-

samuel Annor

EE20 Laboratory Report 2-602

7/27/2005

Title, BJT CHARACTERISTICS AND AMPLIFIERS

Abstract: ]lipolar Junction Transistor, precisely the npn type, can operate in three regions:

e active, saturation and cut-off regions. The active mode occurs when "the bas ­

emitter junction is forward biased and the base-collector junction is reverse

biased" (Hambley, 212). In saturation mode, the voltage across the collector-

emitter junction is approximately a.2V and the transistor goes into cut-off when

current flowing through the collector is approximately zero. Applying a small

change in the base-emitter voltage, VBE, particularly in the active mode- results in

a much larger change in the collector current, ie, this can result in a much larger

voltage. This process is called amplification and the ratio of the output voltage to

the input voltage is the voltage gain. On the other hand, a current gain (S) is the

ratio of the collector current, ie, to the base current, l'B.

Introduction:

rans1stor. Fustly, the data and theThe experiment involved analyzing a 2N3904 t . . .

graph of Ie vs IB and Ie vs. VCE plotted. Then from this graph,the current gain

and the output resistance, that is the Th evenm. res1stance looking from the out.

toward the inside of the circuit. The next part of th .e expenment entailed

0.G-applying an " signal and then measuring the small-signal input resistance, and

the small signal common-emitter current gain. The rest of the experiment

involved design and constructing involving an LED that switches on and off at

5Hz with a diode current of 20mA.

LED Finally, the..beCf circuit was modified by adding DC blocking capacitors. The rest

of this part involved constructing a mirror current circuit involving two identical

transistors and measuring their individual collector currents.

Mostly this experiment involved constructing several 2N3904 transistors and

using the DMM and the Oscilloscope to measure voltages o

and currents and displaying input and out signals

respectively.

100kVBDiscussion: ..

I started the experiment by constructing the circuit in

figure above using the 2N3904 transistor and to measure -=-0 -=-0

the voltage drop across the CE junction with a DMM while VB and Vcc are

varied. Obtaining Vee enabled me to obtain Ie. Writing KVL equations

)I

ield, IB = VB - 0.7 IC = VCC ­ VCE Y lOOk ' RC

This helps to complete the table next page

VB IB Vee VeE Ie

0.79 9E-07 0.2 0.1418 0.0000582

0.79 9E-07 0.4 0.2988 0.0001012

0.79 9E-07 0.6 0.502 0.000098

0.79 9E-07 0.8 0.701 9.9E-05

0.79 9E-07 1 0.907 0.000093

0.79 9E-07 2 1.902 9.8E-05

0.79 9E-07 5 4.89 0.00011

0.79 9E-07 10 9.88 0.00012

0.79 9E-07 15 14.87 0.00013

0.79 9E-07 20 19.86 0.00014

1 0.000003 0.2 0.1144 0.0000856

1 0.000003 0.4 0.1725 0.0002275

1 0.000003 0.6 0.3401 0.0002599

1 0.000003 0.8 0.534 0.000266

1 0.000003 1 0.736 0.000264

1 0.000003 2 1.728 0.000272

1 0.000003 5 4.73 0.00027

1 0.000003 10 9.7 0.0003

1 0.000003 15 14.7 0.0003

1 0.000003 20 19.68 0.00032

2 0.000013 0.2 0.0746 0.0001254

2 0.000013 0.4 0.0973 0.0003027

2 0.000013 0.6 0.1148 0.0004852

2 0.000013 0.8 0.1319 0.0006681

2 0.000013 1 0.1516 0.0008484

2 0.000013 2 0.787 0.001213

2 0.000013 5 3.75 0.00125

2 0.000013 10 8.7 0.0013

2 0.000013 15 13.66 0.00134

2 0.000013 20 18.61 0.00139

3 0.000023 0.2 0.0614 0.0001386

3 0.000023 0.4 0.0797 0.0003203

3 0.000023 0.6 0.093 0.000507

3 0.000023 0.8 0.1046 0.0006954

3 0.000023 1 0.1153 0.0008847

3 0.000023 2 0.1772 0.0018228

3 0.000023 5 2.683 0.002317

3 0.000023 10 7.56 0.00244

3 0.000023 15 12.47 0.00253

3 0.000023 20 17.39 0.00261

4 0.000033 0.2 0.0531 0.0001469

4 0.000033 0.4 0.0691 0.0003309

4 0.000033 0.6 0.0806 0.0005194

4 0.000033 0.8 0.0905 0.0007095

4 0.000033 1 0.0994 0.0009006

4 0.000033 2 0.1388 0.0018612

4 0.000033 5 1.636 0.003364

From this table values I constructed two graphs with the first one being Ie versus

IB with vee being constant: shown in the figure below.

IC vs vCE --..0.79

_1

_11_ _~

__I_~.

~ __l_-6-:-80-3

-B..{)-85­

..8-:-00·

r-+-H-IH....'-,-----,---------r-------I 2

I_~

~~~O-~OOIC

-5 o 5 10 15 20 25

VCE NJ.(. W

The second graph is IC versus VCE with VB being constant. It is also show

below.

IC vs IB at Constant vee 0.006

0.005

0.004

~ 0.003

0.002

0.001

....

/,~

/~ /

/ .---­

~ v - -o

_0.2 - 0.4

0.6

0.8

~1

_2

~5

--10

15

20

o 1 E-05 2E-05 3E-05 4E-05 5E-05

IB

The purpose of plotting this graphs is to enable as to ~ and rOt the small signal

output resistant.

· IC 6.VCESmce /3 = - and ro = I can calculated a few a values of ~ and ra . IB MC

First I calculated for the values of ~ and calculate for those of ra .

At VCE=5 and VB=3 IC =3.5mA and IB= O.023mA

/ /3 = 3'%.023 = 152.17

Similarly, going through the same calculations for the values below the

subsequent values were obtained.

At VCE=5 and VB=5, IC =4.8and IB= O.043A, /3 = 1 ..... 63 c;.

/And finally a VCE=10 and VB=5, IC =5.1m and IB= O.043mA

/3=118.6

so the values of pobtained suggest that p~100 which is the current gain that is

normally used.

Secondly using the ro = 6.VCE values can be obtained for the small-signal inputMC

resistance at some selected values of the collector current.

For Ie = O.UmA, VCE= 4.89 & Ie = O.14mA, VCE=19.86 all obtained at VB=O.79;

ro = 19.86 - 4.89 = O.5MD (.14-.11)m

\J; ) r>.J ........­

Also for Ie = 5.35mA, VCE= .65 & Ie = 4.42mA, VCE=O.58 all obtained at

VB=O.79;

ro ~= 15.13KD (5.35 - 4.42)m

For part 3 of the experiment I apply a signal to the previous DC circuit to obtain

an equivalent AC of the circuit which is shown Fig.l

o

lk OSC

100k

OSC

/~ '" +

~~ R2

50k -=-0

-"-0 f'ig. 1-"-0

e 1 But the value of R2 however is not known so we have to figure out for that.

We do this, by writing a couple of KVL equations for select loops.

From the circuit and choosing the right-most loop;

VCC-RCIC=VCE -eqn 1

I then plotted the equation on the Ie vs VCE graph to get a Qpoint & then

determine ICQ, by obtaining that value I can read off the value of IB.

Next is write out a KVL for the next loop, to obtain

VCC-RI11-RBIB-VBE=o

Since I know the values of VCC=15, Rl=lOOk, RB=lOOk, IB=3311A, nd VBE =O.7V

we can find 11.

15-(3311A xlOOk) -lOOk11-O.7=O

Solving the above will result in 11=110flA

Using a KCL at node, 12=11-IB,

Thus 12=77flA

Final y, writing a KVL loop for the third loop,

R212-RBIB-0.7=O

By substituting the all the known values, as follows,

R2=4/7711A=51kQ

Then I chose relatively large capacitors to make capacitance depletion very low. I

use 50KQ because it is a resister I have readily available by connecting two lOOkQ

in parallel.

This part of the experiment enables me to calculate~, the current gain and the

small signal input resistance.

Gnplll

e I Calculation:

From graph 1,

Vin =1.9Sincut &

Vout = 2Sincut

And fJsmall = Vout x RB = 2Sin OJt x lOOk =

vin RC 1.9SinOJt lk

Rin = (RIll R2)II(RB + rTf) =~

re..-o-U J I. Thus assuming that m«~

v 2 lC..=

Part4:

This part involves designing a BJT circuit that will switch an LED on and off at

5Hz. It is at 5Hz so that the human eye can be able to see the changes in the LED.

The other specification is that the current flowing through the LED must be

20mA when it is on. The circuit is show in fig2.

o

In the design, I refer to the previous Ie vs

YCE graph, look at YCE=2Y because the

voltage required to light an LED is

approximately 2Y and I found out that 2Y

correspond to IC=4.75mA.

However when the LED is in saturation,

/R3

22k

f'ig. 2

Q2N3904

D1

YCC-RCICI-YCE=O, a KYL in of the loops,

will be result in YCC= RCICl because the YCE=O.2Y, which is approximately

zero.

Also IC1=20mA + IC= 20mA + 4.75mA= 24.75mA

There choosing RC=lKQ then YCC= 24.75Y.

Using a ~=lOO, IB =47.5pA since IC=~IB.

Writing a KYL equation in the loop containing RB, I yielded 2-IBRB-O.7=O giving

RB = 27.37K.Q. But since 22k is readily available I will use that instead. Changing

the RB to 22kQ will utter the reason a little bit but still the result will be close. In

other words, the LED should still switch on and off.

Part5:

This part involves removing the LED and adding a blocking capacitor to the

output so that it looks like fig 3.

o

RC

1k

C3

f---qR3 1n

22k Q2N3904'"'v

<0

f' i 9· J

The out of this is a graph that is shown on the oscilloscope and it is that of graph

2 show below.

( -­ls

c Gnp!l3

The graph shows a triangular input signal but the out voltage is saturated at the

negative turn and cut-off at the positive.

From this graph, the voltage gain (Av) can be calculated since Vo=23.9V and

~ us Av = 23.9V =o.~ ---M.-4V _

~ h/J~ r·-....0 24.4._ 2..-L

The last part of this experiment was to construct a given mirror circuit involving

two BJTs illustrated in fig 4.

1k

+ VRef Q2N3904

10V -=­

1k 1k

The result of the mirror circuit is tabulated below.

J Vref lref/rnA Ie/rnA Iref/le

1 0.03 0.03 1

2 0.07 0.07 1

3 0.12 0.12 1

4 0.17 0.17 1

5 0.22 0.22 1

6 0.27 0.27 1

7 0.32 0.32 1

8 0.37 0.37 1

9 0.42 0.42 1

10 0.47 0.47 1

From fig 4 , it seems t ays the base to emitter voltages of the two transistor

are going to be equal because their base-to-emitter current is always going to be

equal, thus their collector currents is going to be equal giving the ratio to be

always 1.

Conclusion: In part one plotting Ie vs. Vee and the Ie vs. 1B, I was able to find

the device characteristics of the 2N3904 transistor that I worked with. From these

graphs I was able to calculate the common-emitter current gain and the collector­

emitter voltage ranged from 111.63 to 152.17 and also obtained the small-signal

out resistance through simple calculations to various values for a few collector

currents. In part three, I converted the circuit to its small current equivalent,

doing so helped to calculate the small-signal input resistance result in an

equivalent resistor at the right side of the transistor when a small equivalent is

drawn which was about 25KQ. The common-emitter current gain was also

calculated and I found out that value was very close the normal value that is

normally used in amplifier circuits, which is about 100. With part 4, I designed

the circuit to the specification given but modified a few of the values of the

resistors to those that I had readily to work with in the actual system. However

the values used were not too far from the original values so the net effect was

small on the entire circuit. Due to the minor effect, the circuit function correctly

by switching the LED on and off noticeably at the 5Hz frequency.

With part 5, I obtained a voltage gain to be about 0.98 and I think that value

would have been higher had the voltage of the out been clipped at the top due to

the transistor being in cut off mode. Also the below the out is clipped at .2 V.

*The cut off occurs at the top because the Ie at that point reaches zero and it

saturates at the bottom because Ib becomes large enough so the circuit's

operation is driven in the region at the upper end of the output load line. *

The last part with involves yielded a ratio which was one in all cases because the

circuit is designed to give identical base currents in the two transistors which is

supplied by one reference current, making ~ always larger than unity then the

collector currents in both transistors almost equal to each other and also equal to

the reference current since the base currents will be negligible and almost zero.