Question

2) For the circuit shown below find values for R1, R2, R3, R4, and R5 so that these criteria are met: - Overall gain G, Vsource to Vload, is nominally 10, i.e. between 9.8 and 10.2 - Input resistance is 10K or greater Assume that the capacitors are large enough so that they can be considered short circuits for AC signals.. R3 Rsource 2000 0 3 Rload - 5000 Vsource

Answer 1

> 3 w ho vo 2k § Rhead = 5k 1. Ry · R2 ? u q al n = a - overall Input gain= 10 Teristance z loka Small Signal model ; - - Voul- no 3 m ube ak Tie Roller mube 2ks il y of tie - Vbes ne fill Be

take P, & R20 = 2 . M. ie Rc 2me P2= 2 Mus Ub = Vs . c ie Us x Rill R Rst Rilla No = Vb VS X Dilma a vs L - 2ktima -KVL) - Now - Vst be t Im Ube Ro = 0 at output Voe ('t gm Rq) = vs Ube a itgm Rq vout = - gm Vbe (23|| R.) Vout= -9ml vs . -gm (R3 ll Rey Vout = It gm Rq) (Rollkel It aim Rq take R3= R2 = 5k Now Boll Re = 2.5ks = o Cgiven) -9m (2.5ke) itom R4 gm = fa la = 40mAlv take Ic= IMA . L Now - qomalu (2.562) 10 It (240mA/uj Rq R4 = 2252 =

. Now Rina Rst ( R 11 (2) || (B+1) (be + Rt) - 25 t (ime) || ( 300+ 1) (2587 228) (3=300) me = vt fic = 2K+ (1ma) || (301) (250) = 250g = 26+6919842 1. = 11.983 kr Rin z non loka = = Now DC Analysis . I lov 3 R3 = 5km Ri=anon 1 Vbe - 3 Ra=2282 Ames Ę RS - ? 10x amu VB = lux R2 Rata amoat Rma - (voltage dewision) -Vet Vher It (Rat R5) EO VB- 50.7 = -Te ima R4 + 25 (If ~ 1C) 4.3 kr

Rg + RS= 4.3k2 R5 = 4.3 kr - 2252 Rs = 4:075 km so finally Ri= R2 = ama R3= 5ks Rq = 2252 R5 = 4.075kr Rin = floagka z loka DC 1 3) if not at the capacitor ca is short fort our desired low frequency, so for gain will reduce small it may freas. Cie low frequency UUS Signals Zc = 2nfo to if c. is we need small . High freal to it. ie Zc zo short