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For example, say that a system requires an input voltage of 5V, however, you only have a source that can provide 1V. The use of a voltage amplifier is obviously the solution to the problem. However, since closed loop systems with negative feedback are ideal for stability etc. and hence are used instead of open loop configurations (except for oscillators), but doesn't a close loop change the dynamics of the entire system? Because the new gain is given as:

enter image description here

So obviously the gain in closed loop is less than the open loop (as expected), but now the device that requires a 5V input cannot run as the output voltage is considerably less (beta is 0.5). I have made a MicroCap-12 simulation to convey what I am trying to ask:

enter image description here

So my question is, how to make thse configurations equivalent such that the closed loop outputs a voltage of 5V? Is it just adding another amplifier to amplify the voltage 1.43V to get to 5V? If so then what is the point of making it closed loop (other than stability)?

Thanks.

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  • \$\begingroup\$ Try R3=1k. But depends on hFE or beta \$\endgroup\$ Jan 17, 2021 at 16:05
  • \$\begingroup\$ I do not understand what you want to achieve. Simply use a modern opamp instead. \$\endgroup\$
    – G36
    Jan 17, 2021 at 16:06
  • \$\begingroup\$ There isn't something in particular that I want to achieve @G36, I am just trying to understand closed and open loop amplifiers, and why bother with closed loop except stability. Thanks \$\endgroup\$
    – Benny
    Jan 17, 2021 at 16:18
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 ok I will try this. Thanks \$\endgroup\$
    – Benny
    Jan 17, 2021 at 16:18
  • \$\begingroup\$ I think, on your side is a small "gap of understanding". ("...closed loop systems with negative feedback are ideal for stability."..). The opposite is true !! \$\endgroup\$
    – LvW
    Jan 17, 2021 at 17:47

1 Answer 1

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So my question is, how to make thse configurations equivalent such that the closed loop outputs a voltage of 5V?

If your gain stage has an open-loop amplification of 5 then, adding negative feedback can never turn a closed-loop gain into something bigger than a gain of 5 it can only reduce it. For instance, with the lower resistor made 0.01 Ω the output isn't quite 5 volts: -

enter image description here

So, make the open-loop gain larger and see what your output voltage becomes. In other words, experiment a bit; you have your simulator set-up and experimenting is so easy given that you are using dynamic DC simulation in micro-cap.

Then make the open-loop gain a lot bigger and see what the output voltage tends to become. What you will find is that as you make the open-loop gain (say) 1 million, the closed-loop gain (with R2 and R3 at 10 kΩ) becomes precisely 2.

Then, you will find that the closed-loop gain is precisely defined by the ratio of R2 to R3 so, if you make R2 four times bigger than R3, your output will be precisely 5 volts: -

enter image description here

Because the formula for an op-amp (with high open loop gain) is: -

$$A_V = 1 +\dfrac{\text{Feedback resistor}}{\text{Grounding resistor}}$$

And, of course, op-amps have a mighty big open-loop gain at DC.

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