Theory
What is a hybrid vehicle?
A hybrid vehicle is a vehicle that uses two or more distinct power sources to move the vehicle.(combines an internal combustion engine and one or more electric motors.
Power train configurations
Parallel hybrid(petrol->transmission->wheels)
(petrol->batteries->motor->wheels)
In a parallel hybrid, the single electric motor and the internal combustion engine are installed so that they can both individually or together power the vehicle.
Series hybrid(12 cylinder -> generator->batteries->motor)
Series-hybrid vehicles are driven by the electric motor with no mechanical connection to the engine.Instead there is an engine turned for running a generator when the battery pack energy supply isn't sufficient for demands.
Fuel cell, electric hybrid
The fuel cell hybrid is generally an electric vehicle equipped with a fuel cell. The fuel cell as well as the electric battery are both power sources, making the vehicle a hybrid.
Fuel cells use hydrogen as a fuel and power the electric battery when it is depleted.
<source from http://en.wikipedia.org/wiki/Hybrid_vehicle>
Safety issues
Wednesday, August 31, 2011
Tuesday, August 23, 2011
Day 10 - Mosfet and fault finding
Theory
MOSFET(metal oxide semiconductor field effect transistor)
Mosfet has higher input impedance. Like bipolar FET's come in two polarities, P-channel and N-channel, corresponding to PNP and NPN bipolar transistors. They also come in enhancement mode and depletion mode types, specifying what happens when the gate voltage is zero. An enhancement mode FET will be turned off with no voltage at the gate; like a bipolar transistor, it requires a bias voltage to turn it on. A depletion mode FET will be turned on with zero gate voltage. The only way to turn it off is to apply a voltage of opposite polarity to the one that will increase current flow.
Symbols
ref: http://www.youtube.com/watch?v=Te5YYVZiOKs
MOSFET(metal oxide semiconductor field effect transistor)
Mosfet has higher input impedance. Like bipolar FET's come in two polarities, P-channel and N-channel, corresponding to PNP and NPN bipolar transistors. They also come in enhancement mode and depletion mode types, specifying what happens when the gate voltage is zero. An enhancement mode FET will be turned off with no voltage at the gate; like a bipolar transistor, it requires a bias voltage to turn it on. A depletion mode FET will be turned on with zero gate voltage. The only way to turn it off is to apply a voltage of opposite polarity to the one that will increase current flow.
Symbols
Signals applied to the Gate.
MOSFETs are used as switches, permitting the microprocessor to turn power on and off to various parts of the circuitry.
Practical
Fault finding
Create an open circuit on the marked "x" as show above the circuit diagram and diagnose the circuit.
Diagnosed results before testing
At node A - All the leds are off because there is no positive connection to the leds.
At node B - Everything is fine as normal conditions except the yellow led always off. Because the voltage of op-amp pin.no 2 is floating(around 0 volts), the output is always positve power supply voltage which is the voltage of op-amp pin.no 4. Therefore it always overrides the yellow led to light off.
At node C - When the sensor input voltage is above 0.63volts, the yellow led doesn't turn off.
At node D - The red led always lights off. Because there is no connected ground to the red led.
At node E - The red led always lights off. Because the voltage of the op-amp pin.no 6 is floating(around 0 volts), the output is always positive power supply voltage which is the voltage of op-amp pin.no4.
At node F - The yellow led always lights off. Because the voltage of the op-amp.no 9 is floating(around 0 volts), the output is always positive power supply voltage which is the voltage of op-amp pin.no8.
At node G and H - The green led is always off. Because the voltage of the op-amp pin.no 13 is always 0 volts, so the output voltage is always connected to the positive power supply voltage.
The yellow led is always on. Because the voltage of the op-amp pin.no 10 is always 0 volts, so the output voltage is always connected to the ground.
The red led is always off except that the sensor input voltage is above 9.1volts. Because the voltage of the op-amp pin.2 and 5 is 9.1volts.
Monday, August 22, 2011
Day 9 - Oxygen sensor tester circuit, DC motor
Theory
Op-amp as a comparator
Sometimes we want to compare one voltage with another to see which is larger. A comparator is similar to an op amp because it has two input voltages and one output voltage. But it differs from a linear op-amp circuit because it has a two state output, either a low or a high voltage.
Some examples are as follows -
IF V1 > V2 then Vout=-VEE
IF V1 < V2 then Vout=+VCC
Ref: http://www.youtube.com/watch?v=icFo5Zeydqg
http://www.youtube.com/watch?v=y0Q0ERSP24A&feature=related
DC motor control by using relays
Measure available voltages at each position
When a switch is at the position of N:
A(0V) B(0V) C(12V) D(12V)
When a switch is at the position of F:
A(12V) B(0V) C(0V) D(12V)
When a switch is at the position of R:
A(0V) B(12V) C(12V) D(0V)
H bridge DC motor control
H bridge is commonly used to drive motors. As shown below circuit, two of four transistors are selectively enabled to control current flow through a motor.
An opposite pair of transistors(Tr1 and Tr3) is enabled, allowing current to flow through the motor. The other pair is disabled and can be thought of as out of circuit.
By determining which part of transistors is enabled, current can made to flow in either of the two directions through the motor.
Practical
Oxygen sensor tester circuit
Component lists
<><><><><><><><><><><>
<Pricelist from Jaycar Electronics>
Calculations
To calculate the resistance of R5, we assume IzRm=5.6mA
By ohm's law, R5= V/I =(11.4-9.1)/5.6mA = 410.71Ω(nearest value 470R)
To calculate the resistance of R7 and R8, firstly we calculate IT.
IT = (9.1-0.63) / 10K = 847uA
Then by ohm's law, we calculate the resistance of R8 and R7.
R8 = (0.63-0.23) / 847uA = 472.25Ω(nearest value 470R)
R7 = 0.23 / 847uA = 271.55Ω(nearest value 270R)
To confirm the result of the calculations, the voltage divider method can be used.
Firstly we calculate the total resistance
RT = R6+R7+R8
RT = 10743Ω
By using the voltage divider,
Vout1 = VIN x R7/RT = 9.1 x 271/10743 = 0.23V
Vout2 = VIN x (R7+R8) / RT = 9.1 x 743/10743 = 0.64V
To calculate the resistance of R2, R3 and R4, we assume the current of LED is 9.5mA.
By ohm's law, R2 = (12-0.6-1.8) / 9.5mA = 1010.53Ω(nearest value 1K)
Same as R2, R4 = (12-0.6-1.8) / 9.5mA = 1010.53Ω(nearest value 1K)
R3 = (12-0.6-1.8-0.6) / 9.5mA = 947.37Ω(nearest value 1K)
Components layout on the board
How the circuit works
Zener diode provides a stable reference voltage which is 9.1V
Capacitor is used for filtering.
To prevent some risks from connecting power supplies in a wrong terminal, a diode D2 is used.
Capacitor is used from filtering.
R6, R7 and R8 is used for voltage divider which provides the reference voltages to the inputs of the quad op-amp.
R5 is used for limiting current through that node.
R2, R3 and R4 is used for limiting current to each leds.
When the sensor input voltage is greater than 0.63V and if there is no connection between a diode D3 and a resistor R3, a yellow led and a green led can be lit on together.
Quad op-amp is used as a comparator.
If the sensor input voltage is less than 0.23V, a red led lights on.
If the sensor input voltage is greater than 0.63V, a green led lights on.
If the sensor input voltage is between 0.23V and 0.63V, a yellow led lights on.
Test Procedure
1. Connect 0.151V to the node no.1 and Check the available voltages at the nodes 2-16 as shown above diagram
2. Connect 0.540V to the node no.1 and Check the available voltages at the nodes 2-16 as shown above diagram
3. Connect 0.780V to the node no.1 and Check the available voltages at the nodes 2-16 as shown above diagram
Availabe voltages at the nodes 2-16
Wire the circuit to an engine which goes into closed loop.(Number 8 engine in 108-1066)
Problems
1. Once the yellow led lights on, it continues lighting even the input sensor voltage reaches 0.63volts. Therefore, I traced the available voltages at node no.7 , 13, and 14. The problem was that the wire at node no.7 was connected to the node no.5. After reconnecting the wire to the right place, the yellow lights off when the input sensor voltage reaches 0.63volts.
2. When I check the voltage drop acorss diode D2, the voltage was 0.05volts. To check the condition of the diode, I turned the power off and measured the voltage by dialing the multimeter to the "diode" position. The voltage was still 0.05volts. So the diode D2 was replaced by new working diode. Then the voltage drop across the diode was 0.7volts.
Reflection
I have learned the following things as I build and test the O2 sensor tester circuit.
1. Zener diode can be used for the voltage reference.
2. Diode can be used for protecting the circuit from the back EMF.
3. To get the different voltages other than the power supply voltage, a voltage divider can be used by connecting the resistors in series.
4. When connecting LEDs in the circuit, a resistor should be connected to the LED(current limiting purpose).Otherwise, the LEDs can be blown out.
5. ICs pins(eg LM324) should not be touched by fingers. ICs should be kept in an anti statistc bag. When handling the ICs, wear an anti-statice wrist band. If there is no anti-static wrist band, try to touch the case rather than ICs pins.
Op-amp as a comparator
Sometimes we want to compare one voltage with another to see which is larger. A comparator is similar to an op amp because it has two input voltages and one output voltage. But it differs from a linear op-amp circuit because it has a two state output, either a low or a high voltage.
Some examples are as follows -
IF V1 < V2 then Vout=+VCC
Ref: http://www.youtube.com/watch?v=icFo5Zeydqg
http://www.youtube.com/watch?v=y0Q0ERSP24A&feature=related
DC motor control by using relays
When a switch is at the position of N:
A(0V) B(0V) C(12V) D(12V)
When a switch is at the position of F:
A(12V) B(0V) C(0V) D(12V)
When a switch is at the position of R:
A(0V) B(12V) C(12V) D(0V)
H bridge DC motor control
H bridge is commonly used to drive motors. As shown below circuit, two of four transistors are selectively enabled to control current flow through a motor.
An opposite pair of transistors(Tr1 and Tr3) is enabled, allowing current to flow through the motor. The other pair is disabled and can be thought of as out of circuit.
By determining which part of transistors is enabled, current can made to flow in either of the two directions through the motor.
Practical
Oxygen sensor tester circuit
Component lists
Name | Qty | Unit Price | Details |
Resistor R5, R8 | 2 | $0.55 | 470R, 1/4W, 5% tolerance |
Resistor R6 | 1 | $0.55 | 10K, 1/4W, 5% tolerance |
Resisotr R7 | 1 | $0.55 | 270R, 1/4W, 5% tolerance |
Resistor R2,R3,R4 | 3 | $0.55 | 1K,1/4W, 5% tolerance |
Op-amp | 1 | $2.50 | LM324, Low power op-amp wide power supply range 3-32V |
Diode D2,D3,D4 | 3 | $0.60 | 1N4001, VRRM=50V, VF=1.1V IF=1A,IFSM=30A,PD=3W |
Zener diode D1 | 1 | $0.66 | 1N4739, 9.1V, PD=1W @25°C VF=1.2V,IzRm=5.6mA |
LED | 1 | $0.30 | 5mm Red,VF=2.0V,IF=20mA |
LED | 1 | $0.30 | 5mm Yellow, VF=2.0V,IF=20mA |
LED | 1 | $0.30 | 5mm Green, VF=2.0V,IF=20mA |
Capacitor | 2 | $0.54 | 0.1uF 50V Electrolytic |
Connectors | 1 | $1.50 | 3pin |
<Pricelist from Jaycar Electronics>
Calculations
To calculate the resistance of R5, we assume IzRm=5.6mA
By ohm's law, R5= V/I =(11.4-9.1)/5.6mA = 410.71Ω(nearest value 470R)
To calculate the resistance of R7 and R8, firstly we calculate IT.
IT = (9.1-0.63) / 10K = 847uA
Then by ohm's law, we calculate the resistance of R8 and R7.
R8 = (0.63-0.23) / 847uA = 472.25Ω(nearest value 470R)
R7 = 0.23 / 847uA = 271.55Ω(nearest value 270R)
To confirm the result of the calculations, the voltage divider method can be used.
Firstly we calculate the total resistance
RT = R6+R7+R8
RT = 10743Ω
By using the voltage divider,
Vout1 = VIN x R7/RT = 9.1 x 271/10743 = 0.23V
Vout2 = VIN x (R7+R8) / RT = 9.1 x 743/10743 = 0.64V
To calculate the resistance of R2, R3 and R4, we assume the current of LED is 9.5mA.
By ohm's law, R2 = (12-0.6-1.8) / 9.5mA = 1010.53Ω(nearest value 1K)
Same as R2, R4 = (12-0.6-1.8) / 9.5mA = 1010.53Ω(nearest value 1K)
R3 = (12-0.6-1.8-0.6) / 9.5mA = 947.37Ω(nearest value 1K)
Components layout on the board
How the circuit works
Capacitor is used for filtering.
To prevent some risks from connecting power supplies in a wrong terminal, a diode D2 is used.
Capacitor is used from filtering.
R6, R7 and R8 is used for voltage divider which provides the reference voltages to the inputs of the quad op-amp.
R5 is used for limiting current through that node.
R2, R3 and R4 is used for limiting current to each leds.
When the sensor input voltage is greater than 0.63V and if there is no connection between a diode D3 and a resistor R3, a yellow led and a green led can be lit on together.
Quad op-amp is used as a comparator.
If the sensor input voltage is less than 0.23V, a red led lights on.
If the sensor input voltage is greater than 0.63V, a green led lights on.
If the sensor input voltage is between 0.23V and 0.63V, a yellow led lights on.
Test Procedure
2. Connect 0.540V to the node no.1 and Check the available voltages at the nodes 2-16 as shown above diagram
3. Connect 0.780V to the node no.1 and Check the available voltages at the nodes 2-16 as shown above diagram
Sensor input voltage | |||
Node number | 0.151V | 0.540V | 0.780V |
2 | 0.944V | 10.14V | 10.11V |
3 | 9.25V | 10.14V | 10.11V |
4 | 11.29V | 11.44V | 11.42V |
5 | 11.29V | 11.44V | 11.42V |
6 | 9.43V | 0.93V | 0.941V |
7 | 9.99V | 8.84V | 9.12V |
8 | 0.052V | 10.75V | 10.85V |
9 | 11.29V | 11.44V | 11.42V |
10 | 9.99V | 10.14V | 0.946V |
11 | 9.99V | 10.14V | 9.28V |
12 | 11.29V | 11.44V | 11.42V |
13 | 0.005V | 0V | 9.81V |
14 | 9.99V | 8.84V | 9.12V |
15 | 9.12V | 9.12V | 9.12V |
16 | 11.29V | 11.44V | 11.42V |
As shown above video, if the wire at the node no.7 and no.14 is connected, the yellow led lights on.
Wire the circuit to an engine which goes into closed loop.(Number 8 engine in 108-1066)
Problems
1. Once the yellow led lights on, it continues lighting even the input sensor voltage reaches 0.63volts. Therefore, I traced the available voltages at node no.7 , 13, and 14. The problem was that the wire at node no.7 was connected to the node no.5. After reconnecting the wire to the right place, the yellow lights off when the input sensor voltage reaches 0.63volts.
2. When I check the voltage drop acorss diode D2, the voltage was 0.05volts. To check the condition of the diode, I turned the power off and measured the voltage by dialing the multimeter to the "diode" position. The voltage was still 0.05volts. So the diode D2 was replaced by new working diode. Then the voltage drop across the diode was 0.7volts.
Reflection
I have learned the following things as I build and test the O2 sensor tester circuit.
1. Zener diode can be used for the voltage reference.
2. Diode can be used for protecting the circuit from the back EMF.
3. To get the different voltages other than the power supply voltage, a voltage divider can be used by connecting the resistors in series.
4. When connecting LEDs in the circuit, a resistor should be connected to the LED(current limiting purpose).Otherwise, the LEDs can be blown out.
5. ICs pins(eg LM324) should not be touched by fingers. ICs should be kept in an anti statistc bag. When handling the ICs, wear an anti-statice wrist band. If there is no anti-static wrist band, try to touch the case rather than ICs pins.
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