I will rate you at the same time l. (a) Design a base-compensated current source
ID: 1847163 • Letter: I
Question
I will rate you at the same time
l. (a) Design a base-compensated current source to give an output current of Io-3 mA. The transistors are matched with parameters B 75, VCC 5 V, -V 5 V, V 80 EE V, V. 25 mV, V 0.3 V, and V 0.65 V at IC 1mA. BE (b) Find CE2 (when Io 3 mA) V (c) Calculate the output resistance, Ro (d) What is the change in lo for a 1 Vchange in Vo? What is the new value of lo? (e) Find the value of Io when Q2 goes into saturation. 2. A Wilson current source drives a 10 KS2 load such that the output voltage is 5V. All transistors are matched with B -20, V 75V, and V 7 TV when Ic 50 LLA. Find: BE (a) the output current, lo. (b) the base-emitter voltages of all three transistors (assume V VBE2 VBE3) (c) the reference current, REF, using the values of Ic2 and I (d) the current transfer ratio, REF (e) the value of the resistance R. 15V 15V 0 KS2 REF Q3 EN Ct 9:00 PM 3/25/2014.Explanation / Answer
let the reverse saturation current for the transistors is Is.
then we know that,
a)
Ic = Is*(exp(q*Vbe/KT) - 1)------------------1
where KT/q = 0.026 V where K is the boltzman constant and T is absolute temperature.
now as Vo = 5 V given so, Iout = (15 - 5)/10K = 1mA.------------2.----------------------------------ans
from equation 1 we can write,
50u = Is (exp(0.7/0.026) -1) or Is = 1.01490*10^-16 A-------------------------3.
now, for transistor 3 we can wire usig Is value above that,
b)
1m = Is(exp(q*Vbe/KT) - 1)) or Vbe = 0.7778 V
so Vbe1,2,3 = 0.7778 V ------------------------------------------ans.
c) Ic2 = 1mA as Vbe is constant and
Ib3 = Ic3/beta = Ic3/20 = 0.05 mA
so,
Iref = Ic2 + Ib3 = 1.05 mA---------------------------------------ans.
d) from the values wehave,
Io/Iref = 1/1.05 = 0.95238 ---------------------------------------ans
e)
the node voltage at the collector of M1 is Vbe1 + Vbe3
or,
Vc1 = 2*0.778 V = 1.556 V.
Iref = 1.05 mA
so, R = (15 - Vc1)/Iref = (15 - 1.556)/1.05m = 12.8038 K Ohm-----------------------------ans.
2.
to get Io = 2.5mA.
u just need to provide the perfect resister value for the Iref branch.
as we have seen Is = 1.0149*10^-16 A.
so, Vbe for Ic = 2.5 m A will be,
2.5 = Ic(exp(q*Vbe/KT) -1) or Vbe = 0.80171 V.
so, now Vc1 = 2*0.80171 = 1.60 V.
so, for approximately 2.5mA current.
R = (15 - 1.60)/2.5m = 5.358 KOhm.
so u just use R = 5.358 KOHm resister to get that current.