Question

Problem 4: The following figure shows a biplolar op-amp circuit, the input differential pair Q,-Q, is loaded in a current mirror formed by Q2-Q. The second stage is formed by the current-source-loaded common-emitter transistor Q. Unlike the CMOS circuit, here there is an output stage formed by the emitter follower Q. The function of capacitor Cc will be explained later in feedback chapter. All transistors have B=100, VBE = 0.7V, and r = 0. (a) For inputs grounded and output held at OV (by negative feedback, not shown) find 0.4 mA the emitter currents of all transistors. 0.5 mA (a) Calculate the gain of amplifier with R, = 1k12. to 02 16 Cc +5 V 0.4 mA 1 mA -5V

Answer 1

ANSWER

Given that

  
given that your 0.504 . > AMA -JV a biplolar op-amp ciraut, the circuit differential pair Q,-Qq is loaded in a carent minor formed by as-Qu. 2nd stage is formed by the current- source - loaded .common remitter transistor Q5.

CMOs circuit output stage formed by emitter follower QG The function of capacitor Cc will be explained All hansistors have 3=100, lyBel0.77, imputs grounded and outputs al or from the perspective signal source, the current flowing into the transistor is the base current. The emitter current of the transistora, is given by I n = 0.4 mA I €1=0.2 mA

we have I ea a Ica. Thus the emitter current of transistor on is. IER = 0.2mA is substitute in the equation. 0.2'7162 IE2 20:2 ma the emitter current I Eg of transistor o is equal to the emitter current feu of transistoray, that is I E B = Feui That is IE3 Itu 20. 2mA the emitter cument I es of transistor az is given by - IES 0 MA -20 MA

The emitter current Jeu, is. czma. Determine the value of a 9 me 9 ml Ich 0.2 MA 25 mv 20-008 X 8 MAIV = 8 malv Determine the value of res. Vyzomr, fer 2.0.5 Tes as my 0.5 MA son res

netermine the value of Res. Ros=(1+Bores ..! Ver=50 (1+10o) (50) B-100 = 101650) - 5050 R Determine the value of An.. A = 9 me Ros Jml 28x103 - (8x10²) (5050) Rota5050 8x 5050 - U10400 = 40.4 Determine the value of Asi . As - BRL in Ri=1ks 5. RL Riko R = 10x103 -100) (10x103). B = 100 LC100) (10x1000)

- (100) (10000) = 000'004 As 3 - 20,000 the overall voltage gain is, A A AS Substitute thee A, 540,4, A5 = 20000 in the above equation. 7.40.4x6-20000) ..2 (2012) (-20000) . LA -40.4000 V/v.