Question

In inverting amplifiers \[{R_{in}} = 1k\Omega \], \[{R_f} = 5k\Omega \]. Calculate the output voltage if input voltage is 50 mV.

Answer 1

Hint: The voltage gain of the amplifier is defined as the ratio of the output voltage to the input voltage. This ratio is also equal to the ratio of the resistance across the amplifier to the resistance at the input of the amplifier
Formula used: In this solution we will be using the following formulae;
\[{A_v} = \dfrac{{{V_{out}}}}{{{V_{in}}}}\] where \[{A_v}\] is the voltage gain of the inverting amplifier, \[{V_{out}}\] is the voltage at the output terminal of the amplifier, and \[{V_{in}}\] is the voltage at the input terminal.
\[\dfrac{{{V_{out}}}}{{{V_{in}}}} = \dfrac{{{R_f}}}{{{R_{in}}}}\] where \[{R_f}\] is resistance across the amplifier, and \[{R_{in}}\] is the resistance at the input terminal of the amplifier.

Complete step by step solution:
Given the input, we are told to determine the output voltage of an amplifier. Generally, an inverting amplifier is used to generate a larger voltage at the output than at the input.
The voltage gain is defined as
\[{A_v} = \dfrac{{{V_{out}}}}{{{V_{in}}}}\] where \[{A_v}\] is the voltage gain of the inverting amplifier, \[{V_{out}}\] is the voltage at the output terminal of the amplifier, and \[{V_{in}}\] is the voltage at the input terminal.
Based on analysis, the voltage gain can also be proven to be equal to be
\[{A_v} = \dfrac{{{V_{out}}}}{{{V_{in}}}} = \dfrac{{{R_f}}}{{{R_{in}}}}\]where \[{R_f}\] is resistance across the amplifier, and \[{R_{in}}\] is the resistance at the input terminal of the amplifier.
Hence, to calculate the output voltage we multiply through by the input voltage as in
\[\dfrac{{{V_{out}}}}{{{V_{in}}}} = \dfrac{{{R_f}}}{{{R_{in}}}}\]
\[ \Rightarrow {V_{out}} = \dfrac{{{R_f}}}{{{R_{in}}}}{V_{in}}\]
Hence, by inputting all known values, we have
\[{V_{out}} = \dfrac{{5k\Omega }}{{1k\Omega }}50mV\]
\[ \Rightarrow {V_{out}} = 250mV\] or \[0.25V\]

Note: For clarity, conversion of units for the resistance to SI is not necessary because the unit cancels out, this means that all the conversion factors will also cancel. However the unit must be the same for the ratio to be done properly.