Upper Cutoff Frequency of a Common-Emitter Amplifier

Electronics IILaboratory #11Upper Cutoff Frequency of a Common-EmitterAmplifierOBJECTIVESThe purpose of this experiment is to measure the upper cutoff frequency of acommon-emitter amplifier due to parasitic capacitances using the high-frequency modelof a BJT transistor. You will be able to demonstrate the effect of Miller capacitance onupper cutoff frequency and learn how to estimate this frequency.EQUIPMENT LIST1. 2N2222 silicone transistor2. 15-V DC Power Supply3. Signal Generator4. Resistors:(1) 100 kΩ, (1) 15 kΩ, (1) 10 kΩ, (2) 5.6 kΩ, (2) 1 kΩ5. Capacitors:(1) 100 µF, (2) 22 µF, (1) 0.0022 µF, (1) 0.068 µF, (1) 100 pF6. Two oscilloscope probes7. 1 Breadboard8. Jumper Cables9. Multimeter10. OscilloscopeDISCUSSIONThe capacitive reactance of a capacitor decreases as frequency increases. Thisfact can lead to problems when a amplifier is used for high-frequency amplification. Antransistor has inherent shunt capacitances between each pair of its terminals. At highfrequencies, these capacitances effectively short the AC signal voltage. Therefore, inhigh-frequency amplifiers, shunt capacitance must be extremely small.In this experiment, artificial shunt capacitances will be installed in the amplifiercircuit because it is extremely difficult to measure the actual parasitic capacitances of thetransistor. Since the artificial capacitances are much larger than the real parasiticcapacitances, the parallel combination of real capacitances and artificial capacitors isapproximately equal to the values of the artificial capacitors. The objective of thislaboratory is to investigate the high-frequency response of the amplifier to gain insightinto the problems associated with the parasitic capacitance and to obtain practicemeasuring the upper cutoff frequency of the amplifier.By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera Rev: 09/16/05

Electronics IILaboratory #12RsVsRB Rpi CA CB RL'The upper cutoff frequency is where the voltage gain of the amplifier drop –3 dBor 0.707 time the midband voltage gain value For the equivalent circuit of a commonemmiteramplifier in Figure 1, the upper cutoff frequency due to input shunt capacitanceC A can be calculated using the following equations:ICQCARrπ's= CV=IBEfTc= R⎛= β⎜⎝ RCQBB( C+ CA// RCBRE'LSβ; V) =T// rVB− 0.7+ R ⋅( β + 1)( 1−A )= RCMR's2π⋅ Rπ;1B; A// R⋅CLVMA= R= 26mV@300⎞⎟;⎠B1// R2KVccR2=R + R1'L− β ⋅ R=rThe equation for calculating Miller capacitance is only valid for the invertingamplifier. Since the non-inverting amplifier is not affected by Miller capacitance, it isoften used in very high-frequency amplification. It is important to notice that A M doesnot take into account the voltage divider between the internal signal-source resistance andthe input resistance of the amplifier. Therefore, the value of A M is not equal to theoverall gain (A S ) from source-to-load, V L /V S .The upper cutoff frequency due to output shunt impedance C B can be calculatedusing the following equation:oπ2By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera Rev: 09/16/05

Electronics IILaboratory #13fCc( CB) =2 π'⋅ RLB= CCE+ C1CB⋅CBAM⋅A−1Provided that f c (C A ) and f c (C B ) are not close in value, the actual upper cutofffrequency of the amplifier is approximately to the smaller of the two frequencies.PROCEDURE1) Using the digital multimeter, measure the β of the BJT transistor and theresistors values of the circuit resistors. Complete table 1.2) To measure the bias current (I cq ) of the amplifier, connect the followingcircuit. Remember this current is DC. Determine the value of the dynamicresistance of the transistor (r π ). Complete table 1.VCC15VM100kohmR15.6kohmRc22uFC2R65.6kohm50mV35.36mV_rms400Hz0DegVs+Vin-22uFC115kohmR21kohmRe100uFCe+VL-10kohmRLBy: Prof. Rubén Flores Flores, Prof. Caroline González Rivera Rev: 09/16/05

Electronics IILaboratory #143) Set the signal generator voltage and frequency to 50 mV peak and 400 Hzrespectively. Make sure that the signal generator termination is 50 Ω.4) To verify that the amplifier is working, measure with the oscilloscope thepeak-to-peak voltages of V in , V L and V s . These values can be used todetermine the midband voltage gainfrom load-to-sourceA M = VLVinand the voltage gainA = V V . Complete table 2.SLS5) To measure the high-frequency response of the common-emitter amplifier,connect the parasitic capacitors (C CB ,C BE ,C CE ) as illustrated in the figure.These capacitors are intentionally made much larger than the normal devicecapacitances for reasons outlined in the discussion section.VCC15V100kohmR15.6kohmRc22uFC2100pFCcb5.6kohmRs+22uFC10.0022uFCce+50mV35.36mV_rms400Hz0DegVsVin-15kohmR20.068uFCbe1kohmRe100uFCeVL-10kohmRL6) Now increase the frequency of the signal generator until the peak-to-peakvoltage of the load decreases to 0.707 times the value at 400 Hz. This is theoverall upper cutoff frequency of the amplifier. (Table 4).7) Return the signal generator frequency back to 400 Hz and start. Increase thefrequency of the generator to each frequency in Table 3 measuring values ofBy: Prof. Rubén Flores Flores, Prof. Caroline González Rivera Rev: 09/16/05

Electronics IILaboratory #15V L at each frequency. These voltage values will be used to plot the highfrequencyresponse of the amplifier.8) To determine the upper cutoff frequency due to the input capacitance C Aremove the output capacitance C CE . Return the signal generator frequencyback to 400 Hz and again start increasing the frequency until the peak-to-peakvoltage of the load decreases to 0.707 times the value at 400 Hz. Thisfrequency is f c (C A ). (Table 4).9) To determine the upper cutoff frequency due to the output capacitance C Bremove the input capacitances C BE and reconnect C CE capacitor. Return thesignal generator frequency back to 400 Hz and again start increasing thefrequency until the peak-to-peak voltage of the load decreases to 0.707 timesthe value at 400 Hz. This frequency is f c (C B ). (Table 4).10) Which of the two cutoff frequencies from steps 8 and 9 determine the overallcutoff frequency of the amplifier (step 6)?11) Using table 2, plot the amplifier frequency response (A V (dB) vs. freq.). The⎛ VdB)= 20 ⋅ log⎜⎝Vvoltage gain in decibels is equal to Av(L. From theingraph identify the overall cutoff frequency. Remember this happen when thegain drops 3 dB from the midband voltage gain in decibels.⎟ ⎞⎠By: Prof. Rubén Flores Flores, Prof. Caroline González Rivera Rev: 09/16/05

Electronics IILaboratory #16DATATable 1: Steps 1 and 2Amplifier’s MeasuredParameters ValuesR 1R 2R SR CR ER LβI CQTable 2: Midband Voltage Gain (f = 400 Hz). Step 4V L (peak-to-peak) A V = VLVinA s = VLVSTable 3: Frequency Response (Step 6)Frequency (Hz) V L (peak-to-peak) s L S4006008001k1.2k1.4k1.6k1.8k2k3k4k5k10k16k20kTable 4: Amplifier’s Cutoff FrequenciesUpper cutoffStep Numberfrequency689A = V V A V = VLVinBy: Prof. Rubén Flores Flores, Prof. Caroline González Rivera Rev: 09/16/05