Physics Courseware Electronics

Transcription

1 Physcs ourseware Electroncs ommon emtter amplfer Problem 1.- In the followg ommon Emtter mplfer calculate: a) The Q pot, whch s the D base current (I ), the D collector current (I ) and the voltage collector emtter (V E ) b) Is the transstor the actve regon? c) alculate the voltage ga on the amplfer assumg the capactors behave lke shortcrcuts at the sgnal frequency. Soluton: The base crcut gves us I : I MΩ µ, so: ollector current: I β I µ 2. 03m nd the collector sde gves: V E kΩ(2.03m) 5. 5V The transstor s the actve regon. To calculate the voltage ga, notce that the dynamc resstance s: r 2. kω d I 11.3µ 21 The sgnal s coupled through a capactor drectly to ths resstance, so: 180, whch means that the collector current s: 2.21kΩ 2.21kΩ Ths current goes through the load resstance parallel wth the collector resstance: R 1 0. Ω 1/ 3.2kΩ + 1/1.2kΩ 872k Vout 180, so 0.872kΩ -71 V 2.21kΩ

2 Problem 2.- a) Fd the value of base resstance (R ) that would put the transstor the mddle of the actve regon. b) alculate the voltage ga on the amplfer. [ssume the capactors behave lke short-crcuts at the sgnal frequency] Soluton: a) To fd the deal value of base resstance (R ) we assume that the voltage collector emtter s half the source voltage, so: 12V V E 6V 2 Ths would put the transstor the mddle of the actve regon. Gven ths, the voltage drop across the collector resstor has to be 6V too, so the collector current wll be: 6V I 1.875m 3.2kΩ I 1.875m Wth ths formaton we can fd the base current: I 10.7µ β 175 To fd R we wrte the base loop equaton: 12V R I + 0.7V R 12V 0.7V 10.7µ 1.054MΩ b) To calculate the voltage ga on the amplfer, we do an analyss: We assume the capactors behave lke short-crcuts at the sgnal frequency, whch gves us the crcut:

3 Our model for the transstor conssts of a dynamcal resstance on the base-emtter sde and a current source on the collector sde. The value of the dynamcal resstance wll be: r 2. kω d I 10.7µ 33 nd the current source wll have a value of β as shown the fgure: Sce the sgnal s connected drectly to the base resstance, the base current wll be: rd Multplyg by beta we get the collector current: β V 175V β Notce that the load resstance and the collector resstance are parallel, gvg an equvalent of: 1 R eq 640Ω kΩ 800Ω The output voltage wll be the collector current tmes ths resstance: 175V 175(640Ω) Vout Req 640Ω 47. 9V So the voltage ga s -47.9

4 Problem 3.- alculate the voltage ga the followg amplfer. lso check that t s workg the lear regon (make sure VE s somethg sensble). Soluton: Frst we calculate the Q pot of the transstor. To do ths: The base loop equaton s 5V 250kΩI + 0.7V I 5V 0.7V 250kΩ 17.2µ Wth ths value we can calculate the collector current: I βi 125(17.2µ ) 2.15m Wth ths value of the collector current, the collector-emtter voltage s V E 12V 2.4kΩ(2.15m) 6.84V Ths s perfectly OK. The transstor s the mddle of the lear regon. Now we can look at the small sgnal crcut. The capactors are gog to behave lke short crcuts at hgh frequency and the base emtter juncton wll be represented by a resstance whose value s: r' 1.45kΩ I 17.2µ Wth these pots consderaton the base loop looks lke ths:

5 The base current wll be: 1.45kΩ The collector loop wll be: Where the collector current wll be: β kΩ The load resstor and the collector resstor are parallel, so they gve an equvalent of: 1 R equvalent 600Ω Ω 800Ω So, the voltage ga wll be: 600Ω 125 V R out equvalent 1.45k Ga Ω 51.7 V V V

6 Problem 4.- alculate the voltage ga of the followg amplfer: Soluton: To solve ths problem we frst fd the Q pot. To do ths we analyze only the D sgnals. In ths case, the capactors behave lke open crcuts so the crcut looks lke ths: We can start analyzg the base loop frst. The voltage base emtter s approxmately 0.7V, so the base current s: I V 0.7V 20V 0.7V R 630kΩ 30.6µ We need ths value to get the ternal, dynamc resstance of the base-emtter juncton, but we should also check that the transstor s the actve regon. So let s see the collector loop: I βi 100(30.6µ0 ) 3.06m V V R I 20V 3.06m(3.3kΩ) 10.1 V E

7 Ths last value tells us that the transstor s operatg the mddle of the actve regon (perfect!). Now we analyze the small sgnal: Ths tme the capactors behave lke short crcuts. From the pot of vew of the base loop, the transstor looks lke a smple resstance wth a value of: r' 816Ω I 30.6µ Q The base loop small sgnal looks lke ths: The small sgnal current ( ) can be calculated now: r' 816Ω The collector current s then: β Ω Now we analyze the collector loop. Recall that the transstor behaves lke a current source: Notce that the load resstor s parallel wth the collector resstor. They together have an equvalent value of: R R L (3.3kΩ)(1k Ω) R out R L // R 767Ω R + R L 3.3kΩ + 1kΩ The output voltage s ths resstance tmes the collector current: Vout R out 100 (767Ω) Ω So the voltage ga s -94.