Linear Inverting Amplifier Help

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SUMMARY

The linear inverting amplifier discussed has a gain of 35 dB, which translates to a voltage gain (Av) of approximately 5.62 V/V. When the input voltage (vi) is -8.0V, the output voltage (vo) is limited to the maximum output rail of +5.0V due to the amplifier's output range constraints. The calculation of gain must use base 10 logarithms, as clarified in the discussion. The amplifier's behavior confirms that when the input exceeds the output range, the output saturates at the maximum voltage level.

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Homework Statement


A linear inverting amplifier has these characteristics:
-5.0V < vo < +5.0V Av=35 dB
The output rails at -5.0V and +5.0V That is, the amplifier outputs no voltage larger in magnitude than 5.0V.

What is the gain in V/V?
What is the output if vi=-8.0V?


Homework Equations


Av = vo/vi
Av = 20log|Av|


The Attempt at a Solution


Av = 35db
35/20 =log|Av| => Av = e^(7/4) V/V
vo = Av * vi => (-8.0V) * e^(7/4)
but since this is out of the -5V to 5V range, it would just be 5V

Does this look right?
 
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Fizzill said:

Homework Statement


A linear inverting amplifier has these characteristics:
-5.0V < vo < +5.0V Av=35 dB
The output rails at -5.0V and +5.0V That is, the amplifier outputs no voltage larger in magnitude than 5.0V.

What is the gain in V/V?
What is the output if vi=-8.0V?


Homework Equations


Av = vo/vi
Av = 20log|Av|


The Attempt at a Solution


Av = 35db
35/20 =log|Av| => Av = e^(7/4) V/V
Watch it! The log is base 10, not base e. And you should be able to write a numerical value for a result.

vo = Av * vi => (-8.0V) * e^(7/4)
but since this is out of the -5V to 5V range, it would just be 5V
Yup. The input already exceeds the output range and the gain is greater than unity.
 

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