Question

g simple difference amplifier
the table below lists the measurement results:
description value ($v_a$ grounded) value ($v_b$ grounded) value ($v_a = v_b$)
calculated gain $g_{a_{calc}} = 10$ $g_{b_{calc}} = 10$ $g_{c_{calc}} = 0$
measured gain $g_{a_{mes}} = 10$ $g_{b_{mes}} = 10.1$ $g_{c_{mes}} = 0.91$

Explanation:

Step1: Identify given output formula

The expected output function is provided as:
$v_o = \frac{R_4}{R_3}v_a - \frac{R_2}{R_1}v_b = 10(v_a - v_b)$

Step2: Verify gain for $v_a$ grounded

When $v_a=0$, $v_o = -\frac{R_2}{R_1}v_b$. Substitute values:
$\frac{R_2}{R_1} = \frac{10\ \text{kΩ}}{1\ \text{kΩ}} = 10$, so $G_{a,calc}=10$

Step3: Verify gain for $v_b$ grounded

When $v_b=0$, $v_o = \frac{R_4}{R_3}v_a$. Substitute values:
$\frac{R_4}{R_3} = \frac{22\ \text{kΩ}}{2.2\ \text{kΩ}} = 10$, so $G_{b,calc}=10$

Step4: Verify gain for $v_a=v_b$

Substitute $v_a=v_b$ into the output formula:
$v_o = 10(v_a - v_a) = 0$, so $G_{c,calc}=0$

Step5: Analyze measured gain deviations

For $v_b$ grounded, measured gain $G_{b,mes}=10.1$ (1% error from ideal). For $v_a=v_b$, ideal gain is 0, but measured $G_{c,mes}=0.91$, which is a common deviation from ideal op-amp behavior (non-zero input offset, finite open-loop gain).

Answer:

1. The calculated gains match the derived ideal values:

• $G_{a,calc}=10$ (when $v_a$ grounded)
• $G_{b,calc}=10$ (when $v_b$ grounded)
• $G_{c,calc}=0$ (when $v_a=v_b$)

1. The measured gains show small deviations from ideal behavior:

• $G_{a,mes}=10$ (matches ideal)
• $G_{b,mes}=10.1$ (1% positive error)
• $G_{c,mes}=0.91$ (non-zero due to real op-amp imperfections like input offset voltage)

1. The ideal output function remains: $v_o = 10(v_a - v_b)$