electrical engineering, need help in PSPICE program

ENGR 301

LAB #3: BJT CHARACTERISTICS AND APPLICATIONS

Date: 12/07/2017

Mark Elazegui

Abdulaziz Alharbi Merewan Jemal

Mark, Merewan, Abdulaziz

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Objectives:

The goal of this experiment was to characterize bipolar junction transistors, known as BJTs. We

investigated the use of BJTs as amplifiers and current sources. We measured the BJTs

characteristics through implementation and proved the measured data with theoretical

calculations and simulations from PSPICE. We then compared our measurements, calculations

and simulations to justify our finding for BJTs characteristics.

Components:

The following components were used for the experiment:

 1 × 2N2222 npn BJT

 1 × 2N3906 pnp BJT

 1 × 1N4733 5.1 V

 1 W Zener diode

 2 × 0.1 μF capacitors

 1× 100 μF capacitor

 1 × 10 kΩ potentiometer

 Resistors: 1 × 100 Ω, 2 × 1.0 kΩ, 4 × 10 kΩ, 1 × 100 kΩ, and 2 × 1 MΩ (all 5%, ¼ W)

MC1

The images in Figure 1 and Figure 2 were captured using the curve tracer shown and used to find

and .

Figure 1. Curve tracer.

Mark, Merewan, Abdulaziz

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Figure 2. Curve Tracer.

MC2

This time we used the circuit of Figure 3 to measure a more accurate value of the early

voltage, . While measuring with the digital current meter (DCM), power was applied and the

potentiometer was adjusted until the current of reached 0.5mA. This biases the BJT at the

point Q(IC, VCE) = (0.5 mA, 5 V). We then shorted out with a wire so as to effect the change

ΔVCE = 5 V, which moves the operating point from Q to Q′. Then recorded the change ΔIC and

computed the following values:

( )( )

Mark, Merewan, Abdulaziz

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Figure 3. Test circuit to find. and .

MC3

We constructed the circuits of Figure 4 for a more accurate estimate of , as well as for finding

and . We assembled the circuit of Fig. 4a and starting out with =10.7 V, adjust for

= 1.0 mA. Next, we turned power off, inserted the DCM in series with the base as in Fig. 4b,

reapply power without changing the setting for VCC, and measure IB. Finally, calculated:

±5

Figure 4. Test circuits to find βF.

MC4

With power off, we connected the digital voltmeter (DVM) in parallel with the base-emitter

junction as in Fig. 4. Reapplied power, and measured and recorded VBE(1 mA). Next, turned

power off and connected the BJT again as in Fig 15a, but with R = 100 kΩ. Reapplied power and

adjusted VCC for IC = 0.1 mA. Then, with power off reconnected the BJT as in Fig. 15c,

reapplied power, and measured and recorded its new base emitter voltage drop VBE(0.1 mA).

Recorded the following values: VBE(1 mA) = 0.625 ± 0.001 V and rewrote the following

equations to find and :

( )

( ( )

)

Mark, Merewan, Abdulaziz

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( )

( ( )

)

We have two unknown variables as well as two equations to solve for the unknown variables, and . We calculated and to be, 0.382 pA ± 0.005 and 26.1 mV ± 0.3, respectively. Our measurements are in a typical range.

MC5

To observe the characteristics of the BJT at the saturation region, we assemble the circuit shown

in Figure 5. By adjusting the 10 K-Ω Pot in the circuit in Figure 5, we can set the BJT to be at the

edge of saturation (EOS). We then measure , , and at the EOS point; = 1.47 ± 0.05

V, = 0.596 ± 0.003 V, and = 102.3 ± 0.3 mA. In addition to the preceding

measurements, we also measure IB and IC to calculate for βF at EOS; IB = 0.0067 ± 0.0002 mA

and IC =1.0450 ± 0.003 mA. Therefore, we calculated βF at EOS as 155 ± 7, which is in good

agreement with our earlier findings.

Figure 5. Test circuit for saturation measurements.

M6

To see the different characteristics of the BJT at different operating point, we readjust VW back

to 10 V. At this particular point of operation, we measured to be 9.92 ± 0.05 V. This

Mark, Merewan, Abdulaziz

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behavior reveals that at deep saturation mode, varied noticeably from VCE(EOS). As for βF

at this operational point, we measured IC and IB and calculated βF as follows:

βF does not vary noticeably as the operating point is shifted from EOS to deep saturation; this

value is still relative to our earlier finding of βF.

C7

We analyzed the circuit of figure 6 by using the NPN BJT large-scale equivalent circuit to

calculate for , , , , and . Assuming is 0V, we know that will be very low and therefore ≈ 0V.

M8

To obtain the BJT gain, AV, we need to assemble the circuit shown in Figure 6. Next, we

connected the waveform generator, and while monitoring it with Ch.1 of the oscilloscope,

adjusted it so that vs is a 10-kHz sine-wave with a peak-to-peak amplitude of 2 V and 0-V DC

offset. Finally, we used Ch. 2 to measure the peak-to-peak amplitude of vo, and then find the

gain Av = vo/vi of your amplifier, where vi = vs/100.

Figure 6. Common-emitter (CE) amplifier.

S9

We simulated the circuit of Step M8 via PSpice (DC as well as AC analysis).

Mark, Merewan, Abdulaziz

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Figure 7. Pspice circuit setup.

= 5.298V

= -277.5uV

= -539.9mV

Figure 8. PSPICE Simulation.

C10

M11: By using the circuit built as in Fig. 17, we switch Ch. 2 back to DC mode. The waveform

generator is switched to a triangle wave from a sine wave. The amplitude is increased to view

when it starts to distort, then clips both at the top and at the bottom. In our case we increased the

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amplitude to the max which was 10V but had to take out the R2 resistor which was 100Ohms as

10V wasn’t large enough.

Fig.17

What causes distortion to occur? Justify the two clippings in terms of the regions operation of

your BJT.

The distortion is occurs due to some un-bypassed external resistance within the emitter circuit

which is usually found in common-emitter amplifiers.

What are the values of the upper and lower clipping?

Upper Clipping: Vpp = 4.56 ± 0.005V

Lower Clipping: Vpp = 6.48 ± 0.005V

C12: The circuit to look at was the circuit of Fig. 18 which is obtained from Fig. 17. The

difference is inserting the emitter-degeneration resistor . The lab then assumes that has a DC value of 0V.

Fig. 18

Mark, Merewan, Abdulaziz

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Find the collector current and predict the value of the small-signal gain which in this case is ( ) where and are respectively given in Eqs. (14) and (15).

0.0009750*10k = 9.75 Amps

( ) (0.0009750)*(10k||10k) = 4.87329

M13: We turned off the power and assembled the circuit of Fig. 18. Then applied power of 10V

and adjusted the waveform generator. The waveform generator is setup so is 10-kHz sine- wave of 0.2-V peak-to-peak amplitude.

Measured Gain -

2.16/0.2 = 10.8

Mark, Merewan, Abdulaziz

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How does it compare with the predicted value of Step C12?

The predicted value and the measured value of gain had a difference of 1.05 with the measured

value having more gain than the predicted value.

How does it compare with the rule-of-thumb value of ? The comparison is the same as above.

 Common-Collector Amplifier C14: The lab assumes has a value of 0V by values in schematic of Fig.19. By using the large- signal model to find the DC collector current .

Fig. 19

Predict the values of:

Small Signal Gain - ( (

( ) )) = 0.9375

Output Resistance - = 296.141Ω

M15: We turned off the power and assembled the circuit of Fig. 19 without connecting at first. Then mounted a 0.1-µF power-supply capacitors in close proximity to the circuit. Power is

now applied and the waveform generator is adjusted to be set with as a 10-kHz sine-wave with 0V DC and the peak amplitude 3.

Measured Gain -

6.84/6.55 = 1.0442

How does it compare with the predicted value of step C14?

The measured value of gain was higher in than the predicted value with a difference of 0.1067 so

it wasn’t a huge difference.

Mark, Merewan, Abdulaziz

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What happens if you now connect the load to your circuit? After we connected the nothing change as the potential across the node was the same.

Is loading noticeable?

Loading was not noticeable because with the signal that appeared only the AC value was used

and the DC was blocked which ensured that there was no high and low limits.

Current Sources

M16: In this part we used the 2N3906 BJT’s and used the Curve Tracer. The POLARITY knobs

were set up to “-“ (PNP) both in BASE STEP GENERATOR and the COLLECTOR SWEEP

controls.

Estimate the following:

Early voltage - - = 3.89V

Current gain at and

What is the value of ? = 1/12.435=80411Ω

C17: Using Eq. (20), predict the Line Regulation and the Load Regulation of source at

Line Regulation – ( ) = 2.25 * 10^-6

Load Regulation – =

( )

M18: With the power off we assembled the circuit of Fig. 20. Then power is applied and by

using the Ammeter as the Load, the pot is adjusted for .

Line Regulation: Supply voltage is raised from 10V to 15V to produce ∆VCC = 5 V.

Load Regulation: Ammeter is still acting as Load. The circuit is break at the collect C with 5kΩ

resistor is series between ammeter and collector.

C19: Sketch the Norton’s equivalent as seem by the load in Fig. 20. Then, based on the

measurements above give its element values and , and comment.

Mark, Merewan, Abdulaziz

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Conclusion

For this experiment, we observed the characteristics of BJT. We looked at the

Forward-Active region where we measured different values of the BJT such as the

early voltage VA, -ꞵ and the Is saturation current. For the Saturation Region, we adjusted the wiper voltage in our circuit to see when the BJT starts to saturates. We

also built different BJT amplifiers We predicted the gain of these circuits by using

large signal model and small signal model. We compared these values to our

simulated circuits in LTspice by calculating the gain Vo/Vi.

Mark, Merewan, Abdulaziz

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