Oxygen sensor circuit

Today in practical session we started working on building our next project, which is an oxygen sensor. The first step was to calculate the value of the resistors to go into the circuit.

Calculations

The first resistor I calculated using ohms law was R5. To get this value I had to use the formula R = V/A which is 12V – 0.6 (voltage drop across D2) – 9.1 (Zener-diode D1) equals 2.3v. The Amperage is shown on the sheet, which is 5.6mA that I have to convert to Amps so it would be 0.0056A. Therefore R = V/A = 2.3/0.0056 = 410.71 ohms. So for R5 I would use a resistor around this value.

The next resistor I calculated was R8. So R = V/A. the voltage I will be using is the voltage drop across this resistor and to work this out I had to minus 0.63v from 0.23v which were the available voltages before and after R8. Therefore the answer is 0.40v. The current is calculated using the formula A = V/R but before we can move on I have to calculate the current flow by using the values of R6. 9.1v is the available voltage before R6 and 0.63v is the available voltage after R6 (9.1 – 0.63 = 8.47v) so the voltage drop across this resistor is 8.47v, which is used for the voltage to calculate the current. The value of R6 is already known and is 10K. Therefore the current flow is A = V/R = 8.47/ 10000 = 0.000847A. So back to working out the value of R8 we now have the current to finish of the formula R = V/A. so 0.40/ 0.000847 = 472 ohms. So for R8 I will be using a resistor around this value.

To calculate the value of R7 I used the formula R = V/A. The available voltage is 0.23 before R7 and 0v after it so the voltage drop will be 0.23v and the current will be the same as R8, which is 0.000848A. So the value of R7 will be 0.23/0.000848 = 271ohms.

Now I will be calculating the values of R2, R3 and R4. So to calculate R2 I need to use R = V/A. First I need to work out my voltage I will be using and to do this I need to minus 0.6v which is the voltage drop across D2 IN4001 diode and 1.8v which is the voltage drop across LED1. I need to minus this from 12v so (0.6 + 1.8 – 12) = 9.6v. The current is already shown on the sheet and is 9.5mA that I have to convert to Amps which is then 0.0095A. So now I can calculate the value of R7. R = V/A = 9.6/0.0095 = 1010ohms.

The next resistor is R3. I will be using R = V/A to calculate its value. First I need to workout the voltage drop across this resistor. So there’s 12v coming in and 0.6v drops across D2 and another 0.6v drops across D4 and 1.8v drops across LED 5 and added up equals (0.6 + 0.6 + 1.8 – 12) = 9v. The current remains 9.5mA (0.0095A). So the value of R3 is R = V/A = 9/0.0095 = 947ohms. I don’t know if there is a 947ohm resistor, so Il be using the nearest 1 to this value which would probably be 1K or 1000ohms.

The final resistor to calculate is R4.  Its just like above were I have to add up the voltage drops before the resistor. So 12 – 0.6 + 1.8 = 9.6v.  The current is still the same in this part of the circuit, which is 9.5mA. Therefore R = V/A = 9.6/0.0095 = 1010ohms.

Component list

After calculating the values of R5, R8, R7, R2, R3 and R4 I had to show the tutor my workings of how I got these values. Once the tutor checked it I then moved on to build the circuit on a breadboard. The components I used are

Three 1N4001 Rectified diodes for D2, D3 and D4, according to the relevant data sheet these diodes have a maximum Pd of 2.5W, max current is 1A  @ 75°C and max reverse voltage of 100V.

One 9v1 zener diode for D1

Two 0.1 uF capacitors for C1 and C2

Three LED’s, one red for LED1 (I will use this to show when I’m running lean), one yellow for LED5 (to show when I’m running in between lean and rich) and one green for LED6 (for when I’m running rich or when accelerating because green represents GO)

A 380ohm resistor for R5

A 10,000 or 1Kohm resistor for R6

A 470ohm resistor for R8

A 270ohm resistor for R7

Wires


Op Amp


Below are pictures of the circuit i built on bread board:

After building this circuit on bread board i then tested it and it worked. I will go into more detail with a video i will be recording next week because i had no time in class to do it on friday. I also finished building this circuit on the green board which I still need time to record a video and show how the circuit works. But for now I will include a topic from ''subjects to study'' from moodle to add to my blog.

Zener Diodes

A zener diode is just like a normal rectifying diode that only allows current to flow in one direction from anode to cathode (forward bias) when a voltage of 0.6v is reached to open the gate and allow electrons to flow. But with a zener diode current can also flow in reverse direction (reverse bias) when the voltage is above a certain value, for example 6volts. This is called the break down voltage known as the zener voltage.

A zener diode can be used to make a simple voltage regulation circuit. The output voltage is fixed at the zener voltage of the zener diode used and so can be used to power devices requiring a fixed voltage. If you have a regulated fixed voltage - say 12 Volts from a power supply, and you want to power something requiring a lower voltage, it is possible to simply place a zener diode in series with the load device. You would choose a diode with a zener voltage equal to the supply voltage minus the voltage drop across the load.

For example, if you have a 1W 6V lightbulb to power from a 12 Watt regulated power supply, a 6.2V zener diode could be placed in series with the bulb giving the bulb 5.8V, or you could overpower the bulb a little using a 5.6V zener diode and dropping 6.4V across the bulb. Heat would be generated in the zener diode so it is essential to calculate the power lost in it so a suitably rated diode could be chosen.

Below is a picture of a zener diode I got of the internet.

http://www.reuk.co.uk/What-is-a-Zener-Diode.htm