EE 321 Lab 10 Fall 2004EE321 – Lab 10Bipolar Junction Transistors, Part IIn this lab we will investigate how a bipolar junction transistor (BJT) can be used to amplifysignals.1. First, use your multi meter to measure the base-emitter and base-collector junctions of yourtransistor (Figure 1 and 2). When you put your meter in “Diode” mo de, the meter willindicate the voltage for 1 mA current through a forward-biased diode, and will indicate 0if the diode is reversed biased. Test this on a diode then on a 2N3904 transistor. Recordyour readings. This is a useful way to check for bad transistors. This is better than usingthe Ohm-Meter for measuring the forward and reverse resistances of the junctions, as themeasuring voltage and current are not known.EBCFigure 1.CBEFigure 2.BEC10K1KVOVSVCC= 10V2N3904Figure 3.2. The basic principle of a BJT is that the base-emitter voltage vBEcontrols the collector currentiCwhen the transistor is biased in its ‘active’ mode (b-e junction forward biased, c-b junctionreverse biased).• Construct the circuit (Figure 3). Lay your circuit out neatly, using the bus lines for VCCand ground.• Connect a sine wave to vSand display vBEand vCin on your oscilloscope.• Adjust input sine wave, vS(the DC bias (offset) and amplitude) so that vBEvaries fromabout 0.0 to about 0.73 volts. Be sure to measure vBEat the base of the transistor.Make vBElarge enough so that vOgoes from 10 V to 0 V.• Both vBEand vOwill look like clipped sine waves. Where do they clip?3. Note that vC= VCC− iCRC. Use this to determine where the transistor is in cutoff (iC= 0)and where it is in saturation (vCE< 0.7). (In between, the transistor is in its active region.)• Display vCvs. vBEusing the x-y mode. Using this curve, plot iCvs. vBE. This curveshould look like Figure 5.16 of Sedra and Smith. Note the three regions of the plot.• For what value of vBEdoes the transistor start to turn ‘on’? (Collector current start toflow?) Does this make sense?• When vBEis too large, the collector current will drop vCbelow vB, thus forward-biasingthe collector-base junction and driving the transistor into saturation. For what valuesof vBEand iCdoes this occur?1EE 321 Lab 10 Fall 20044. The transfer characteristic is linear if vBEis restricted to small changes about a bias point,as in Figure 5.49 of Sedra and Smith.• Reduce the p-p signal amplitude and change the offset until only a small part of thetransfer curve going from about 3.5 mA to 6.5 mA is left (fiddle with the signal generatoruntil the output is a sine wave that goes between 6.5 V and 3.5 V. Start with an inputamplitude of 0.1 V and offset of 0.5 V). Now you should find that vCchanges by 3 Vp-p about the bias point at IC= 5 mA.• Sketch the operating range on the plot of part 3 above.• In the time domain, display vBEand vCon your scope. What is the voltage gain vc/vbefor the small signals? (Be sure to measure the ‘input’ vbe after the 10 kΩ resistor.)• How does the gain compare with theoretical value found in the prelab?• How could the gain be made larger? (There are two ways.)5. Increase vBEand note the distorted nature of vC. Increase vBEfurther until the output just”clips” on top and bottom. (Re adjust the DC level so that the clipping is symmetrical).• Sketch a clipped waveform.• Note the DC levels of the clipping, and whether the clipping is due to saturation orcutoff.6. The output voltage can be linearized over its entire range by using the input signal to con-trol the base current rather than the base voltage. This is because the collector current isproportional to the base current, i.e. iC= βiB. We will show this in the following way:First we will bias the transistor in its active state by supplying a current to the base throughRBfrom +VCC(see Figure 4). This provides a DC base current IB= (VCC− 0.7V)/RB,which gets amplified by the transistor to give a DC collector current IC= βIB. Becauseβ varies from transistor to transistor you will have to select the value of RBto obtain thedesired IC.• The desired current for ICis 5 mA• Begin by assuming β = 100 and calculate RB.• Use this RBin the circuit, measure IC, and then get a better estimate of β.• Repeat this step until IC= 5 mA ± 0.5 mA.• What value of RBdid you need, and what value of β does this imply?• Compare β to the value from the datasheet. datasheet. (In the datasheet, this DC β iscalled hF E.)VCC= 10VRBBEC1KVO2N3904Figure 4.10KVSVCC= 10VBECVO2N39041uFRB1KFigure 5.2EE 321 Lab 10 Fall 20047. Now add a signal component to the base current through a coupling capacitor, as shown inFigure 5, which does not affect the DC bias. Note that the input voltage has been convertedto a current by resistor RS.• Increase vsuntil the output clips, and back off until the clipping just stops.• What is the maximum output voltage swing• Is the output linearly related to the input. vs? (Adjust the scope — position, sensitivity,invert — until the input and output line up. If they line up and the output is a biggercopy of the input they are linearly related.)• What is the voltage gain, vout/vs?Pre-Lab1. From the datasheet for the 2N3904, find the following:• The maximum collector current.• The maximum voltage between the collector and emitter.• The DC current gain (HF Eor β) when IC= 10 mA.• vCEwhen the transistor is saturated.2. Find the gain for the amplifier in Figure 3. The gain can be determined from these equations:vout= −icRC, ic= gmvbe, and gm= IC/VT, with IC= 5 mA.3. In Figure 4 find RBso that IC= 5 mA (assume a β of 100). What would VObe?For this same circuit, how would you find β if you were given RBand