I was already told that the base of the transistor was the middle and then. i then set my multimeter to diode check so i could check the resistance and then placed my positive prong in the middle leg of the transistor (base) and my negative prong on one of the outer legs(emitter or collector) , i got low voltage which tells me that i had the polarity around the right way, this means that my transistor was a PNP type. If i had my prongs of my multimeter around the other way i would of got a reading of OL (over load). If OL was my reading this would indicate that it is a NPN type transistor.
Below is a graph showing my resistances when my multimeter is in diode test mode.
| Transistor number | Vbe | Veb | Vbc | Vcb | Vce | Vec |
| NPN | 0.719 | OL | 0.722 | OL | OL | OL |
| PNP | OL | 0.722 | OL | 0.719 | OL | OL |
If i wanted to check which of the outer legs was the emitter and which one was the collector, one of the outer legs of the transistor would have a slightly higher resistance, this leg will be the emitter. If you do not have multimeter you can always check the information sheet of the transistor on the internet.
For the next experiment i was asked to make a compound circuit on a breadboard using a NPN transistor and 2 resistors (1000ohms, 10000ohms) and a 15 volt supply.
I was calculated the voltage between the base and the emitter of the transistor. my reading was 0.781V because it take about 0.7V to push through a diode.
I then measured the voltage between the base and the collector and got 0.051V, this was a very small voltage because the resistors are hooked up in a series circuit and this means that all of the voltage gets used, this is why i got a very minimal voltage at the collector.
In the graph below i was then asked to explain it:
the cut off region (B) means the transistor is fully off and not working and not voltage is crossing though it.
the saturated region (A) means the transistor is fully open and active. there is also around 0.5 voltage drop across the collector and emitter.
the power dissipated by the transistor at Vce is:
Pd = Vce x Ic
3 x 13 = 39
Pd = 39W
to work out the Beta of the transistor at Vce 2,3 and 4 volts:
B = Ic/Ib
2 volts:
20/0.8 = 25
3 volts:
15/0.6 = 25
4 volts:
5/0.2 = 25
Beta = 25
For experiment number 8 i was asked to make a circuit on a breadboard using a LED (light emitting diode), a 470ohm resistor, a NPN(negative, positive, negative) type transistor and another resistor which i was asked to change after every test to get different readings.
Rb | Vbe | vce | Ib | Ic |
| 47k ohms | 0.67V | 0.12V | 10mA | 4.0mA |
| 220k ohms | 0.68V | 0.53V | 10mA | 50.27mA |
| 270k ohms | 0.67V | 0.99V | 10mA | 53.13mA |
| 330k ohms | 0.67V | 1.36V | 10mA | 78.98mA |
| 1M ohm | 0.63V | 2.50V | 10mA | 99.83mA |
the voltage drop across the base and emitter stayed the same because it take roughly 0.6 volts to push the diode in the transistor. the voltage drop across the collector and emitter was low to begin with but slowly increased as the resistor got bigger, this happened because there was more resistance in the circuit which leaves more voltage for the other consumers in the circuit. the amps at the base got stayed the same because the resistor stayed the same and the base is a different of a different circuit to the emitter and collector so the change in resistor wont effect the current at the base. The current at the collector got higher as the resistor got bigger, this happened because the larger the resistance, the more force needed to push though the circuit.
I was then asked to plot the points for amps at the collector and the voltage drop across collector and emitter:
The load line in the graph is telling us that when the volts across the collector and the emitter is at a certain voltage, if you look up to the load line, The amps at the collector will be at a certain amperage and then on the opposite side of the graph it will show the amps at the base aswell.
References:
I got the picture of the transistor from www.tradekorea.com
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