Timeline for Why doesn't voltage drop across this resistor when transistor is off?
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| Aug 9, 2018 at 17:09 | history | edited | D.A.S. | CC BY-SA 4.0 |
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| Aug 9, 2018 at 17:03 | comment | added | Big6 | @user334137 Usually the next stage will have a "high" input impedance such that the voltage doesn't drop too much. If the next stage has a high input impedance (compared to Rd), then the effect of the loading is not significant. You are correct, if the loading of the next stage is considerable then there will be a drop in Vo, but chances are the next stage will have a high impedance. | |
| Aug 9, 2018 at 16:50 | comment | added | wanderingmathematician | Thanks for your answer @sixcab. I follow what you are saying, but I am not sure why the book would be assuming no circuit connected at V_out. This section of the book is talking about integrated circuits (and said this circuit was commonly used in IC's after 1980). It would therefore seem safe to assume that there will always be another circuit attached so that V_out of this circuit is V_in of some other circuit. Do we know that this won't change the circuit behavior "too much"? That is, do we know that this voltage inversion switching will still work if we add another circuit onto V_out? | |
| Aug 9, 2018 at 16:28 | history | edited | Big6 | CC BY-SA 4.0 |
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| Aug 9, 2018 at 16:22 | history | answered | Big6 | CC BY-SA 4.0 |