The current i flowing into the op amp input is given by i = ν d/ R i = 10 μV/1 MΩ = 0.00001 μA = 10 pA.
10 5 = 10 μV, which is a very small voltage and is usually neglected.įrom (6.3), we have that ν i = (− R 1/ R F)ν o = (−1/100)(−5) = 50 mV.Assuming ν o = − 5 V, we obtain for the differential voltage ν d = −5 V/− A = 5/5 Any figure displaying a load line such as 4.13 b or 4.17 c shows that ν o,max ≈ ν powersupply. The maximum magnitude of the output voltage ν o is limited by the supply or battery voltage (it usually is less by about 2 V). Ideally, a high-impedance voltage source should work into an infinite-impedance voltage amplifier.
In addition, a feeble source might not be able to provide the large currents which a low input impedance requires. This is a rather low input impedance (in practice the term impedance is used when addressing any kind of input resistance), not suitable when a high-impedance source is to be connected to the input terminals, as then only a small portion of the source voltage would be driving the amplifier. The input resistance (it is the resistance that a source would see when connected to the ν i input terminals) is simply R = ν i/ i 1 = R 1 = 1 kΩ. The minus sign in the gain expression implies that the amplified output signal is 180 ° out of phase with the input signal. The gain of the op amp with external circuitry depends only on the external resistors that are connected to the op amp.
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