1. The signal conditioner -- changes the voltage Amplify Attenuate Filter

2. Electrical Drawings Symbols Wires are straight lines usually horizontal and vertical Connection points are shown as circles on the end of a wire: Ground is a common connection point from which most voltages are measured: Shown as either a small triangle: or as a set of lines forming a triangle: Resistors are shown as zigzag lines, vertical or horizontal:

3. Operational Amplifiers -- a.k.a. “op-amps” Practical signal amplifiers are frequently constructed from inexpensive, integrated circuit “chips” called operational amplifiers. The circuit symbol for an op-amp is a triangle (see Figure 3.10a). + V p  VoVo  Vn Vn V+ VV

4. Operational Amplifiers -- a.k.a. “op-amps” A circuit containing an op-amp can be used to amplify a weak signal from a transducer. Can we get something for nothing? No! There are two power supply connections, marked V+ and V . These connections are often not shown on circuit diagrams. + V p  VoVo  Vn Vn V+ VV

5. Operational Amplifiers -- a.k.a. “op-amps” The common connection point (ground) at the bottom of the diagram can also be shown as a wire running from left to right. + V p  VoVo  Vn Vn

6. Operational Amplifiers -- a.k.a. “op-amps” The common connection point (ground) at the bottom of the diagram can also be shown as a wire running from left to right. + V p  VoVo  Vn Vn The input voltages (V n and V p ) are applied between two input terminals (labeled + and  ) and ground. The output voltage (V o ) appears between a single output terminal and ground.

7. Operational Amplifiers -- a.k.a. “op-amps” Properties The op-amp is sometimes called a differential amplifier because its output equals its internal gain times the difference between the voltages at the + and  terminals. The internal gain is denoted by the lower case g. V o = g(V p  V n )Eq. (3.10) + V p  VoVo  Vn Vn

8. Operational Amplifiers -- a.k.a. “op-amps” Properties (continued) The internal voltage gain is is very high ã (usually g > 100,000). As a result, V n  V p. Stated another way, the voltage between the + and  input terminals  0. + V p  VoVo  Vn Vn V 

9. Operational Amplifiers -- a.k.a. “op-amps” Properties (continued) The resistance between the input terminals (the input resistance) is very high, usually  1M  As a result, the current entering the  input I n  0. Also, the current entering the  input I p  0. + V p  VoVo  Vn Vn I n  0 I p  0

10. Practical Amplifier Circuits Using Op-Amps Practical amplifier circuits can be constructed by connecting other components (e.g., resistors) to an op-amp. Recall that the power supply connections are usually not shown in circuit diagrams. The gain of a practical amplifier circuit can be calculated by using the previously described properties of an “ideal” op-amp.  Voltage between the + and  input terminals  0.  Current into (or out of) the + and  input terminals  0.

11. Practical Amplifier Circuits Using Op-Amps A simple noninverting amplifier using an op-amp can be constructed as follows: +Vi+Vi VoVo R1R1 R2R2 We will now analyze this circuit (Figure 3.11). Objective: Find gain G in terms of R 1 and R 2.

12. Noninverting Amplifier Using an Op-Amp Apply KCL at junction B: I 1 = I 2 + I n But I n  0, so … I 1 = I 2 +Vi+Vi VoVo I n  0 I p  0 R1R1 R2R2 I1I1 I2I2 B

13. Noninverting Amplifier Using an Op-Amp Apply KVL around loop A: I 1 R 1  0 + V i = 0, so … V  I1I1 I2I2 +  A +Vi+Vi VoVo R1R1 R2R2 + 

14. Noninverting Amplifier Using an Op-Amp Apply KVL around the outer loop: +Vi+Vi VoVo R1R1 R2R2 I1I1 I2I2 +  + 

15. Noninverting Amplifier Using an Op-Amp We now have three equations: Solve these for V o in terms of V i, R 1 and R 2 :

16. Noninverting Amplifier Using an Op-Amp Continuing... So the gain of this noninverting amplifier is... … a Positive Number!

17. Noninverting Amplifier Using an Op-Amp Example: +Vi+Vi VoVo R 1 = 1000  R 2 = 9000 

18. Inverting Amplifier Using an Op-Amp As in Figure 3.13 (and your homework): VoVo R2R2 R1R1   ViVi   For this circuit: … a Negative Number!