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# Design of a Single Transistor DC Current Source

Lab 4–Design of a Single Transistor DC Current Source

Perform lab in class during the week of Feb. 18-22, 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)

Wire Jumper Kit

LED Collection

Problem Definition and Background Theory

Goal: Design a current source which provides a fixed DC current through a variable load resistance, RL.  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:

VCC = DC power supply

RL= Load resistance of the current source

IB = DC base current

IC = DC collector current

IE = DC emitter current

VB = DC base voltage

VC = DC collector voltage

VE = DC emitter voltage

VBE = DC base-emitter voltage

VCE = DC collector-emitter voltage

In active operation, the transistor has the following properties:

VBE = 0.7 volts and IC =βIB

where β is the DC current gain of the transistor.

Kirchhoff’s Current Law gives us several current equations:

IE =IB +IC         I1 =IB +I2

Kirchhoff’s Voltage Law gives us several voltage equations:

VB =V+VBE            VC =V+VCE        VCC =VL + VC

Ohm’s Law gives us several resistor equations:

VE =IE RE                                            VL =IC RL         VB =I2 R2

Power equations include:

PT =VCE IC = power dissipated in the transistor in Watts

PRE=VE IE = power dissipated in RE in Watts

PRL =VL IC= power dissipated in RL in Watts

PR1=(VCC -VB)I1 = power dissipated in R1 in Watts

PR2 =VB I2 = power dissipated in R2 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, VCC.  This fixed current value is the same as IC , the DC collector current.

• Derive values for R1, R2, RE, 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 ChallengeDesign 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 IC , the DC collector current.  We will leave IC 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 R1, R2, RE, in terms of VCC and IC. 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 VE, allowing for a wide range of VL.  Next we calculate the emitter voltage from KVL:   V=VB -VBE.

Assuming active operation, we have volts and

(2)

Substitute (1) into (2) to get:

(3)

Next we manipulate equation (3) and the KVL equations:

VC =V+VCE                     VCC =VL + VC                                                        (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 IC as RL changes.  This is a good thing.  Notice that we are allocated about 2/3 of VCC as the range of .  This is also pretty decent.   As we increase RL 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______________

1. (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

1. 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
2. a) as a function of ,  , and β.

1. 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 .

1. c) as a function of ,  , and β.

1. (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:

PT(max)          PREPRL(max)         P1             PR2

PT(max) =maximum power dissipated in the transistor

PRE= power dissipated in RE

PRL(max) =maximum power dissipated in RL

P1 =power dissipated in R1

PR2 =power dissipated in R2

…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 Pre-Lab.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 biasedR = reverse biased

Vc = Vcc-V_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 = _____(6.7-6.56)/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:

1. In step 8, you heated the transistor with a soldering iron and observed the collector current. Recalling from your knowledge of the properties of semi-conductors, what is the effect on conductivity when a semiconductor is heated?

What effect will this have on the transistor currents in a transistor circuit?

1. 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?

1. Plot a current source regulation curve consisting of measured ICvalues on the y-axis and measured VLvalues on the x-axis. Use Excel or MATLAB.  On the graph, draw a vertical line that separates the active region from the region of saturation.

1. 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?

1. Generate an LTspice simulation for RL= 0W. From the simulation, record all DC currents, and voltages in Table 3.  Paste a screen shot of the simulation below:

1. Repeat step 6 for RL=RL(max), as calculated in the Pre-lab. From the simulation, record all DC currents and voltages in Table 3.

1. Repeat step 6 for IC=9 mA. From the simulation, record all DC currents and voltages in Table 3.

1. Table 1: Compare the load resistor power (at VL= 14 V) with the prelab results.  How close are the two?

1. 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 prelabRLmax 20 9 V V V mA V mA mA mA Ω

Last Updated on March 4, 2019

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