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I am trying to find Vout in the following image

enter image description here

What I have found from searching a bit is that the op-amps v- and v+ are supposed to be equal to each other.

$$ V^- = V^+ $$

In the case above I run in to the issue that what ever the op-amps range(-15v to 15v) is equal to Vout. This means that the statement above is not true.

If someone could give a break down of this circuit, it would be much appreciated.

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  • \$\begingroup\$ Vout is high (at the positive rail). \$\endgroup\$ – Andy aka Oct 1 '15 at 7:10
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The inputs of an op-amp with negative feedback are assumed to be equal to each other. Without it, the assumption that \$V^+ = V^-\$ produces an unstable answer.

The circuit you have is using the op-amp with positive feedback and thus \$V^+\$ doesn't have to equal \$V^-\$.

Instead, it is an analog comparator with hysteresis. When \$V^+ >V^-\$, the output swings to \$V_s^+\$, otherwise the output swings to \$V_s^-\$.

the \$12 \Omega\$ resistor is there to add hysteresis. This prevents the output from switching right at the \$V^+ = V^-\$ mark; instead, it must cross a certain threshold beyond this mark to switch. More information about hysteresis can be found in this article from Analog Devices.

As a first step, find \$V^-\$. Then test the two possible cases for \$V_{out}\$. One of these configurations will produce a consistent answer with the operation of the analog comparator. If both configurations are consistent or close to consistent within the hysteresis threshold voltage, then you need to know the initial output voltage otherwise \$V_{out}\$ can be either \$V_s^+\$ or \$V_s^-\$ and there's no way to tell which it is.

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