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I am looking at this simplified schematic for a linear regulator. The way this works is the error amplifier drives the base of the power transistor through the current amp. This in turn varies Vce to eventually stabalize Vcc. ,

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I am wondering how its possible for the error amplifier to be working in linear mode without feedback. Is this just a very simplified schematic that doesn't show the feedback involved? Or possibly could it be an op amp with small open loop gain?

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    \$\begingroup\$ It does have feedback -- through the current amplifier, then the transistor and then the voltage divider formed by Ra and Rb. \$\endgroup\$ – Null Apr 11 '17 at 16:49
  • \$\begingroup\$ Ok I see that. I got focused on the op amp itself and didn't see the big picture feedback loop. Thanks! \$\endgroup\$ – Mtk59 Apr 11 '17 at 16:54
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The error amplifier is in a feedback loop. You might miss it if you are looking for local feedback directly between the input and output of the error amplifier, because the loop extends through the current amplifier, the pass device, and the resistive divider formed by RA and RB. As long as this larger loop is stable and has sufficient gain, it will correctly bias the error amplifier in it's linear mode.

Multi-stage feedback loops can employ a variety of combinations of local feedback (around a single stage) and global feedback (around the greater loop) to achieve their requirements for transfer function, stability, and gain. The analysis of some configurations can get very complicated indeed, but the same principles apply as with a simple feedback loop around a single op amp.

In this case, analysis is relatively simple. The current amplifier and pass device both operate (presumably) at approximately unity voltage gain. Essentially, they just buffer the voltage at the output of the error amplifier to VCC. So the first-order small signal analysis proceeds as if RA were simply connected from the output of the amp to it's inverting input.

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