EE100 07S Mid-Term 2
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E6.1 (a) The frequency of v in (t ) = 2 cos(2 2000t ) is 2000 Hz. For this frequency H (f ) = 260 o. Thus, Vout = H (f )Vin = 260 o 20 o = 460 o
and we have v out (t ) = 4 cos(2 2000t + 60 o ). (b) The frequency of v in (t ) = cos(2 30
WriTe a currenT equaTion aT The Top node:
_ _ V50) Cal/(U)
2 exp( 3:") 7 A) + 6' df
SubsTiTuTe The parTicular soluTion suggesTed in The hinT:
2 exp[3f') E g exp(3f') 3A6 exp(3r)
Solving for A and subsTiTuTing values of The circuiT param
P1.19 Q=currentxtirne=(3 arrperes)x(2 seconds)=6cou|orrbs
The magnitude of the energy Transferred is
Energy : Ql/ : (6) ><(12) : 72 joules
Notice that fab is positive. If the current were carried by positive charge,
it would be entering terminal :17. Thus
*P5.83. 3. Find the Thévenin and Norton equivalent
circuits for the circuit shown in Figure P5.83.
b. Find the maximum power that this circuit
can deliver to a load if the load can have any
complex impedance; c. if the load is purely
135.3. The voltage v(t) 2 10511105030 V appears
across a 20-9 resistance. Sketch v(z) and p(t)
to scale versus time. Find the average power
delivered to the resistance.
P5.10. A current i(t) = 2005(1000m) A ows
through a 5-9 resistance. Sketch i(t) and pm
Helpful things to remember from EE100
From this you can prove the effective resistance of parallel and series
combinations of resistors:
A common circuit is a voltage divider:
You should be able to prove t
Homework Assignment # 6
Due date: 2-27-14
Please attempt the following problems from the text:
"Error in book. The second rectified half-wave is 0 at 125 ms, *not* 150 ms."
Solutions for Exercises
E9.1 The equivalent circuit for the sensor and the input resistance of the amplifier is shown in Figure 9.2 in the book. Thus the input voltage is
v in = v sensor
Rsensor + Rin
We want the input voltage with an intern
E11.1 (a) A noninverting amplifier has positive gain. Thus v o (t ) = Avv i (t ) = 50v i (t ) = 5.0 sin(2000t ) (b) An inverting amplifier has negative gain. Thus v o (t ) = Avv i (t ) = 50v i (t ) = 5.0 sin(2000t ) E11.2
RL Vo 7
E12.1 (a) vGS = 1 V and vDS = 5 V: Because we have vGS < Vto, the FET is in cutoff. (b) vGS = 3 V and vDS = 0.5 V: Because vGS > Vto and vGD = vGS vDS = 2.5 > Vto, the FET is in the triode region. (c) vGS = 3 V and vDS = 6 V: Because
E10.1 Solving Equation 10.1 for the saturation current and substituting values, we have
iD exp(vD / nVT ) 1
10 4 = exp(0.600 / 0.026) 1 = 9.502 10 15 A Then for vD = 0.650 V, we have
= 0.6841 mA
iD = I s exp(vD / nVT ) 1 = 9.502
One of the greatest achievements of electrical
engineering has been the development of the
operational amplifier (op-amp). In a negative
feedback configuration the op-amp is able to
provide linear, distortion free signal a
Nonlinear characteristics of a real op amps.
a) Output voltage swing
As I mentioned before each real op amp has a
saturation output voltage which is usually around
2volts below dc voltage applied to pins 7 and 4.
If we apply 15 the saturation v
We studied inverting and non-inverting amplifiers.
But sometimes in the first glance it is not clear
that whether the amplifier is inverting or noninverting. Following examples show these cases.
Last time we talked about how to find equivalent
resistance. In practical cases, we measure the
voltage difference between the terminals in
question and the total current entering or leaving
either terminal. Then Req = vab/itotal.
Lecture 7 Notes
When a variable current passes through
A coil of wire, it produces a variable magnetic field inside the
coil. This changing magnetic field in turn, induces an extra
current in the coil that according to Lenzs Law opp