Electronic Devices and Circuits

ETE 305

Course Description:

Frequency dependent models for BJT and FET amplifiers, frequency effects upon gain and input-output impedance of single and multistage BJT and FET amplifiers, Bode plots, differential amplifiers.

# Lab1- Biasing and Hybrid-Pi Model

## Prelab:

 IB= 6.67mA R1= 169.5 kΩ RE= 3 kΩ R2= 61.67kΩ RC= 6 kΩ

## Lab:

### Part 1: DC

 Resistor Values (Actual) Voltage Values (Actual) R1= 159.9 KΩ VCC= 15.01 V R2= 62.2 KΩ VBE=  0.66531 V RC= 6 KΩ VRC= 6.40 V RE= 3.02 KΩ

Q-Point

 VCEQ 5.3 V IC VRCRC=6.407 V6 K=1.07mA

 𝛃DC ICQIBQ=1 mA6=164 𝛃AC ΔICΔIB=1.8 Div x 0.2 mA2㎂=180

### Part 1: AC

 Values for Components ƒ= 750 Hz C1= 10 µF Vs(p-p)= 30 mV C2= 10 µF VCC= 15.04 CE= 220 µF

Measured Values

 IC= VRCRC=5.84V6K=0.97mA Vi(p-p)= 29.9mV Vo(p-p)= 51.9mV rm= Vi(p-p)Vs(p-p)-Vo(p-p)Rs=29.9 mV30-29.9mV2k=598KΩ Avs=Vo(p-p)Vs(p-p)=51.9mV30mV=1.73 Vo (with Ce)= 4.04V Avs(with Ce)=Vo(p-p)Vs(p-p)=4.0430mV=134

Calculated Values

 ri = 𝛃RE= 180(3.02K)= 543.6 kΩ rm=ri ll RB= 543.6k ll 44.78 k= 41.37 kΩ r𝜋= 𝛃rm= 180(41.37k)= 7.45 MΩ

## Post-lab:

This lab had two parts a AC and DC portion. We see that the capacitors are open in DC operation. We Find the Q point which is the transistors operating point. We than find the β using a curve tracer, because our β is actually different than what we used when we calculated the values for the prelab we see a noticeable difference in the values. In the second part we apply a AC signal to the transistor. We noticed that are values again are different with those that are calculated in that since our β is different than the actual it causes the gain to change. Using the oscilloscope, we measured the VI VO values, with and without a bypass capacitor.

## Screenshots:

Without Ce capacitor                                                                   Measured Rin

With Ce capacitor

# Lab 2 – Common-Emitter Amplifier with unbypassed Re

## Prelab:

 Ic= 1mA VE = 3V R1= 150k Ω Vce= 6V VB= 3.7V R2= 59k Ω Vcc= 15V β= 150 RE = 3k Ω RC = 6k Ω rb = 44.6k Ω rπ = 6.09M Ω rm = 40.6k Ω ri = 453k Ω ro = ∞

## Lab:

Actual Values

 RS= 9.814 kΩ RC= 6.07 kΩ RE1= 1.01 kΩ CE= 220 uF R1= 166.51 kΩ C1= 10 uF RE2= 2.2 kΩ 𝑓= 750 Hz R2= 59.77 kΩ C2= 10 uF RL= 9.789 kΩ VS= 85 mV

RE1 and RE2 Unbypassed

Measured Values

 Vs 84 mV (see figure 1) VCEQ 9.09 V Avi = 1.5 Vi 200 mV (see figure 1) ICQ 609.56 ㎂ Avs = 3.57 VL 300 mV (see figure 2) rin = 25k Ω Ais = 2.59

RE2 Bypassed

Measured Values

 Vs 84 mV (see figure 3) VCEQ 9.09 V Avi = 1.67 Vi 200 mV (see figure 4) ICQ 609.56 ㎂ Avs = 5.95 VL 300 mV (see figure 3) rin = 22.71k Ω Ais = 2.3

## Post-lab:

Are results had some issues in places and none in others. Due to that the calculated and experimental values varied greatly and not at all. We got similar results when Re2 was bypassed when compared to the calculated values. This is most likely due to that the calculated values are not standard resistor values and so we had to choose resistors that were close to the value needed.

## Screenshots:

figure 1                                                figure 2

figure 3                                                figure 4

# Lab 3 BJT Single-Stage Amplifier Design

## Prelab:

• Find the values for R1 , R2 , RC , and RE
 No bypass Avi = -1.24rin = 45.23kΩRB = 45.23kΩ ri =  759.5kAis = -48.23 RE & Re2 bypass Re = 0Avi = -144.23 Avs = -52.24Ais = -83.29 Re2 Bypass Re = 1kAvi = -365ri = 257.5k rin = -2.9Ais = -50.66 No Bypass Re1 & Re2 Bypass Re2 Bypass Avs -1.004 -52.24 -2.9 Avi -1.24 -144.23 -3.65 rin 42.69k 5.68k 38.47k Ais -48.23 -83.29 -50.66 Re1 = 1k Re2 = 2k

## Lab:

Actual values

 RS= 988.3 Ω C1= 10µF RE= 816.4 Ω R1= 11.972 kΩ CE= 220µF RL= 996.2 Ω R2= 15.768 kΩ 𝑓= 5 kHz VS= 500 mV

Measure Q-Point { ICQ, VCEQ }

 VCEQ 9.5 V ICQ VRERE=6.3 V816.4 Ω =7.716 mA

Measure Vi , VL, VS

 Vs 500 mV (see figure 1) Vi 420 mV (see figure 2) VL 409 mV (see figure 1)

Calculate Avs , Ais , rm

 Avs VLVS=409 mV500 mV = 0.818 rm VSIS-Rs=500 mV80.95 µA-988=5.188kΩ Ais ILIS=410.56µA80.95µA=5.07

## Post-Lab:

1. Calculate rX
 rx 𝛃(rLAV-rl-rm)= (180)(448.690.818-448.69-3.369)= 17.36 kΩ
1. There was some usual variation between the measured values and the calculated values due to that of the actual components. Mainly the difference was caused by the Beta of the transistor where we assumed that beta was 150 when our actual beta was not. Thus giving us the most error within out data.
2. As always calculations vary than actual results, but in order to get less error it would be better to use more accurate numbers for calculations such as resistor value and transistor beta.

Figure 1

Figure 2

# Lab 4 Single-Stage Amplifier Design (C.B.)

## Pre-lab:

 VTH = 5vRTH = 3.75kIB = 5.22µAIc = .856mArm = 30.37Ωrπ = 5.4627k Ωrin = 30.1867 Ω rL = 476.19 ΩAv = 15.68Avs = 0.8rc = 909.09 ΩAv = 29.96Ais = 0.878

## Lab:

Actual Values

 RS= 53 Ω C1= 1µF R1= 15 kΩ CE= 1µF R2= 5.5 kΩ 𝑓= 1 kHz RC= 10.2 kΩ VS= 100 mV RE= 5.5 kΩ RL= .98 kΩ

Measured values

 Vs(p-p) 98 mV VCEQ VRC= 8.61 V Is(p-p) 1.28 mA Vi(p-p) 30 mV ICQ 8.44 mA IL(p-p) 1.22 mA VL(p-p) 1.2 V AVS 122.4 VL1 4.8 V AIS 0.953 VL2 30 mV RIN 23.44Ω ROUT 159 KΩ

## Post-Lab:

The values gathered from pre-lab is what we used to determine the values for the circuit. Due to that they actual values were different than the calculations would give us a typical error. However, the measured values turned out to be pretty off. The reason for this I am still not sure of either we did the calculations wrong or we measured/ read the measurements incorrectly.

# Lab 5

## Pre-lab:

No prelab

## Post-Lab:

We could not get the circuit to work at first, at least we thought we couldn’t. We realized that the circuit was working it was just that are transistor was below cutoff frequency and so the transistor was not active. As we began to increase the frequency the numbers started to get better and mare more sense. We found our midband frequency was about 500kHz after taking enough data points and making the bode plot.

# Lab 6 – Common-Emitter Amplifier Frequency Response

## Pre-lab:

 rB= 50 kΩ rM= 1.73 Ω ri= 3.28 kΩ AVS= -2.6 IB= 100µA rπ= 260 Ω rin= 3.08 kΩ 𝑓 1= 1Hz IC=  15mA ro= ∞  Ω AV= 3.45

## Lab:

Change C and measure VDS to calculate ID. Plot      Change VDD and measure VDS to calculate ID. Plot

Use Curve Tracer to find IDSS  = 10.2mA

## Post-Lab:

Find ID  = ID=IDSS[1-(-RsID)/Vp]^2 = .05mA

Our results are quite the same with the theory of amplifier response and the low frequency characteristics of a common-emitter BJT amplifier. We had a bit of trouble at the beginning but we managed to get it to work in the end.