P 6.4-1 Determine the node voltages for the circuit shown in Figure P 6.4-1.
Answer: va = 2 V, vb = – 0.25 V, vc = – 5 V, vd = – 2.5 V, and ve = – 0.25 V
Figure P 6.4-1
From the op-amp rules, we know
𝒗𝒃 = 𝒗𝒆
We cannot use node d for voltage node equation directly as the op-amp is an unknown
current source or sink, the equation above is the result of the presence of the op-amp.
Now we have two voltage sources so we have two equations that are
𝒗𝒂 = +𝟐 𝑽
𝒗𝒄 = −𝟓 𝑽
Now we can write voltage node equations for 𝒗𝒃 , and 𝒗𝒆
𝟏
𝟏
𝟏
𝟏
𝟏
𝟎 = 𝒗𝒃 (
+
+
) − 𝒗𝒂 (
) − 𝒗𝒄 (
)
𝟐𝟎 𝒌𝛀 𝟒𝟎 𝒌𝛀 𝟒𝟎 𝒌𝛀
𝟐𝟎 𝒌𝛀
𝟒𝟎 𝒌𝛀
and
𝟏
𝟏
𝟏
𝟎 = 𝒗𝒆 (
+
) − 𝒗𝒅 (
)
𝟗 𝒌𝛀 𝟏 𝒌𝛀
𝟗 𝒌𝛀
Using the first of these two node equations we can plug in the known values of 𝒗𝒂 and 𝒗𝒄
𝟏
𝟏
𝟏
𝟏
𝟏
𝟎 = 𝒗𝒃 (
+
+
) − 𝒗𝒂 (
) − 𝒗𝒄 (
)
𝟐𝟎 𝒌𝛀 𝟒𝟎 𝒌𝛀 𝟒𝟎 𝒌𝛀
𝟐𝟎 𝒌𝛀
𝟒𝟎 𝒌𝛀
Multiply by 40 kΩ
𝟎 = 𝒗𝒃 (𝟐 + 𝟏 + 𝟏) − 𝒗𝒂 (𝟐) − 𝒗𝒄 (𝟏) = 𝟒𝒗𝒃 − (𝟐 𝑽)𝟐 − (−𝟓 𝑽) = 𝟒𝒗𝒃 − 𝟒 + 𝟓
𝟒𝒗𝒃 = −𝟏 𝑽
𝟏𝑽
= −𝟎. 𝟐𝟓 𝑽 = 𝒗𝒆
𝟒
From our op-amp equation we know 𝒗𝒆 = 𝒗𝒃 now we can use the other node equation to
find 𝒗𝒅 .
𝟏
𝟏
𝟏
𝟎 = 𝒗𝒆 (
+
) − 𝒗𝒅 (
)
𝟗 𝒌𝛀 𝟏 𝒌𝛀
𝟗 𝒌𝛀
𝒗𝒃 =
𝟏
𝟏
𝟏
𝒗𝒅 (
) = 𝒗𝒆 (
+
)
𝟗 𝒌𝛀
𝟗 𝒌𝛀 𝟏 𝒌𝛀
𝒗𝒅 = 𝒗𝒆 (𝟏 + 𝟗) = 𝟏𝟎 𝒗𝒆 = 𝟏𝟎(−𝟎. 𝟐𝟓 𝑽) = −𝟐. 𝟓 𝑽
𝒗𝒂 = +𝟐 𝑽
𝒗𝒃 = −𝟎. 𝟐𝟓 𝑽
𝒗𝒄 = −𝟓 𝑽
𝒗𝒅 = −𝟐. 𝟓 𝑽
𝒗𝒆 = −𝟎. 𝟐𝟓 𝑽
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This page last updated on February 18, 2021