MASSACHUSETTS INSTITUTE OF TECHNOLOGY 22 071 6 071 Introduction to Electronics Signals and Measurement Spring 2006 Laboratory 9 Dependent sources Op Amps Current Source We have been introduced to dependent sources and have analyzed the following circuit The dependent source in the above circuit is a voltage controlled voltage source whose output is proportional to Vin with a constant of proportionality A 1 Additionally we know that resistance Rin is large M in practical devices and resistance Rout is small 10s The large value of the resistance Rin implies that the amount of current flowing through Rin is very small zero in the ideal case If we assume that the input voltage Vin corresponds to an output of a transducer then we see that this circuit does not demand the delivery of power from the sensor circuit In other words our transducer device whose goal is to just give an electrical signal that is related to some physical quantity is buffered protected isolated from the demands of the measuring circuit The part of the circuit enclosed by the dashed lines is a device called the Operational Amplifier op amp The schematic of an op amp is 22 071 6 071 Chaniotakis and Cory Spring 2006 Page1 VDD Vout Vin VSS The terminals labeled VDD and VSS are the positive and negative power supply connections for the device In this class we will use the LF356 op amp The 356 is a good cheap easy to work with op amp well suited to most basic circuits More expensive op amps tend to sacrifice opamp ideality finite input current stability power supply reject etc for specific purposes high speed low noise etc and can be much more difficult to work with The data sheet for the 356 is available in the class web site The pinout of the 356 is the same as that of another widely used op amp the 741 and is shown below We will use this op amp to construct and characterize a current source 22 071 6 071 Chaniotakis and Cory Spring 2006 Page2 Current Source and Power Transfer Let s investigate the following circuit Vout As you recall from the class notes Vout Vin R RL and the current through resistor RL R is I Vin R Notice that the current I provided by this circuit does not depend on the value of the resistor RL the load resistor The amount of current flowing through the load is determined by the value of resistor R and the input voltage Vin Build the circuit shown above The circuit to the left of the vertical line AB provides the input voltage Vin for your current source For the voltages VDD and VSS use 15V and 15V respectively You have just built a fundamental current source as indicated below VDD 20k VB 5V Load Load I RL VSS I 22 071 6 071 Chaniotakis and Cory Spring 2006 RL R Page3 Perform the appropriate measurements calculations and complete the table below Vin 1 1 5 5 5 RL R 1k 1k 10k 1k 1k 1k 1k 10k 10k 1k Vout measured I calculated I measured Now that we have constructed a current source let s use it to investigate the simple current divider and related power transfer issues We will simply replace the load resistor RL with two resistors connected in parallel as shown on the circuit below One of these resistors R2 is the variable 20k resistor For R1 use a resistor in the range 1k to 10k VDD 20k VB 5V Vin I VSS R1 I R2 R Select appropriate values for all relevant circuit parameters in order to generate a current of 10mA List the parameters and their values 22 071 6 071 Chaniotakis and Cory Spring 2006 Page4 Next let s make the connections indicated in the schematic shown below ACH0 ACH1 VDD 20k VB 5V Vin I VSS ACH0ACH1ACH2 R1 I Vout R2 R ACH2 The channel ACH2 measures the current I while ACH0 and ACH1 measure the voltage Vin and Vout respectively From the class web site download and open the virtual instrument Current Divider 01 Vary the value of resistance R2 by trimming the potentiometer Plot and comment on the power dissipation characteristics of the current divider 22 071 6 071 Chaniotakis and Cory Spring 2006 Page5 Answer the following 1 Give expressions for the current through resistors R1 and R2 2 Give expressions for the power dissipated in resistors R1 and R2 3 From your experimental results you should observe that as R 2 0 the power dissipated in R1 PR1 and the power dissipated in R2 PR2 0 Explain this result 4 Why is there a maximum in PR2 but not in PR1 22 071 6 071 Chaniotakis and Cory Spring 2006 Page6 Equivalent Circuit for extra credit We will next perform a set a simple measurements in order to determine the Thevenin equivalent circuit seen by resistor RL VDD 20k VB 5V Vin I A R1 VSS RL B I R In order to determine the Thevenin equivalent circuit we must calculate measure the equivalent resistance RTH and the equivalent voltage VTH The resulting equivalent circuit across terminals A and B will have the form VTH is the open circuit voltage across terminals AB and RTH the equivalent resistance is the ratio of VTH to the short circuit current I SC through AB the Norton current V RTH TH I SC 22 071 6 071 Chaniotakis and Cory Spring 2006 Page7 You are free to use any values for the resistors R1 and R provided that you do not violate the specifications of the operational amplifier Schematically the measurements are performed as indicated below VDD 20k VB 5V VDD 20k VB Vin A VSS R1 5V VTH Vin A VSS R1 B R Isc B R Draw a graph of the IV characteristic curve of your equivalent circuit Indicate the values of ISC and VTH on the above graph Determine the operating point for RL 1k RL 10k 22 071 6 071 Chaniotakis and Cory Spring 2006 Page8