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Rex N. Fisher                                                                               Computer Science and Engineering Department, BYU-Idaho

 Example Lab Report

 

This is an example lab report.  It has five sections, with each one beginning on a new page.  The long horizontal bars represent page breaks.  Specific explanations about each section can be found here. 

 

 

Laboratory # 5

 

Design for a Specific Q Point

 

 

Rick Sparks

 

ECE 355


 

Introduction



Purpose

 

   1.  Practice designing a transistor circuit that meets certain specifications.
   2.  Verify that transistor circuits with negative feedback are not sensitive to changes in beta.
   3.  Troubleshoot problems, without the teacher's help.

 

Equipment

 

   1.  Three 2N3904 NPN transistors
   2.  One 1800 ohm resistor 
   3.  One 330 ohm resistor
   4.  One 820 ohm resistor
   5.  One 270 ohm resistor
   6.  One +20 V DC power supply
   7.  One prototyping board
   8.  22 AWG wire, cut to length and stripped on both ends 

   9.  One DVM with test leads

 

Procedure

 

Design a circuit with a 2N3904 NPN transistor that has Ic = 10 mA and Vce = 10 V. The allowable tolerance of this circuit is ±10%. The circuit must use a single 20 V power source. Prove that the Q point is independent of beta by testing the circuit with three different transistors.

 

   1. Use the "Approximation Method" to calculate the resistor values, starting with the desired Ic and Vce values.

 

             -  Select the desired value for Vb and calculate the required values of R1 and R2 in the voltage divider network.

             -  Select actual resistor values closest to the ones calculated. Determine the expected Vb using the actual resistors for R1

                and R2.

             -  Using the value of Vb from step 1B, calculate Ve and the value of Re required to make Ie = 10 mA.

             -  Select the actual resistor value for Re that is closest to the one calculated. Use this value to determine the expected Ie.

             -  Assume that Ic = Ie. Record this value of Ic as "expected data".

             -  Assume that Ic = Ie. In order for Vce to be 10 V, Vc must be 10 V higher than Ve. Using the expected Ic, calculate the required

                value of Rc.

             -  Select the actual resistor value for Rc that is closest to the one calculated.

             -  Use this value to determine the expected Vce. Record it as "expected data".

 

   2. Construct the circuit and measure the values of Ic and Vce, using three different transistors. Record the values as "measured

       data".

             -  Measure Vce directly with a voltmeter.

             -  Determine Ic by dividing the measured voltage across Rc by the measured resistance of Rc.

 

Schematic Diagrams


Schematic Diagram

 

 

 

Experiment Data



 

Calculations for voltage-divider network:

    Assume Vcc = 20 V; Ivd = 10 mA; Vb = 3 V

    R2 = 3 V / 10 mA = 300 ; use 330

    R1 = 17 V / 10 mA = 1.7 k; use 1.8 k

    Expected Ivd = 20 V / (330 + 1800 ) = 9.85 mA

    Expected Vb = 330 x 9.85 mA = 3.25 V

 

Calculate Ve, Re, Ie, and Ic:

    Assume Ie = 10 mA

    Ve = 3.25 V - 0.7 V = 2.55 V

    Re = 2.55 V / 10 mA = 255 ; use 270

    Expected Ie = 2.55 V / 270 = 9.44 mA

    Expected Ic = Ie = 9.44 mA (It meets spec: Ic = 10 mA ± 10%)

 

Calculate Rc, Vc, and Vce:

    Assume Vcc = 20 V; Vce = 10 V; Ve = 2.55 V; Ic = 9.44 mA

    Vrc = 20 V - 2.55 V - 10 V = 7.45 V

    Rc = 7.45 V / 9.44 mA = 789 ; use 820

    Expected Vrc = 9.44 mA x 820 = 7.74 V

    Expected Vce = 20 V - 2.55 V - 7.74 V = 9.71 V (It meets spec: Vce = 10 V ± 10%)

 

 

Expected Results (Calculations):

 

                                                       Vce                                           Ic                                         

                                  9.71 V (Vce = 10 V ± 10%)     9.44 mA (Ic = 10 mA ± 10%)

 


Actual Results (Measurements):

 

Transistor #                                 Vce                                            Ic                                         

        1                       9.8 V (Vce = 10 V ± 10%)       9.9 mA (Ic = 10 mA ± 10%)

        2                       9.6 V (Vce = 10 V ± 10%)       9.7 mA (Ic = 10 mA ± 10%)

        3                       9.7 V (Vce = 10 V ± 10%)       9.8 mA (Ic = 10 mA ± 10%)

 

Discussion & Conclusions 



The measured values were well within the 10% tolerance allowed by the design specifications.  They were also within 2% of the calculated values.  This demonstrates that the "approximation method" used to design this circuit works well when the bias current through R1 and R2 is much larger than the base current.  A ratio of 10:1 or greater for these two two currents works best.  If R1 and R2 had been 10 times larger, a more accurate calculation method would have been required.

 

There was a slight difference in measured values for the three transistors.  This is because the beta of each one is quite different.  This shows that a large difference in beta corresponds to a very small difference in the Q point for this circuit configuration.  Although this voltage-divider-biased circuit (a form of emitter-biasing) is very stable over a wide range of beta, the difference in beta does have some effect.

 

This circuit uses the emitter resistor as a source of negative feedback.  This results in good Q point stability.  Increasing the amount of current through the voltage-divider network makes it even more stable, because the percentage of current lost to Ib becomes smaller, and there is less of an effect on Vb.  Beta affects Vb by changing Ib, which changes the amount of current through the voltage-divider.

 

This circuit also requires only one power supply, while some other emitter-biased circuits require two voltage sources.  It does waste power in the voltage-divider network, however.  It would probably not be a good choice if the power source was a battery or solar cell.