ETE 305 LAB collection

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 2

Lab 2 – Common-Emitter Amplifier with unbypassed Re 5

Lab 3 BJT Single-Stage Amplifier Design 8

Lab 4 Single-Stage Amplifier Design (C.B.) 12

Lab 6 – Common-Emitter Amplifier Frequency Response 17

IB= 6.67mA | R1= 169.5 kΩ |

RE= 3 kΩ | R2= 61.67kΩ |

RC= 6 kΩ |

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 |

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 |

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.

Without Ce capacitor Measured Rin

With Ce capacitor

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 = ∞ |

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 |

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.

figure 1 figure 2

figure 3 figure 4

- Find the values for R1 , R2 , RC , and RE

No bypass | Avi = -1.24 rin = 45.23kΩ RB = 45.23kΩ | ri = 759.5k Ais = -48.23 | ||

RE & Re2 bypass | Re = 0 Avi = -144.23 | Avs = -52.24 Ais = -83.29 | ||

Re2 Bypass | Re = 1k Avi = -365 ri = 257.5k | rin = -2.9 Ais = -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 |

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 |

- Calculate rX

rx | 𝛃(rLAV-rl-rm)= (180)(448.690.818-448.69-3.369) = 17.36 kΩ |

- 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.
- 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

VTH = 5v RTH = 3.75k IB = 5.22µA Ic = .856mA rm = 30.37Ω rπ = 5.4627k Ω rin = 30.1867 Ω | rL = 476.19 Ω Av = 15.68 Avs = 0.8 rc = 909.09 Ω Av = 29.96 Ais = 0.878 |

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 |

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.

No prelab

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.

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 |

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

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.