{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Suhas 14

# Suhas 14 - EE 3221 Electronics II Laboratory Suhas Patel...

This preview shows pages 1–7. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE 3221 Electronics II Laboratory Suhas Patel Evan Trapani (Partner) 437-93-4360 June 28, 2004 Lab 5 (Experiment 14) Characterization and Design of an A C Variable-Gain Amplifier EE 3221 Suhas Patel June 28, 2004 Objective: To design and study the performance of a typical single-stage ampliﬁer, with particular attention to a stable dc operating point and a desired ac gain. Circuit Diagrams: Italicized Questions: 1.) What values did you choose? _) V+=15V R. = 1k potentiometer t v— = -1.45 V What is the dc voltage at the base? 9 2.5 mV. And how is it determined? 9 With no input, we attached the voltmeter at the base to measure the voltage. Theoretically, the voltage at the base can be determined by first finding the emitter current, which is approximately equal to the collector current. Then using the IR drop across the emitter resistor and the forward voltage of the pn junction, we can find the voltage at the base. If the value of Beta is known, one can use the collector current to calculate the base current and then determine the voltage at the base by computing the IR drop across the base resistor. 9 We biased our transistor such that our collector current was 0.75 mA, which is approximately also the emitter current. This current causes the emitter to be at approximately ~ — 0.7 V. Assuming the forward voltage across the pn junction to be 0.7V, the base would be at about 0 V potential, which is comparable to our measured voltage of 2.5 mV. What value did you choose for the capacitor? 9 1 UP. What is its impedance? —> 12c; = 1 /(2ﬁ*10000*10’6) = 15.9 Ohms. 2.) What is the value of R1 9 982 Ohms, which is a little less than the expected value of lk. Explain any difference from your measured value. 9 -10 = Gain = - Rc / Re. This would mean that our Re or R1 should be lk. Our measured R1 was 982 ohms, which is very close to lk. The difference is due to a couple of reasons. First, the formula we used to determine the value of R1 that would yield a gain of —10 is an approximation of the small / signal model, according to which the Gain = — (BRc) / (Re*(B + 1)); I/ therefore, our measured resistance for R1 should indeed be less than the M one indicated by the approximation formula. Secondly, the capacitor doesn’t 0D completely short out R2. Although relatively negligible, we have to take into »' account the equivalent impedance of the capacitor in parallel with R2. Since this J equivalent is actually nonzero, R1 is slightly less than lk. 3.) At what frequency below l0 kHz does the gain fall from —10 to —7? —> 81 Hz. 9w 1 47 P If necessary choose a new value for C; what is your new value for C? 9 2 uF. Estimate it theoretically from your measurement ole. 9 f=1/(2n*982*2 x106): 81 Hz. 4.) Redesign the circuit in ﬁgure 1 so that it has a gain of -2, by replacing the potentiometer with a ﬁxed resistor and removing the capacitor. .9 V+ =l0V 1111: = a: T Elli: 4.71:. l ' V— = What is the new dc collector voltage? 9 4.70 V, which is approximately half the positive supply voltage. 5.) What is the value of RL? W @ 9 10k, same as Re. What is the value of your Cc? 9 0.47 uF. The 3 dB point is at 15 Hz, which is less than 100 Hz as the design required it to be. PSPSCE 6 Cup"! 3.0671): \/ , 36» E30 Nsn’. >ocmskum AAmupmV> \Amuqmv>vmo D Scan 2%: N58; 253 3% 2:: r. __ i - 1 .- H I _ - L. _. ,. 1 +1 3:. 'PSP'ICE GENEMTED PLoT I‘WALYsIs w "THeee Avie Two BREAY-PO‘LMT FREQUEchgs 0N nus 91.01: 1146:0025, , ﬂa‘m: ‘ Ho) “nuts :5 P: BANDPAss FREQUENCY Résmmsb BAN-owier ls RQUQHLY \0 kHz, The LoxJéﬂ BQCH‘POINT F‘Réqoeruo.’ IS an)va By I \ x|4H—z' 2111‘ch QC) 7 7.1: (o.-£r1a¢)(lok+ 10k) ' —-—> 1% “AND THC HXQMQ SRCArvoxiur FREQ“:ch IS GIVE/U 13y ‘ l ‘ —————-————— :3 mm; 11c“, (an RC) 2 21v (3:06'AF)00‘< Wok) Hewce/ +Me owefl BReAwowv raeauewc‘f Is SET BY Cg) mun me 6 Rm 5 p LoweR is PEPENDENT on cc. 'THE GﬁIN 3:5 AppgoXthT-ELY 1 FOR F‘ZQuemcIe'g ggrwu—w £0 dAA/b FHI _ THe‘ @IN IS Arre/uvATED FOR ALL OTHGQ FQEQuemcxes_ ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

Suhas 14 - EE 3221 Electronics II Laboratory Suhas Patel...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online