Lab Report 6

Lab Report 6 - Scott Smith ENEE306-0104 12 December 2005...

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Scott Smith ENEE306-0104 12 December 2005 Lab Report 6 Design and Build Your Own Op-Amp Introduction to Op-Amps The ideal op-amp has some good characteristics. Input impedance is supposed to be infinite. Output impedance is supposed to be zero. The output is the amplification of the difference of the input signals. This property is called common mode rejection. Any noise or DC bias that is common to both signals will be “filtered out” at the output. As will be shown later in this report, the op-amp has infinite open loop gain, v A . The gain can be controlled with a feedback circuit that sends a portion of the output voltage back to the input. The different types of feedback will be discussed in this report. The Current Mirror Figure 1 Op-amps use a current mirror, shown in figure 1, as an active load. The current mirror is composed of two matched transistors, 1 Q and 2 Q , with the bases connected. 2 1 BE BE V V = (1) Since the current at the collector of a BJT is represented as T BE V V S C e I I = (2) Then the collector currents will be the same since the base-emitter voltages are the same. 2 1 C C I I = (3) R 1 R 2 V CC Q 2 Q 1

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Smith, 0104 This is the point of the current mirror. 1 C I is set to a certain value by biasing 1 Q . Then 2 C I becomes the same value no matter what value is chosen for 2 R as long as the BJT’s do not saturate. To set the value of 1 C I , use equation (4). 1 1 1 R V V I BE CC C - = (4) Experiment: Simple BJT Current Source Step 1 : Use the value = k R 10 1 and V V CC 7 . 10 = with figure 1. Verify that mA I C 1 2 2245 for = k k R 5 ; 1 ; 100 2 . R 2 (Ω) V R ( V ) I C 2 ( mA ) 100 0.1205 1.205 1000 1.205 1.205 5000 5.670 1.134 Table 1 The values for 2 C I are within 2% of 1 mA . This is an acceptable verification. Step 2 : Design a current mirror which acts as a constant source providing 2 mA . Keep the supply voltage at 10.7 V . Using equation (4), = - = k mA V V R 5 2 7 . 0 7 . 10 1 (5) We constructed and tested the circuit figure 1 with equation (5) and measured the current to be 2.48 mA . This means we have an error of 2.4% which is acceptable. Step 3 : Try to determine a method to measure the output resistance of the current source. The method we decided to use is as follows. Vary 2 R and measure 2 C V . Then calculate the current through 2 R in each case. Then the output resistance is the change in voltage over the change in current. 2 2 C C out I V R = (6) 2
Smith, 0104 R 2 (Ω) C 2 ( V ) I C 2 ( mA ) 100 10.48 104.8 1000 8.20 8.2 Table 2 = = - - = 6 . 23 6 . 96 28 . 2 2 . 8 8 . 104 2 . 8 48 . 10 mA V mA mA V V R out (7) This measurement of output resistance confirms that the output impedance is very small. This measurement is consistent with theory. The ideal op-amp has output impedance of zero, but there is no such op-amp in real life. Step 4 : Design and build a differential amp with single ended gain of approximately 100. Start by designing a current mirror that provides 2 mA . Then choose appropriate values for C R to give the required gain.

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This note was uploaded on 05/04/2008 for the course ENEE 306 taught by Professor Goldsman during the Spring '04 term at Maryland.

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Lab Report 6 - Scott Smith ENEE306-0104 12 December 2005...

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