Formal Write Up - Transistors

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Analog Circuits – Transistors

Jamie A. Benitez

Partner: Shantel Adams

December 10, 2009

Physics 243

Abstract: Imagine the control one would have over technology, if they

were able to apply a little voltage or current, and create a huge magnitude

of what they put in. That imagination became reality when the transistor

was invented. With this, now simple to use, device, we are able to

amplify and control current greatly. These devices are allowing

technology to expand exponentially. In this lab experiment, we utilize

two different transistors: the NPN (2n3904) and the PNP (2n3906). We

want to test the different theories such as, the three different regions of a

BJT, the proportionality of currents, and gain. Determining the function

of the circuits and figuring out the β value for the NPN are other goals.

At the end of the experiment, the theories are proven to be correct with

the use of different components and trials. Our goals were reached and

we were able to calculate the desired values. An overall knowledge of

BJTs will widen.



I. Theory

Bipolar Junction Transistors (BJTs) are vital components to the technology we use today. They

have come a long way since the middle of the century. Decreasing in size – from large objects to

miniscule chip components, and increasing in efficiency, these transistors are only a fraction of potential

for what is stored in their future. BJTs consist of two different types of junctions: the NPN and the PNP

junctions. They are comprised of two diodes with the opposite direction, giving their title and structure;

refer to Figure 1. The difference in structure between these two transistors causes the NPN to “source”

current, while the PNP “sinks” current [2]. Both of these BJTs have three terminals: Base, Collector, and

Emitter. The way these terminals are arranged with resistors and/or other components will determine the

function of that particular circuit. Figures 2, 3, and 4, will display examples of simple, yet differently

orientated, circuits with diverse types of gains, i.e. voltage and/or current.

The NPN junction is the more commonly used transistor. NPN junctions have mobile electrons

due to N-type doping[1]. The mobility of these electrons cause current to flow, rapidly. Much more

rapid than the mobility of “holes” where electrons could fill – causing a similar but, slower electron

movement. This is much more efficient than working with the PNP junction which undergoes P-type

doping. In this junction, our goal is to be able to control the current that flows from the collector to the

emitter. The general circuit symbol, Figure 5, will allow a better opportunity to understand the flow of

current. The base current is what primarily controls the current that flows from the collector to the

emitter. In the simple common-emitter configuration, Figure 3, the iC (current coming from collector)

wants to pass directly to the emitter, which VC must have a greater voltage than VE [2]. The transistor

does not allow any current to flow unless a small current flows from the base. The transistor must be

turned “on”, or in the active region - which will later be discussed, by hitting a voltage (V BE between the

base and emitter) of 0.7V before any action will occur. This 0.7V is because there is a voltage drop

between the base and emitter for Silicon devices due to the input characteristics of an NPN Transistor are

of a forward biased diode [2]. Once turned “on”, the small iB (current from base) will allow the

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potentially large iC to flow rapidly into the emitter. Since iC is controlled by iB (current from base), then

we know they are proportional, which led to,

iC = β iB [1]. (1)

The exact proportion for this equation is determined by the β (current gain) which each transistor has

fixed – roughly between 100 and 200 (unit-less). While the current from the base controls the current

from the collector from reaching the emitter, the current from the base travels to the emitter along with

the collector’s current, developing the next equation,

iE = iB + iC [2]. (2)

These two equations are only functional when in the Active region.

An important aspect of the transistor is the three regions it works in. They are the Off region,

Active region, and Saturation region. Figure 6 illustrates the three regions. When the base-emitter diode,

or VBE, does not reach 0.7V, the transistor is in the Off region. When in the Off region, there is no current

flow, iE = iB = iC = 0. The Active region is our favorable region. Once the diode is turned “on”, the

currents can flow as previously stated. While in the Active region, we obtain more equations than just (1)

and (2). Referring back to Figure 3, there is an input signal (V in). This forms the iB that travels across the

resistor (R B), connected in series with the input signal and to the base. Since the voltage of the emitter

(VE) is 0, then VBE will be the same as the voltage (VB) across R B. This forms the equation,

VBE = VB = VIN - iBR B [1]. (3)

(with VBE = 0.7)

Similarly, when the collector is the subject, the power to the transistor (VCC) forms iC that flows across R C,

which is connected in series with the power supply and the collector. The equation,

VOUT = VC = VCC - iCR C [1] (4)

is then produced. When combining equations (1), (3), and (4), we get our last equation,

VOUT = VCC -β (R C/R B)(VIN – 0.7) [1]. (5)

The -β (R C/R B), in that equation, is the slope (gain) of the Active region that is demonstrated in Figure 6.

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The last region is the Saturation region. This is where the transistor hits its maximum capability. Once in

saturation, even when more VIN is supplied, VOUT remains small at about 0.2V (transistor saturation) [1].

After Saturation is when the transistor reaches the constant Active region as illustrated in Figure 7. It is

constructive to work in the Active region as close to the Saturation region as possible. This is helpful

because after the Active region, the transistor will come to a new region where it does not work – the

Breakdown region, Figure 8.

The PNP junction works completely opposite to the NPN junction. The emitter now has a

voltage source attached to it instead of being grounded [2]. Unlike the NPN, this junction has mobile

“holes”, as declared previously, rather than those “extra” electrons. Since the emitter is now transmitting

a current to the transistor, and the current from the emitter to the collector are “holes”, the VIN needs to be

negative [2]. Also, current flows out of the base and the collector, Figure 9. Even though the PNP

junction is the reverse of the NPN junction, equations (1) and (2) still hold true.

II. Experiment

Now that we are knowledgeable in the Bipolar Junction Transistor topic, I will discuss the

procedure of the lab my partner and I performed. To begin, we constructed two separate circuits: circuit

one using an NPN, and circuit two using a PNP. Circuit one consists of the common-emitter set-up

discussed earlier, except we added a resistive load, which had one end connected to the junction of R C and

the collect, and the other end connected to ground, Figure 10. We used R C = 1k Ω and R L = 2 k Ω , and

conducted two tests: using R B = 10 k Ω in one and 121 k Ω in the other. VL was measured as a function

of VBB between 0V ≤ VBB ≤ 6V. When in between 0V 1V, 0.1V intervals were recorded, and when in

between 1V 6V, the intervals were in 0.25V steps. These values are recorded in Tables 1 and 2. VBB

was then replaced with a 1kHz sinusoid and we captured V OUT and VIN on the oscilloscope as images;

Figures 12, 13, and 14 were using R B = 121 k Ω and Figures 15 and 16 were using R B = 10 k Ω . Next,

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circuit two was assembled, Figure 11. VL is again measured across R L, which we used a 1 k Ω , 2 k Ω , and

10 k Ω . The data for the VL and iL is on Table 3.

III. Conclusion

After the experiments were finished and all the data was collected, many earlier theory statements

became visible. While looking at Table 1, we can witness the three regions that the transistor works in.

At the start, while VBB increases, VL barely varies. It is only when VBB is about 0.6~0.7V that VL begins

to change. This illustrates the Off region of the transistor. VL begins to change when the transistor

reaches the required voltage of ~0.7V and turns “on”, as the theory described, and the Active region is

then visible. As we continue along the Active region, VL decreases dramatically up until VBB gets to

1.25V. VL drops to 0.265V (and decreases to 0.076V) which is very close to the saturation voltage of

~0.2V, previously acknowledged. We are now in the Saturated region in this circuit. Similarly, with the

121 k Ω data, we are able to see the Off and Active region, but we do not reach Saturation in this

experiment. When VBB reaches the 0.6~0.7V range again, the transistor turns “on”, except this time we

are in the Saturation region for a much longer voltage supply because of the magnitude of our R B. Now

we want to find the β value. Using the gain note from equation (5), we set that equal to ∆ VOUT/∆ VIN.

To find ∆ VOUT/∆ VIN, we must pick two voltages from Table 1. I chose to use VBB = 0.80V and 0.90V.

This gives us a β value of: 102.1. Now using Table 2, I chose VBB = 2.75V and 3.00V, using the same

equations, our value of β is: 114.708. These b values are very similar. Experimental errors should be

taken into consideration for the different values. Voltage measurements would alter while a constant

voltage was actually supplied – caused a lack of accuracy.

Replacing VBB with the 1kHz sinusoid and generating Figures 12 to 16, more inferences can be

made. Looking at Figure 12, our VOUT and VIN are exactly out of phase. This demonstrates a gain

(∆ VOUT/∆ VIN) of -1. Figure 13, reveals saturation when we changed the amplitude. When using the

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sinusoid for the 10 k Ω , Figure 15 exemplifies a gain of -10, which is close to the 12.1x greater gain

calculated and expected.

Circuit 2 involved connecting a LED (Light Emitting Diode) into the circuit. Each of our three

trials allowed the LED to light up. Also to note, as resistance (R L) increased, the voltage (VL) increased

proportionally. We were able to see that circuit 2 was a constant current source.

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References

1. J. R. Cogdell, Foundations of Electrical Engineering, 2nd ed. (Prentice Hall, 1996).

2. “Electronics Tutorial about Bipolar Junction Transistors.” Electronics-Tutorials.

<http://www.electronics-tutorials.ws/transistor/tran_1.html>.

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[2]Figure 1. The basic structures of PNP and NPN transistors.

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[2]Figure 2. The Common Base Amplifier Circuit.

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[2]Figure 3. The Common Emitter Amplifier Circuit.

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[2]

Figure 4. The Common Collector Amplifier Circuit.

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[2]

Figure 5. Circuit Symbol in NPN junction.

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[1]

Figure 6. VOUT vs. VIN graph with labeled regions.

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[1]Figure 7. Active region showing comparison iB and iC with increasing VCE. The almost vertical

regions on this graph representation the Saturated region.

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[1]

Figure 8. Breakdown region visible on graph. Shows how too much voltage after a certain point

can destroy the transistor.

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[2]Figure 9. A PNP Transistor Circuit

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Figure 10. Hand-drawn sketch of the first circuit used in the experiment.

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Figure 11. Hand-drawn sketch of the second circuit used in the experiment.

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Figure 12. Shows gain of -1 using 1kHz and R B = 121k Ω.

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Figure 13. Saturation with amplitude change using 1kHz and R B = 121k Ω.

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Figure 14. Change offset towards negative using 1kHz and R B = 121k Ω.

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Figure 15. Shows gain of -10 using 1kHz and R B = 10k Ω.

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Figure 16. Greatly increased offset and amplitude using 1kHz and R B = 10k Ω.Starts to show truncation.

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Table 1. Circuit 1 Table 2. Circuit 1

using R B = 10k Ω. using R B = 121k Ω.

RL (kΩ ) VL (V) iL (mA)

1 0.983 0.983

2 1.984 0.99210 9.93 0.993

Table 3. Circuit 2 data collection.

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VBB VL

0 7.983

0.1 7.983

0.2 7.983

0.3 7.983

0.4 7.983

0.5 7.982

0.6 7.9620.7 7.905

0.8 7.826

0.9 7.747

1 7.663

1.25 7.415

1.5 7.182

1.75 6.958

2 6.713

2.25 6.46

2.5 6.218

2.75 5.9663 5.729

3.25 5.472

3.5 5.227

3.75 5.005

4 4.764

4.25 4.511

4.5 4.277

4.75 4.042

5 3.809

5.25 3.5745.5 3.346

5.75 3.11

6 2.892

VBB VL

0 7.989

0.1 7.988

0.2 7.988

0.3 7.989

0.4 7.989

0.5 7.987

0.6 7.9370.7 7.429

0.8 6.521

0.9 5.5

1 4.43

1.25 2.005

1.5 0.264

1.75 0.138

2 0.158

2.25 0.143

2.5 0.132

2.75 0.1233 0.116

3.25 0.11

3.5 0.105

3.75 0.1

4 0.097

4.25 0.093

4.5 0.09

4.75 0.087

5 0.085

5.25 0.0825.5 0.08

5.75 0.078

6 0.076



Appendix

Calculating β:

Using R B = 10k ΩVOUT VOUT VIN VIN

6.521V 5.5V .80V .90V

∆ VOUT ∆ VIN

1.021V -0.1V

- ∆ V OUT = -β (R C/R B) = Gain

∆ VIN

- 10.21 = - β (1k Ω /10k Ω )

β = 102.1

Using R B = 121k ΩVOUT VOUT VIN VIN

5.966V 5.729V 2.75V 3.00V

∆ VOUT ∆ VIN

.237V -0.25V

- ∆ V OUT = -β (R C/R B) = Gain

∆ VIN

- 0.948 = - β (1k Ω /121k Ω )

β = 114.7

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Formal Write Up - Transistors