I would be at a loss to explain how this circuit works. The Emitter-Base of the transistor is used as the feedback circuit. Can somebody put into 100 words or less the details of why this works?
I can see how the input voltage is developed across the sensing resistor. I recognize this as an inverting amplifier. The more positive voltage is applied to the inverting input so the output is going to go lower, increasing conduction in the transistor, thus giving a higher voltage out.
???? How do you calculate gain when the feedback is part of the transistor?
When I try to explain an op amp operation to myself like this I try to explain that the voltage applied to the non-inverting input drives the output to the point where the inverting input equals the non-inverting input. In this case the more negative voltage applied to the non-inverting input drives the output low. But the E-B junction is not a fixed resistance. I get lost right there.
This circuit isn't covered in the op amp tutorials here.
I can see how the input voltage is developed across the sensing resistor. I recognize this as an inverting amplifier. The more positive voltage is applied to the inverting input so the output is going to go lower, increasing conduction in the transistor, thus giving a higher voltage out.
???? How do you calculate gain when the feedback is part of the transistor?
When I try to explain an op amp operation to myself like this I try to explain that the voltage applied to the non-inverting input drives the output to the point where the inverting input equals the non-inverting input. In this case the more negative voltage applied to the non-inverting input drives the output low. But the E-B junction is not a fixed resistance. I get lost right there.
This circuit isn't covered in the op amp tutorials here.
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