14.4 CHAPTER 14. ELECTRONICS
Extension: Calculating the gain of the amplifier in Figure 14.16.
- The input resistanceof the operational amplifier is very high. This means
that very little current flows into the inverting input of the op-amp. Ac-
cordingly, the current through resistor R 1 must be almost the same as the
current through resistor R 2. This means that the ratio of the voltage across
R 1 to the voltage across R 2 is the same as the ratio of the two resistances. - The open loop gain A of the op-amp is very high. Assuming that the out-
put voltage is less than afew volts, this means that the two input terminals
must be at very similar voltages. We shall assume that they are at the same
voltage. - We want the outputvoltage to be zero if the input voltage is zero.As-
suming that the transistors within the op-amp arevery similar, the output
voltage will only be zerofor zero input voltage if V+is very close to zero.
We shall assume that V+= 0 when the trimming potentiometer is cor-
rectly adjusted. - It follows from the last two statements that V−≈ 0 , and we shall assume
that it is zero. - With these assumptions, the voltage across R 2 is the same as Vout, and the
voltage across R 1 is the same as Vin. Since both resistors carry the same
current (as noted in point 1), we may say that themagnitude of Vout/Vin=
R 2 /R 1. However, if Vinis negative, then Voutwill be positive. Therefore
it is customary to write the gain of this circuit as Vout/Vin=−R 2 /R 1.
Exercise 14 - 6
- What are operational amplifiers used for?
- Draw a simple diagram of an operational amplifier and label its terminals.
- Why is a trimming potentiometer needed when using an op-amp?
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