \] Since \(V_{REF}\) is fixed, the positive input voltage is constant.
Step 1: Write the output equation
The circuit is an inverting amplifier with a reference voltage. The output equation is: \[ V_{OUT} = V_+ - \frac{R_F}{R_{IN}} (V_{IN}-V_+) \] Let: \[ K=\frac{R_F}{R_{IN}} \] Therefore: \[ V_{OUT}=V_+-K(V_{IN}-V_+) \]Step 2: Apply first condition
Given: \[ V_{IN}=0.1V \] \[ V_{OUT}=1V \] Therefore: \[ 1=V_+-K(0.1-V_+) \]Step 3: Apply second condition
Given: \[ V_{IN}=1V \] \[ V_{OUT}=6V \] Therefore: \[ 6=V_+-K(1-V_+) \]Step 4: Subtract the two equations
Subtract equation 1 from equation 2: \[ 6-1 = -K(1-V_+) + K(0.1-V_+) \] \[ 5 = K(0.1-1) \] \[ 5=-0.9K \] The magnitude of the gain is: \[ K=\frac{5}{0.9} \] \[ K=5.56 \]Step 5: Find resistor ratio
Since: \[ K=\frac{R_F}{R_{IN}} \] Therefore: \[ \boxed{ \frac{R_F}{R_{IN}}=5.56 } \]Important Understanding
\(V_{REF}\) only shifts the DC output level. It does not change the slope:
\[ \frac{\Delta V_{OUT}}{\Delta V_{IN}} \]The resistor ratio is determined by the change in input and output voltage.
\[ \frac{R_F}{R_{IN}} = \frac{\Delta V_{OUT}} {\Delta V_{IN}} \]
Final Answer:
\[ \boxed{ \frac{R_F}{R_{IN}}=5.56 } \]
\[ \boxed{ \frac{R_F}{R_{IN}}=5.56 } \]