ECE 5040Prof. Alyosha MolnarHomework 5Tri CaoDate:
12/05/2014
Problem 1 Recordinga. We haveBecause Vn* = 4uV, BW = 10kHz and
Vod = kT/q = 26mV, we have Imin = 0.42uA.
b. Total power consumption is: (Ibias + 2I)*VDD
c. If NEF = 2.5 and we have Vn*ideal = 4uV then Vn* = 10uV.Using
the equation in part a, we can find Ibias = 67.96 nA.
d. We haveand then 73.62nA.
e. From that, overall NEF would be 2.4 (with Vn*ideal =
4uV.)
Problem 2 Stimulationa. Sketch
We want Vds for each transistor is larger than 400mV, then VDD
should be 1.6V because we have 2 transistors in the way from Vout
to VDD, and 2 transistors from Vout to VSS.
b. Assume that the cathodic pulse is only 71.25uA (-5%) and the
anodic pulse is 31.5uA (+5%), then the difference in charge is:With
Cj = 10nF, we can find V:
c. Maximum current the reset transistor must be able to handle:
ts = 300us.
Problem 3 System Powera. The system has 500 recording electrodes
and 50 stimulation electrodes.During 1 second, each stimulator
pulses 100 times. Assume that each pulse lasts 500us, then in 1
second, the stimulator uses 100uA for 0.05 sec and 1uA for 0.95
sec. In average, we can say that the stimulator uses about
5.95uA.
We assume that each amplifier uses a 73.62 nA bias current. Then
we have total current = 20uA + 50*73.92nA + 50*5.95uA = 0.32mA.From
that we can find total power = 1.6V * total current = 0.51W.
b. Using bridge rectifier requires VAC to be 1.4V larger than
VDC. Hence, VAC = 3V.
We have ZL = V/I = 3/0.32mA = 9375 .
c. We have V2 = V1 . k . sqrt(L2/L1)With L2 = L1 = 1uH, k = 0.1,
V2 = 3V, we find V1 = 30V.
d. In the non-ideal case, we have
We can find Rs2 = wL/Q = 2*pi*400MHz*1uH/20 = 125.6
Also, ZL = 9375, L2 = L1 = 1uH.
Therefore, V1 = 37.97V.
We notice there is a 21% loss in the coil. Total power =
