314 EECS Fall 2007 HW 02 Solutions
Homework 2 Problem 1
Part1: 1) The problem statement gives us the following information: R R X " P ! K! 2 Vout " a ! b!
(R In addition, we know because of voltage division that Vout " VS & X &R ' P the given information for R X , you get: ( 1 %( RP % Vout " VS & # & R #& 2 ! K! # $ ' P $' ( 1 K! % Vout " VS & ! # &2 R # P $ ' KVS 1 Vout " VS ! ! 2 RP By matching coefficients, we can see that: 1 a " VS " 6V 2 KV S 3 V V " " 0.12 b" 25 deg deg RP
% # . If you substitute # $
Page 1 of 12
EECS 314 Fall 2007 HW 02 Solutions
2)
Plot of Vout vs. Alpha 16 14 12 10 8
12V
Vout (V)
6 4 2 0 -2 -4 -80
0V
-60
-40
-20
0 alpha (degrees)
20
40
60
80
x=linspace(-75,75,5000); y=6+ 0.12*x; plot(x,y); xlabel('alpha (degrees)'); ylabel('Vout (V)'); grid;
3 ! , the output voltages 25 exceed 0V and 12V at -50 and +50 degrees respectively. Since this isn't possible, it implies that with angles lower than -50 degrees, the output voltage is 0V and with angles greater than 50 degrees, the output voltage is 12V.
For the plot, note that if you just plot the equation Vout " 6 !
3) If we rearrange Vout " 6 !
,
3 25 ! , we get: ! " )Vout + 6* . 25 3
25 50 ! 4 " )4 + 6* " + - +16.67, 3 3
!7 "
25 )7 + 6* " 25 - 8.33, 3 3
,
4) .! "
25 .Vout 3 25 .! " ).010* " .083333, 3
Page 2 of 12
EECS 314 Fall 2007 HW 02 Solutions
5) We can rewrite the error above as: .! "
.Vout KVS . We also know that b " . b RP .Vout .Vout / RP When you substitute, you get: .! " " . By inspection, you can KVS K / VS RP see that .! will decrease when R P is decreased, when K is increased, or when VS is increased.
Part2: No, you cannot use the device above. The range of the output voltage exceeds the desired range. 1 When ! " 0 , Vout " VS . Since we know that we want Vout " 2.5V when ! " 0 , we 2 have: 1 Vout " VS " 2.5 2 VS " 5V This means that you must decrease VS Next, when ! " 50 , we would like to have Vout " 5V . KVS 1 Vout " VS ! ! 2 RP 1 5K 5 " )5* ! )50* 2 RP K 1 " RP 100 Therefore, as long as you keep the above ratio, there is more than one solution. For example, you could change K to 3 k0 / deg and R P to 300 k0 . However, if K is increased, so must R P and if K is decreased, so must R P .
Page 3 of 12
EECS 314 Fall 2007 HW 02 Solutions Problem 2
Potentiometer RP is what effectively limits the range of detection for the circuit. The adjustable tap on the potentiometer gives fractional resistance Rx = RP / 2 + K !. Because the range is limited as Rx ! (0, RP), this puts ultimate limits on the angle variable !, given as:
0 Rx Rp Rp 0 K Rp 2 Rp Rp 2K 2K 40 40
The actuator is turned on when Vout = Vs. The comparator will make the output voltage Vout = Vs whenever the potential V2 at the non-inverting (+) input terminal is greater than the potential V1 at the inverting () terminal. V1 and V2 are given by two voltage dividers:
V1 V2 Vs Vs Rx Rp R4 R3 R4
So the actuator turns on when:
V1 V2 Rx R4 Rp R3 R4 1 K R4 2 Rp R3 R4 1 2 2.5 k 200k 10 250k 150k 250k
And conversely, off when ! > 10. Of course, both on/off angle ranges remain within the above 40 limit.
Page 4 of 12
EECS 314 Fall 2007 HW 02 Solutions Problem 3
1.
As stated, the resistance RT is decreases by 25 k! at the elevated temperature, so the thermistor has negative temperature coefficient (NTC).
2.
The actuator is turned on when when the MOSFET switch closes; this occurs when the gate voltage is above the stated threshold VT = 3.5 V. The comparator will make the output voltage Vout (connected to the MOSFET gate) equal to Vs whenever the potential V2 greater than V1. Since Vs = 10 V, this is sufficient to turn on the actuator. V1 and V2 are given by two voltage dividers:
V1 V2 Vs Vs RT R1 R3 R4 RT R4
Conveniently, R1 = R3, so the on/off transition point occurs whenever RT = R4 = 89k!.
3.
Knowing that the thermistor is linear, RT is given by:
RT 120k 45k T 100
So solving gives the on/off transition temperature:
RT R4 45k T 100 89k
120k T
68.89
4.
High temperatures will decrease RT, causing voltage V1 to decrease, causing the comparator to raise Vout, turning on the MOSFET and the actuator. Logically, rising temperatures should activate a cooler.
Page 5 of 12
EECS 314 Fall 2007 HW 02 Solutions Problem 3
5.
Solve for RT at 55 F to find the necessary R4:
RT T 55 F 120k 45k 55 100 95.25k
6.
Vs was not used in any of the RT calculations, so no change is necessary.
Page 6 of 12
EECS 314 Fall 2007 HW 02 Solutions Solutions to problems 4, 5 Revised Problem 4
Part 1 The output voltages changes from VOUT,HIGH to VOUT,LOW when VIN = V+,HIGH. We can use the equation for V+,HIGH given in the big picture: = +, = , 2 1 + 1 + 2 1 + 2
Plugging in VOUT,HIGH = 10V, VIN = 6.05V, VREF = 6V, R2 = 2k: 6.05 = 10 Solving gives R1 = 158.0 k. 2 1 + 6 1 + 2 1 + 2
To find V+,LOW just use the similar equation given in the big picture. +, = , Substituting values gives: +, = 0 Part 2 use Simply the same method as in part 1 to calculate the resistance for the new conditions: 3.05 = 10 2 1 + 3 1 + 2 1 + 2 2 158 + 6 = . 158 + 2 158 + 2 2 1 + 1 + 2 1 + 2
Solving gives R1 = 278.0 k. The same resistor cannot be used.
Part 3 Simply calculate the resistance given the new conditions: 3.15 = 10 2 1 + 3 1 + 2 1 + 2
Page 7 of 12
EECS 314 Fall 2007 HW 02 Solutions Solutions to problems 4, 5 Revised
Solving gives R1 = 91.33 k. The resistance cannot be tripled to get the desired effect. In fact, it should be much less than before.
Part 4
See part 3.
Problem 5
The equation for the RX with the given information included is: = 200 + 1/ 40 = 140 2
Since the voltates over RX and R4 should be equal, both resistors should divide the same amount of the total voltage, VS. Therefore: 4 = 3 + 4
Plugging in values, and cancelling out the source voltage.... 140 4 = 200 100 + 4 Solving gives R4 = 233.3 k. Also, using voltage division: 2 = 4 = 8.4 3 + 4
Part 2 Increasing the angle of the arm will increase the resistance RX, and thus will increase the voltage V1, which is the input voltage to the Schmitt trigger. So we require that the Schmitt trigger switches from high to low when RX reaches the appropriate value determined by the stopping angle.
Page 8 of 12
EECS 314 Fall 2007 HW 02 Solutions Solutions to problems 4, 5 Revised
First calculate the threshold RX. = 200 + 1/ 42 = 142 2
Now calculate the input voltage to the Schmitt trigger, which is found by voltage division: 1 = 142 = 12 = 8.52 200
This is the input voltage at which the output will switch from high to low, so it must be equal to V+,HIGH. VREF is the V2 that was solved for before. +, = , 8.52 = 12 Therefore R1 = 4.35M 2 1 + 1 + 2 1 + 2
150 1 + 8.4 1 + 150 1 + 150
Now calculate V+,LOW. +, = 0 2 4.35 + 8.4 = 8.12 1 + 2 4.35 + 150
This corresponds to the VIN at which the arm is at the start angle. Using the voltage division formula again, we can find the resistance RX at which this occurs. 1 = 8.12 = 12 Therefore RX = 135.3k Now solve the arm angle equation 135.3k = And so finally, = . 200 + 1/ 2 200
Page 9 of 12
EECS 314 Fall 2007 HW 02 Solutions
Problem 6
I1k
1&2. Write node voltage equation for the non-inverting input terminal and simplify it using the GR#1:
I 5k = I 20k " 5 ! V+ V+ ! 0 . Thus, V+ = 4 (V) = 5k 20k
3. From GR#2, V- = V+ = 4V = V1 4,5 & 6. Write node voltage equation for the inverting input terminal and simplify it using the GR#1:
I 10 k = I 1 " 6 ! 4 4 ! Vout = . Thus, I1 = I10k = 0.2mA and Vout = -6 (V) 10k 50 k
(!6) ! 0 = !6mA 1k 8. KCL: I 1 + I 2 = I 1k " 0.2mA + I 2 = !6mA . Thus, I 2 = ! 6.2 mA
7. I 1k =
Page 10 of 12
EECS 314 Fall 2007 HW 02 Solutions
Problem 7 Part 1
Vout ,1 = Vs R2 '' R4 , R1 + R2 ' ' R3 + R4 R2 ( R3 + R4 ) / R2 + R3 + R4
where R2 ' ' = R2 || ( R3 + R4 ) =
Vout ,1 = Vs
R2 R4 R1 ( R2 + R3 + R4 ) + R2 ( R3 + R4 )
Problem 7 Part 2
Using GR#1, we know I1 = I2. R2 Thus, V + = Vs R1 + R2 Using GR#2, we know V+=V-.
Vout = V! = Vs
R4 R3 + R4
R2 R4 R1 + R2 R3 + R4
Problem 7 Part 3
Plug in R1=10 R2=20 R3=1 R4=1
Vout,1 = (6/7)V = 0.923077 V, and Vout,2 = 4V.
Comment by Alexander Ganago:
As stated in lectures, node voltage equations are the most powerful tool for solving circuit problems. In Part 1 of this problem, for the node A, where R1, R2 , and R3 are connected, we obtain:
! ! Page 11 of 12
!
EECS 314 Fall 2007 HW 02 Solutions
VA " VS VA " 0 VA " 0 + + =0 R1 R2 R3 + R4 $1 1 1 ' 1 VA # & + + ) = VS # R1 % R1 R2 R3 + R4 ( VA = VS # 1 R1 1 1 1 + + R1 R2 R3 + R4
Also, for the output node,
!
VOUT " VA VOUT " 0 + =0 R3 R4 $1 1 ' 1 VOUT # & + ) = VA # R3 % R3 R4 ( 1 R3 R4 VOUT = VA # = VA # 1 1 R4 + R3 + R3 R4
The last equation ends with the familiar formula for voltage division (note how it is derived from the node voltage equation).
!
By comparison, observe that the equation for
VA
has an extra term
1 R3 + R4
in
the denominator; this term represents the loading of the first voltage divider by the second voltage divider (an extra path for current in parallel to R2). In Part 3 of this ! problem, you should notice that the output voltage is grossly decreased due to this ! extra term. Addition of the buffer, or voltage follower, eliminates the extra path for current and algebraically the extra term in the equation.
Page 12 of 12
EECS 314 Lecture notes
Note that the equation for node A above was written in the assumption that the entire current that flows through R3 also flows through R4, in agreement with the circuit diagram. If we do not make this assumption, the equation for node A should be written as
VA " VS VA " 0 VA " VOUT + + =0 R1 R2 R3 $1 1 1 ' VOUT 1 VA # & + + ) " = VS # R3 R1 % R1 R2 R3 (
The equation for the output node remains as above
!
VOUT " VA VOUT " 0 + =0 R3 R4 $1 1 ' 1 VOUT # & + ) = VA # R3 % R3 R4 ( VOUT = VA # 1 R3 1 1 + R3 R4 = VA # R4 R4 + R3
The expression for VOUT is substituted into the equation for VA Of course, the final result remains the same.
!
Alexander Ganago