modified 7 years ago

Question about EC Op Amps

01:21:15

My question: What is the point of the Op Amp value that can be adjustable(100kV/V)? 

You set the gain with R1 and R2 in a inverting amplifier with the formula Gain=R2/R1+1. In this setup the gain is 3 which led me to the conclusion of Vout(Blue wire)=Vin(Power Source)×Gain.

I'll await your response.

published 7 years ago

abobaker

7 years ago

Do you know how negative feedback works ?

crake

7 years ago

open loop gain

kiani

7 years ago

The op amp has extremely high gain as indicated ( 100kv/v). So negative feed back is used ti rrduce that high gain, to a reasonable value,, so the output is not cliped by the limited supply voltage of the opamp ( +,-15v in this case). Otherwise yiu put 1 v in and the opamp output voltage wants ti gi ti 100 v. So through negative feed back this high gain available can be adjusted to lower gains as required.

kiani

7 years ago

Correction.... You put in 1 v and get 100kv out ( 100kv/v)..... This high gain is the reason for very high input impedance of the opamp,, and the low output impedance.

kiani

7 years ago

https://www.electronics-tutorials.ws/

eekee

7 years ago

it's useful when you want to represent a whole amplifier as a single component within a larger circuit: just turn the gain down to that of the amplifier. the triangle doesn't really represent op-amps as such, just amplifiers in general.

eekee

7 years ago

i guess it could also be used to explore one of the weaknesses of real op-amps, especially with very small signals, but few of the other weaknesses are simulated... *shrug*

sshsslfun

7 years ago

Thanks for the help everyone!!

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