MOSFET. (Metal Oxide Semiconductor Field Effect Transistor) vs BJT (Bi-polar junction transistor)
-The MOSFET is used just like any other transistor where it can be used as a switch in a circuit or used to amplify voltage. just like the Bipolar Junction Transistor (BJT) it has three main terminals. Where the terminals in the BJT are called Base, Collector and Emitter. The names change in the MOSFET, the terminals are the Gate, Drain and the Source. Also the difference between the BJT and the MOSFET is the BJT requires high current to the Base of the transistor to allow voltage to flow from the collector to the emitter where in the MOSFET it requires voltage at the Gate to allow voltage to flow from the Drain to the Source.
There are two types of different channel MOSFET's, An N-Channel Mosfet, P-Channel Mosfet.
The N-Channel Mosfet acts like a NPN bipolar junction transistor where the base of the BJT needs a positive feed to turn on the N-channel Mosfet requires positive voltage at the Gate to turn on. The greater the Gate voltage the increases the size of the channel created.
The P-Channel MOSFET acts like a PNP BJT where the PNP requires a negative feed to the base, so it requires zero voltage to allow voltage to flow the collector to the emitter the MOSFET requires zero voltage at the Gate to allow voltage to flow from the drain to the source but requires positive voltage to switch it off.
These MOSFET's can either be of a Depletion type where it requires voltage at the Gate to turn on or Enhancement type where the voltage at the Gate is used to turn the MOSFET on.
Schematic diagrams of MOSFET's and BJT's
G=GATE
D=DRAIN
S=SOURCE
S=SOURCE
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| P-Channel N-Channel Depletion Type MOSFET's |
 P-Channel N-Channel Enhancement Type MOSFET's |
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| Op-Amp It has two input terminals and an output terminal and a negative and positive voltage rail. The positive input is known as the inverting terminal and the negative known as the non-inverting terminal and basically the op-amp amplifies the difference between the to inputs. It can also be used as a comparator in a circuit and the output will be dependant on the difference of the two inoputs and the voltage at the rails. Below is a diagram of an intergrated circuit chip with four op-amps. I will be using a similar chip to create an oxygen sensor circuit. |
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| Wiring diagram for an oxygen sensor . |
r5= 12v (power supply)-9.1v(Zener diode)-0.6(rectifying diode)/5.6mA
= 2.3v/0.0056A (5.6mA)
= 410.7 ohms
Then going on to find the resistor values for r7 and r8 i have to find the amperage that will flow through as it is a parallel circuit and amperage may vary. As r6 is already given as 10k (10,000 ohms) and the available voltage after the zener diode is 9.1volts and the available voltages given after each resistor
A=voltage/resistance
= 9.1v-0.63v-0.23/10,000ohms
=8.24v/10,000ohms
= 0.000824A (0.824mA)
Now having the amperage we can find r8 which is resistance = voltage/ amperage and since we have two available voltage points
r8 = 0.63v-0.23v/0.824mA
= 0.4/0.000824
= 485.4 ohms
r7 = 0.23v/0.824mA
= 0.23v/0.000824A
= 279.1 ohms
Now finding the resistor values r2,r3,r4 which lead to the output pins of each op-amp and after the LED. Taking note that the diode creates a voltage drop of 0.6v and the LED's use 1.8volts and require a minimum of 9.5mA to operate according to the datasheet, again using the ohm's law resistance = voltage/ amperage and instead of working with the minimum amperage i decided to go with 10mA, therefore
r2 = 12v-0.6v-1.8v/10mA
= 9.6v/0.01A
= 960 ohms
r3 = 12v-(0.6x2)-1.8v/10mA
= 9v/0.01A
= 900ohms
r4 = 12v-0.6v- 1.8v/10mA
= 9.6v/0/01A
= 960ohms
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