E6.1

Since the differential pair bases are supplied with difffernet voltages, one transistor is ON and the other is OFF.

Therefore the emitter voltage ve is (0 + 0.7) = 0.7 V. This leaves Q0 OFF and Q1 ON.

Emitter-base voltage of Q0 = 0.2 V. Therefore, it is in cutoff.

Current flowing through R1 = (5 -0.7)/ 1K = 4.3 mA.

Current flowing in R2 =0 since Q0 is in cutoff. Therefore vc1 = vee = - 5V.

Current in R3 is 4.3 mA. Therefore Vc2 = -5 +(4.3 mA * 1 K) = -0.7 V.

E6.2

i_E1 = 0.99 I. Also i_E1 + i_E2 = I; Therefore i_E2 = 0.01I.

i_E1/i_E2 = exp((v_B2 - v_B1)/V_T) = 99; V_T = 25 mV;

Differential signal = v_B2 - v_B1 =  V_Tln(i_E1/i_E2) = 25 ln (99) = 114.8 mV

E6.3

Collector current (bias) = 1 mA/2 = 0.5 mA.

re = 1/gm = VT/Ic = 25 mV/0.5 mA = 50 ohms.

Total resistance between the bases of the transistors is (on emitter side) = 2(re + Re) =

The input differential resitance = (B+1) 400 = (100 +1) 400 = 40 Kohms.

Voltage gain = vo/vs = vo/(vb1-vb2) *(vb1-vb2)/vs

Worst-case common mode gain of two collectors are accurate to 1%. Worst case error occurs when RC1 is 1.01Rc and RC@ is 0.99RC.

Common mode gain = Acm= 10k * 0.02/[150 + 50 + 400K] = 0.02/40 = 0.5 mV/V

CMRR = 20 log(Ad/Acm) = 20 log(80000) = 98.06 dB

VA = 100 V; ro = VA/IC = 100/0.5 mA = 200 K.

Ricm = 0.5ru || (B+1) R || (0.5ro*(B+1)

ro = 200 K; R = 400K ; ru = 200 M

Ricm = 100 M || 40 M || 10 M = 7.41 M

P6.1

I = 1 mA; B = 100;

Voltage at the emitters = -2 - 0.7V = -2.7 V

Ic = 0.5 mA;

Vc1 = Vc2 = 5 - (0.5 mA)* 3 K = 3.5 V

Accurate values of Ve = -2 - 0.7 +V_Tln(1/0.5) = -1.3 + 25 mV * 0.69 = -1.317 V

Vc1 + Vc2 = 5 - (100/101)*0.5 * 3 = 3.515 V

P6.5

(a) Largest CM signal that can be applied at the inputs of BJT with Vcb = 0 is the voltage that prevents the diff amp transistors from entering saturation. i.e, =Vcm = Vc = Vcc - I/2 Rc

(b) To steer current entirely on one side of the differential pair, the differential input voltage can be obtained from

Note that one collector will see a positive voltage change whereas the other will see an equal negative voltage change.

(c) Vcc = 5 ;

To allow a common mode signal of + 5 V, the value of IRc/2 = 2V. (2.5,  if we assume that collector-base junction will be forward biased (transistor entering saturation) with Vbc = 0.5).

Negative supply voltage should be at least -3 -0.7 = -3.7 V.

(d) B = 100; Ib,max = 2 uA.

Ic, max = 100* 2uA = 200 uA;

Rc = 2/200 uA = 10K.

P6.8

For large values of B, the output voltage is given = (iE1 - iE2)

For (VB1 - VB2) = 5 mv and with VT of 25 mV, gain an be obtained in terms of IRc as shown below.

The common mode voltage can be obtained in terms of the differential gain. 

The table given below lists the values of gain and corresponding values of common-mode input voltage. It is clearly seen that increasing values of differnetial gain results in reduced range of common-mode input voltage.

P6.11

(a) Q1 is cutoff; Therefore, vo = VOH = Vcc - 0.7 - (10 mA /100) * 1 K = 4.2 V

Including variations in Vbe, VOH = 4.2 - VT ln(10) = 4.144 V

(b) Q1 is carrying all the current. Therefore current through Rc2 = 1 mA + (10 mA/100) = 1.1 mA.

The output voltage vo = VOL = Vcc - 0.7 - 1.1 mA * 1K = 3.2V

Including Vbe variations, VOL = 3.15 V.

(c) For Q1 to to conduct 1% of I, Vb2 - Vb1 = VT ln(99) = 115 mV.

Therefore vi = VIL = VR - 0.115 = 3.525 V

(d) For Q1 to conduct 99% of I, Vb1-Vb2 = 115 mV.

Therefore vi = VIH = VR + 0.115 = 3.755 V.

P6.12

I = 200 uA;

B = 200;

Ic = 100 uA;

gm = 100 uA/25 mV = 0.004 S

re = 1/ gm = 250 ohms

Input differential resistance = 2*(B + 1) re = 100 Kohms.

P6.16

Gain, G = 4/0.2 = 20

Vbe,max = 5 mV; Rid > 80 Kohms; B >= 200.

To limit Vbe drop, external resistance (RE1 = RE2 = RE) must be added as shown in above schematic.

With 200 mV appled between the inputs, a 5mv drop should occur across the two base-emitter junctions.

Therefore re/(re + Re) = 2*5mv/200 mv = 1/20 = 0.05.

Re = 19 re.

Rid,min = 2* B * (re + Re) >= 80 Kohms or (re + Re) >= 200 ohms;

Gain = 20 = Rc/(re + Re) or Rc = 20*(re+Re).

One possible solution, re = 10 ohms; Re = 190 ohms; Rc = 4 Kohms;

re = 10 ohms => I /2 = 25/ 10 = 2.5mA

I = 5 mA

P6.21

Current through Re = [0 - (-5) -0.7]/4.3 K = 1 mA;

Current in each emitter = 0.5 mA;

re = 25 mV/0.5 mA = 50 ohms;

Gain = Ad = vo/vd = (RC/re)/2 = 2 K /(2 * 50) = 20 V/V

(b) Common mode gain = Acm = vo/VCM = RC/[re + 2*RE] = 2 K/[8.65 k] = 0.231

(c) CMRR = 20 log(Ad/Acm) = 38.75 dB.

(d) vo = 0.0231sin(2II x 60t) + 0.1 sin(2II x 1000t)

P6.26

Ad/2 = 100.

Rid >= 10 K

Acm < 0.1

I = 2 mA

B >= 100.

Ie = I/2 = 1mA;

re = 25 mV/ 1mA = 25 ohms

Rid >= 10 k = 2 *B*(re + Re); Therefore an extrernal RE of at least 25 ohms must be added.

Ad = 200 >= Rc/(re + RE) or Rc>= 200*50 = 10 Kohms.

To allow for +2 V swing VCC >= 7V and select VEE = -VCC.

Acm = RC/[re +RE + 2ro] < 0.1

2ro = 10K/0.1 - 50 = 100Kohms

The output resistance of the bias source, ro >= 50 Kohms.