Lab 4–Design of a Single Transistor DC Current Source
Perform lab in class during the week of Feb. 1822, 2019
Group Lab Report
Purpose: To measure and analyze the characteristics of current sources.
To measure and analyze the characteristics of a voltage divider used for biasing transistor circuits.
Equipment: Dual DC power supply
2N3904NPN transistor
To Be Determinedresistors
10 kW potentiometer
Two Digital Multimeters (DMM)
Breadboard
Wire Jumper Kit
Connecting leads
LED Collection
Problem Definition and Background Theory
Goal: Design a current source which provides a fixed DC current through a variable load resistance, R_{L}. The current will remain fixed over a voltage range of 0 volts to some defined maximum DC voltage.
The basic layout of the Transistor Current Source Circuit is shown in Figure 1. The parameters are:
V_{CC} = DC power supply
R_{L}= Load resistance of the current source
V_{L }= DC load voltage
I_{B }= DC base current
I_{C }= DC collector current
I_{E }= DC emitter current
V_{B }= DC base voltage
V_{C }= DC collector voltage
V_{E }= DC emitter voltage
V_{BE }= DC baseemitter voltage
V_{CE }= DC collectoremitter voltage
In active operation, the transistor has the following properties:
V_{BE }= 0.7 volts and I_{C }=βI_{B}
_{ }
where β is the DC current gain of the transistor.
Kirchhoff’s Current Law gives us several current equations:
I_{E }=I_{B }+I_{C }I_{1 }=I_{B }+I_{2}
Kirchhoff’s Voltage Law gives us several voltage equations:
V_{B }=V_{E }+V_{BE }V_{C }=V_{E }+V_{CE }V_{CC }=V_{L }+ V_{C }
_{ }
Ohm’s Law gives us several resistor equations:
V_{E }=I_{E }R_{E }V_{L }=I_{C }R_{L }V_{B }=I_{2 }R_{2}
Power equations include:
P_{T }=V_{CE }I_{C }= power dissipated in the transistor in Watts
P_{RE}=V_{E }I_{E }= power dissipated in R_{E} in Watts
P_{RL }=V_{L }I_{C}= power dissipated in R_{L} in Watts
P_{R1}=(V_{CC }V_{B})I_{1 }= power dissipated in R_{1} in Watts
P_{R2 }=V_{B }I_{2 }= power dissipated in R_{2} in Watts
The equivalent resistance looking into the base is
The design process starts by choosing a DC fixed current for the current source and a DC voltage for the power supply, V_{CC}. This fixed current value is the same as I_{C }, the DC collector current.
The design tasks are to:
 Derive values for R_{1}, R_{2}, R_{E}, to produce the desired fixed current value.
 Compute the transistor and resistor powers to make sure that the transistor and resistors can handle the power expected.
Design Challenge Design a DC current source using the basic circuit in Figure 1. Constraints: You are limited to a fixed DC power supply voltage. You must use only one BJT transistor.
The customer would like a fixed current value of I_{C }, the DC collector current. We will leave I_{C }as a variable for now.
Desirable characteristics: We would like the current source to operate at full current over a wide range of voltage as possible.
The design tasks are to: 1) Derive values for R_{1}, R_{2}, R_{E}, in terms of V_{CC} and I_{C}.
2) Compute the transistor and resistor powers to make sure that the transistor and resistors can handle the power expected.

We begin by setting
(1)
This will produce a relatively low V_{E}, allowing for a wide range of V_{L}. Next we calculate the emitter voltage from KVL: V_{E }=V_{B} V_{BE}.
Assuming active operation, we have volts and
(2)
Substitute (1) into (2) to get:
(3)
Next we manipulate equation (3) and the KVL equations:
V_{C }=V_{E }+V_{CE }V_{CC }=V_{L }+ V_{C } (4)
_{ }
to get:
(5)
Next we compute the maximum allowable load voltage. This occurs when is at a minimum value. is minimized when the transistor is operating in the saturation region. In the saturation region, both PN junctions are in forward mode and volts. To summarize, the maximum load voltage is:
_{ }
where volts. We substitute volts to get:
(6)
We now have a current source, equivalent to figure 2, in which the current is fixed at I_{C} as R_{L} changes. This is a good thing. Notice that we are allocated about 2/3 of V_{CC} as the range of . This is also pretty decent. As we increase R_{L} from 0 Ohms, the load voltage will change from 0 volts to as defined in equation (6).
The maximum load resistance comes from Ohm’s Law:
(7)
Figure 1 –Transistor Current Source Circuit. 
Figure 2 –Equivalent Circuit

Prelab – Design Calculations for the DC Current SourceName______________
 (10 pts) Using the background information in part 1, and the approximation that , derive the following equation for as a function of , , and β. Hint: Substitute , and into
 Using the answer to question 1 and the background information in part 1, in which the base voltage is designed for , and is designed for , derive equations for
 a) as a function of , , and β.
 b) (10 pts) as a function of , , and β using . The goal of a stable current source is to have a stable base voltage, in which the transistor does not load down the circuit. If we remove the transistor, the base voltage should not change. To do this, we set to target approximately a tolerance of 5%, we set .
 c) as a function of , , and β.
 (50 pts) Given the answers to question 2, design a 10 mA transistor current source with a supply voltage of volts, and β =150. Show all calculations for the following quantities:
P_{T(max) }P_{RE}P_{RL(max) }P_{1 }P_{R2}
P_{T(max) }=maximum power dissipated in the transistor
P_{RE}= power dissipated in R_{E}
P_{RL(max) }=maximum power dissipated in R_{L}
P_{1 }=power dissipated in R_{1}
P_{R2 }=power dissipated in R_{2}
…and draw a schematic diagram of the final circuit with all currents and node voltages indicated on the sketch.
Lab Instructions
Team members:_________________________________
Part 1 – Construct Current Source Circuit
 □On the breadboard, construct the 10 mAvoltagedivider transistor circuit that you designed in the PreLab.For this initial step, use a load resistor of RL= 0W.
With RL = 0 W, what should VC be? ________
What should VB be approximately? ________ (refer to the prelab)
What should VE be approximately? ________ (refer to the prelab)
 □Turn on the power supply and set the voltage to 20 volts and observe the current value displayed on power supply. There should be a source current in the mA range.
If the source current is 1.000 Amps, then there is a short circuit connection… if the source current is 0.000 Amps, then there is an open circuit … In either case, turn off the power supply, debug the circuit, and try again.
When the source current is in the mA range, record the source current, measure the transistor voltages and record below:
Source current = __0.011A____________
VB = ___6.56V_______ , VC = ____19.8V______ , VE = ______5.83V____
 □Compare the measured VBand VEabove with your Prelab calculations. They should differ by about 0.7 volts. If they are way off, then you will need to debug the circuit. Debugging tips:The high sides of RLand R1 should both be about 20 V.So you can use a DMM set up as a DC voltmeter to check these test points. The low sides of R2and RE should be connected to ground, so the voltages at these points should both be 0 V. You can use a voltmeter to check these test points. If any of these four voltage test points are off, then there is probably a bad connection somewhere. To find a bad connection, use the voltmeter to trace the faulty test point back to the power supply. If this still does not help, then ask the instructor for assistance.
 □ Now that the circuit is operation properly, set up the DMM to measure collector current, IC.
Measured IC = ____9.52______ mA
What should IC be according to the prelab? _____10_____ mA
 □Show the circuit to the instructor before continuing.
Instructor sign off: ____RDA 2/19/19__________
Part 2 – Testing of the Current Source
A current source should be able to deliver a fixed current over a range of load resistance. In this section, the BJT current source is tested to determine the range of operation and % regulation of the current source.
 □Fill in Table 1 with measurements of VB, VE, VC, and ICfor values of VLranging from 0 to14 volts. You will need to adjust the variable resistor to change the value of the load voltage. For the last row of table 1, adjust the variable resistor until the collector current drops to 9 mA.
Table 1 – Measured and Calculated Results of Current Source
Measured  Calculated  Junction Bias  
VCC  VB  VE  VL  IC  RL  PT  PL  VBE  VBC  VCE  BE  BC 
20  6.56V  5.83
 0  9.52  0  0  
20  6.56v  5.8  2  9.62  
20  6.5V  5.86  4  9.61  
20  6.55  5.85  6  9.59  
20  6.55  5.84  8  9.56  
20  6.55  5.84  10  9.56  
20  6.54  5.82  12  9.53  
20  6.38  5.66  14  9.23  
20  6.26  5.54  14.1  9  
20  5.74  5.01  14.7  8  
V  V  V  V  mA  W  mW  mW  V  V  V  F or R?  
F= forward biased R = reverse biased 
Vc = VccV_L
P_T = VCE * IC
VCE= VC – VE
 □ You should notice that the collector current (IC) remains fairly steady as RL(and VL) increases, but then at some point IC drops significantly. How high does VLget before the current start to drop significantly?
VLmax (experimental)= _____14____. How close is this to the prelab calculation? ___13.8___________
Pull out RL and measured its resistance. RLmax (experimental)= __1.85Kohm_______.
How close is this to the prelab calculation? __________1.384kohms____
 □ Calculate the indicated quantities in Table 1, based on the measured values. Fill in Table 1 with these calculations.
 □ Replace the potentiometer with an LED. Observe the brightness. Fill in Table 2 for 1 LED.
 □ Repeat for 2, 3, 4, 5, 6, as many LED’s in series as needed until the brightness diminishes. Fill in the rest of Table 2.
Table 2 – Test of Current Source using Series LED’s as the Load
VCC  # of LED’s in series  VL  Brightness is the same or getting dimmer? 
20  1  2.8  same 
20  2  5.7  same 
20  3  5.7  same 
20  4  9.46  same 
20  5  11.3  same 
20  6  13.1  same 
20  7  14.9  same 
20  
20 
 □ Pull out the transistor and measure the voltage at the base test point.
VB with transistor removed = ___6.7V_______ ,
Calculate the percent change in VBbetween no transistor and transistor in with RL = 1kW.
% change = _____(6.76.56)/6.7*100=________
 □ Insert the 1kW resistor for RLand apply a hot soldering iron to the transistor and observe the collector current.
Does the current increase or decrease when the transistor is heated? ________increase___________.
When you remove the soldering iron, does the current increase or decrease? ________decrease___________.
Post Lab Questions:
 In step 8, you heated the transistor with a soldering iron and observed the collector current. Recalling from your knowledge of the properties of semiconductors, what is the effect on conductivity when a semiconductor is heated?
What effect will this have on the transistor currents in a transistor circuit?
 From the data in Table 1, for what range of load voltage was the transistor in saturation?
For what range of load voltage was the transistor in active operation?
Would you say that the transistor is a more stable current source in active mode or in saturation?
 Plot a current source regulation curve consisting of measured ICvalues on the yaxis and measured VLvalues on the xaxis. Use Excel or MATLAB. On the graph, draw a vertical line that separates the active region from the region of saturation.
 One way to have a stable base voltage is to create a voltage divider such that REQ»(b+1)REis much larger than R2, so that the parallel combination of REQand R2 is approximatelyR Assuming b = 150, calculate REQ for the circuit used in this lab. Does this result suggest a stable base voltage?
 Generate an LTspice simulation for R_{L}= 0W. From the simulation, record all DC currents, and voltages in Table 3. Paste a screen shot of the simulation below:

 Repeat step 6 for R_{L}=R_{L(max)}, as calculated in the Prelab. From the simulation, record all DC currents and voltages in Table 3.
 Repeat step 6 for I_{C}=9 mA. From the simulation, record all DC currents and voltages in Table 3.
 Table 1: Compare the load resistor power (at VL= 14 V) with the prelab results. How close are the two?
 Table 1: Compare the transistor resistor power (at VL= 0 V) with the prelab results. How close are the two?
Table 3 –Current Source Simulation in LT Spice
Simulation Results  Calculated  
VCC  VB  VE  IC  VC  IR1  IR2  IRE  RL  PL  PT  VL  
20  0  
20  prelab RLmax
 
20  9  
V  V  V  mA  V  mA  mA  mA  Ω  