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What is meant by inverting amplifier? | ||||||
asked | 1631 views | 1 answers. | |||||||
Explanation of inverting amplifier working principle and circuit diagram using op-amp. | |||||||
InvertingAmplifierOp-ampCircuit | |||||||
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| The basic OP-AMP inverting amplifier is shown in below circuit.
The input voltage [$:]V[/:$] in is applied to the inverting input through the input resistor [$:]R_{in}[/:$]. The non inverting input is grounded. The feedback resistor [$:]R_f[/:$] is connected between the output and the inverting input. Since the input impedance of an op-amp is considered very high, no current can flow into or out of the input terminals. Therefore [$:]I_{in}[/:$] must flow through [$:]R_f[/:$] and is indicated by [$:]I_f[/:$] (the feedback current). Since [$:]R_{in}[/:$] and [$:]R_f[/:$] are in series, then [$:]I_{in}[/:$] = [$:]I_f[/:$]. The voltage between inverting and non-inverting inputs is essentially equal to zero volt. Therefore, the inverting input terminal is also at 0 volt. For this reason the inverting input is said to be at virtual ground. The output voltage ([$:]V_{out}[/:$] ) is taken across [$:]R_f[/:$]. So, [$$:]I_f = - \frac{V_{out}}{R_f}[/:$$] But [$:]I_{in}[/:$] = [$:]I_f[/:$]. So, [$$:]\frac{V_{in}}{R_{in}} = - \frac{V_{out}}{R_f}[/:$$] Rearranging the equation, [$$:]\frac{V_{in}}{V_{out}} = - \frac{R_{in}}{R_f}[/:$$] The voltage gain of an inverting amplifier can be expressed as, [$$:]A_v = - \frac{R_f}{R_{in}}[/:$$] The amplifier gain is the ratio of [$:]R_f[/:$] to [$:]R_{in}[/:$]. Finally, the output voltage can be found by, [$$:]V_{out} = - \frac{R_f}{R_{in}}V_{in}[/:$$] The output voltage is out of phase with the input voltage. | |||||||||||
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