Identifying ,Testing and Combining Resistors...First two or three bands may be the number to write down
- Next band is the Multiplier(howmany zeros to add to the number)
Gold multiplier makes one decimal places smaller.
Last band to right may be tolarance values.
Notice the examples below.. Brown, Orange,Orange,Govld=1,2,100,5%=1200 Ohms 5%, or 1.2K Ohms.
Obtain 6 resistors of different values. I am going to determine their value two ways
Use the colour code to calculate the value of the resistor.
• Include the maximum and minimum tolerance value of each resistor
• Then measure the resistor value with a multimeter.
I choose two resistors and record their individual ohm resistance value measured with a multimeter: Brown, Black,Orang,Gold / brown ,Black ,Black,Purple,Yellow.....
Resistor 1. _21.93 Ohms_ / Resistor 2. _09.96 Ohms__
- As these two resistors put together in series (end to end, one right after another) calculated and then measured their combined values.
Calculated value 1 and 2 in series: _31.89 Ohms_ .- Measured value 1 and 2 in series: _28.5OHms_.
- As these two resistors put together in parallel (connect both ends when they are side-by-side). Calculated and then measured their combined value.
Calculated value 1 and 2 in parallel: _10.06 Ohms_ .- Measured value 1 and 2 in parallel: _9.96 Ohms_ .
In a Series circuit the totale value of recistor 1 and resistor 2 is similer to the of thier Ohms values added to gether. . In a parrolel curcuit the total resistanc is the smallest recistor.
Experiment NO. 2
Components: 1 x diode, 1 x LED
Using a multimetor identify the anode and the cathode of the diode and LED.
| Voltag drop in fowerd biased | Voltage drop in reverse biased direction | |
| LED | 1.750 Ohms | 0.0 Ohms |
| Diode | 0.55Ohms | O.o Ohms |
- In the diode there is a line on one side wich is cathode.
- In the LED bulb there is one shot side the other longer short side is negative(-).
Components: 1 x resistor, 1 x diode. 1 x LED.
Exercise: For Vs=5v, R=1kOhmme, D=1N4007 Build the fallowing circuit on a bread bord.
Calculate: first the value of current flowing though the diode, now measure and chek your answer.?
Working: Calculated:- I=(Vs-Vd)
- I=5v-0.6
- I=4.4Ma
- 4.6Ms//
Is the reading as you expected; explan why or why not.....
The current flowing threw the diode is similer to the calculated amount. 0.2ms diffrenc witch could of occerd becouse the current has to travell threw the wiring, the power suply current could be diffrent too..Calculated the voltage drop across the diode, now measure and chek your answer?
Calculate the voltage drop across the diode, now measure and chek yor answer?.
Clculated:
- Vd=Vs-I x R
- Vd= 5v - 4.4 x 1k
- Vd=600ma.(0.6A)
- Vd=0.6v
What i sthe maximum value of the current that can flow through the given diode?
30A at 75oC.
For B=1kOhme. the maXimum value of Vs so that the diode operates in a safe region.
1000Ohme.
Replace the diode by an LED and calculate the current, then measure and chek your answer?
Clculated:
- I=Vs-Vd/R
- I=5v -1.75
- I=3.25/1000
- 0.00325A (3.2ma)
Measured:
- 3.26ma
What is observed .. explane Briefly.
There is a 0.1 ma diffrenc wich could of coursed by factors such as the elngth of
the wires connected, or the power suply it self could be putting out a slight diffrenc in the Vs..
Experiment no.3
Components:2 x resistors, 1 x 5V1 400mW Zener diode (Zd).Exercise: Obtain a bread board,suitable compones from your tutor and build the following circuit.
for R=100Ohmsd and RL=100Ohme, Vs =12v
- Wht is the value of Vz?: 5V//
- Vary Vs From 10V to 15V.
- What is the value of Vz: 4.68V//
The Zener diode uses 5volts to open so the rest of the volts can pass through. Eg: 12Vs passes threw the diode, 5v hold the ciruit open and lets 7 passes threw, /if 10volts is the spupply 5Volts holds the Zener diode open while the rest of the 5 volts passes threw the circuit.
What could this cuicuit be used for.?
Voltage regulator.
Reverse the polarity of the zener diode.What is the value of Vz? make a short comment why you had that reading.6V becouse ifyou reverse the polarity of the zener diode the volts go through the 100Ohm resistor first then to the zener diode.Experiment no:4 components:1 x resistors, 1 x5V1 400mW Zener diode, 1 x diode 1N4007.
Exercise:Obtain a bread board, sutable components from your tutor and build the following cuicuit. Vs=10 and 15V, R=1K Ohme
Readings below..Voltag: 10V 15V
Volt Drop V1 _4.9V_ _5.03V__
Volt DropV2 _0.6V_ _0.6V_
Volt Drop V3 _5.5V_ _5.5V_
Volt DropV4 _4.9V_ _8.7V_
Volt DropV5 _0.004v_ _0.005v_Explanation of what i have observed off the readings:
10Volts threw the resistor to the zena diode the Voltage drop is what is expected. Ones Voltag passes therw the zena diode the votage drops aproxcimatly by 5 Volts, this is becouse the Zena diode takes 5 volts to operate, letting the 1N4007 diode is passing 5volts threw. The resistor measherd with a multimetor resists 4.9V(1K, o.5Volts).When 15 Volts applyed to the circuit, the major diffrence i knowtised was that the (V4) is almost the dubbled than when 10 volts were applyed. I think this is becouse the Zena diode only takes 5 to open letting 10 pass threw, and the resistor regulates the Vs in the circuit.
Experiment No5.
THe Capacitors.
In a way, a capacitor is a little like a battery. Although they work in completely different ways, capacitors and batteries both storeelectrical energy. If you have read How Batteries Work, then you know that a battery has two terminals. Inside the battery, chemical reactions produce electrons on one terminal and absorb electrons on the other terminal. A capacitor is much simpler than a battery, as it can't produce new electrons, it only stores them. Inside the capacitor, the terminals connect to two metal plates separated by a non-conducting substance, or dielectric (nonelectric).You can easily make a capacitor from two pieces of aluminum foil and a piece of paper. It won't be a particularly good capacitor in terms of its storage capacity, but it will work.
Non_electrolytic capaciytio.
Fairl small capacitance - normally about 10pF to 1mF
No polarity requirement- they can be inserted ether way into a cuicuit.
Can take a fairly high v
oltage.
Cuircuit for experiment 5:
note: you will need to adjust the time bace to enable you to observe the pattern.
| Cuircuite NO: | Capacitence (uF) | Recistance (KOhms) | Calculated time (Ms) | Observed time |
| 1 | 100 | 1 | 0.5 | 0.6sec |
| 2 | 100 | 0.1 | 0.05 | 100ms |
| 3 | 100 | 0.47 | 0.23 | 2.5s |
| 4 | 330 | 1 | 1.63 | 500ms |
working for calculated time ...
Formula:T=5 x R x C..
Formula:T=5 x R x C..
Capacitence:100 Recistence: 1
Capacitence: 100 . recistence:0.1
Capacitence:100 recistence: 0.47
Capacitence:330 recistence:1
how to work out the alculated time.
formula:T=5 x r x C
how dose the change in the recistor change the charging time?
- 1& 0.1K Ohme recistor took up to 500 ms to charge the capacitor
- the 0.47 recistor took abit longer, 2.50 sec to charge.
- the last cuircuit with the 330 capacitor took500msto charge . Higher the recistor is slower the capacitor charges.
How dose the canges in the capacitor change the carging time.
- Bigger the capacitor is the longer it will take to charge , as saying that if the resistor is a low Ohmes resistor the capaciotr will take less time to charge.. As shown in the results above.
Experiment No:6
Meter chek of a transistor
Bipoler trancistor are constructed of a three-layer semiconductor "sandwich" ether PMP or NPN .as such, transistors register as two diodes connected back-to-back when tested with a multimeter's "diode chek" function as illustrated in figure below. Low voltage readings on the base with the black negetive (-) leads correspond to an N-type base in a PNP transistor. On the symbole, the N-type emitter crrospond to "pointing"end of the base emitter junction, the base. the P-type emmiter crosponds to pointing end of the base emitter junction the emitter
BIPOLER JUNCTION TRANSISTORS.
Hear i'm assuming the use of a multimeter has a diode test function to chek the PN junctions. if your meter has a designated "diode'chek " function, and the meter will display the actual fowerd voltage of the PN junction and not just wether or not it conducts current. meter readings will be exactly opposite, of course, for an NPN transistor. with both PN junction facing the other way.
Low voltage eadings with the red(+) lead on the base is the "opposite" condition for the nPn transistor. if a multimeter with a "diode chek" function is used in this test, it will be found that the emitter-base junction possesses a slightly higher foward voltage drop than the collector base junction. this foward voltage fifference is due to the disparity in doping concentration between the emmitter and collector region of the transistor, the emitter is a much more heavily doped piece of semiconductor material than the collector, causing its juncton with the base to produse a highyer forwed voltage drop. Knowing this, it becomes possible to determine which terminal is which on an unmarked transistor.
tis is important becouse transistor packaging, unfortunatly, is not standerd. all bipolar transistors have thrr treminals, of course, but the positions of the three terminals on the actual physical package are not arranged in any universal, standerdised order. supoose a tecnical findes a bipoler transistor and proceedes to measure voltage drop with a multimeter set on the " diode chek" mode. measering between paires of terminals and recording the values displayed by the meter, the technician obtaines the data in figure figuer below
Unknown bipolar transistor. Which terminals are emitter, base, and collector?
-meter readings between terminals.
The only combinations of test points giving conducting meter readings are terminals 1 and 3
(red test lead on 1 and black test lead on 3), and terminals 2 and 3 (red test lead on 2 and black
test lead on 3). These two readings must indicate forward biasing of the emitter-to-base junction
(0.655 volts) and the collector-to-base junction (0.621 volts).
Now we look for the one terminal common to both sets of conductive readings. It must be the
base connection of the transistor, because the base is the only layer of the three-layer device
common to both sets of PN junctions (emitter-base and collector-base). In this example, that
terminal is number 3, being common to both the 1-3 and the 2-3 test point combinations.
METER CHECK OF A TRANSISTOR
Those sets of meter readings, the black (-) meter test lead was touching terminal 3, which tells us
that the base of this transistor is made of N-type semiconductor material (black = negative).
Thus, the transistor is a PNP with base on terminal 3, emitter on terminal 1 and collector on
terminal 2 as described in Figure below
• E & C reverse: 1(+) and 2(-): “OL”
• E & C reverse: 1(-) and 2(+): “OL”
• E & B forward: 1(+) and 3(-): 0.655 V
• E & B forward: 1(-) and 3(+): “OL”
• C & B forward: 2(+) and 3(-): 0.621 V
• C & B forward: 2(-) and 3(+): “OL”
-meter readings between terminals.
The only combinations of test points giving conducting meter readings are terminals 1 and 3
(red test lead on 1 and black test lead on 3), and terminals 2 and 3 (red test lead on 2 and black
test lead on 3). These two readings must indicate forward biasing of the emitter-to-base junction
(0.655 volts) and the collector-to-base junction (0.621 volts).
Now we look for the one terminal common to both sets of conductive readings. It must be the
base connection of the transistor, because the base is the only layer of the three-layer device
common to both sets of PN junctions (emitter-base and collector-base). In this example, that
terminal is number 3, being common to both the 1-3 and the 2-3 test point combinations.
METER CHECK OF A TRANSISTOR
Those sets of meter readings, the black (-) meter test lead was touching terminal 3, which tells us
that the base of this transistor is made of N-type semiconductor material (black = negative).
Thus, the transistor is a PNP with base on terminal 3, emitter on terminal 1 and collector on
terminal 2 as described in Figure below
• E & C reverse: 1(+) and 2(-): “OL”
• E & C reverse: 1(-) and 2(+): “OL”
• E & B forward: 1(+) and 3(-): 0.655 V
• E & B forward: 1(-) and 3(+): “OL”
• C & B forward: 2(+) and 3(-): 0.621 V
• C & B forward: 2(-) and 3(+): “OL”
BJT terminals identified by meter.
Note that the base terminal in this example is not the middle lead of the transistor, as one
might expect from the three-layer “sandwich” model of a bipolar transistor. This is quite often
the case, and tends to confuse new students of electronics. The only way to be sure which lead
is which is by a meter check, or by referencing the manufacturer’s “data sheet” documentation
on that particular part number of transistor.
Knowing that a bipolar transistor behaves as two back-to-back diodes when tested with a
Diode test function is helpful for identifying an unknown transistor purely by meter readings.
It is also helpful for a quick functional check of the transistor. If the technician were to measure
Using the Diode test function in any more than two or any less than two of the six test lead
combinations, he or she would immediately know that the transistor was defective (or else that it
wasn’t a bipolar transistor but rather something else – a distinct possibility if no part numbers can be referenced for sure identification!). However, the “two diode” model of the transistor fails to explain how or why it acts as an amplifying device.
| Transistor number | Vbe | Veb | Vbc | Vcb | Vce | Vec |
| NPN | 0.825 | OL | 0.821 | OL | OL | OL |
| PNP | OL | 0.828 | OL | 0.822 | OL | OL | <><>Diode test (v) meter readings
Transistor as a switch
Components: 1 x Small Signal NPN transistor, 2 resistors.
Exercise: Connect the circuit as shown in Fig 12 and switch on the power supply.
Connect the multimeter between collector and emitter.
Note the voltage reading and explain what this reading is indicating.
*0.85 Vbe =the cuircuit is saturated. it takes 0.85 volts to flow threw the emitor..
Connect the multimeter between base and emitter.
Note the voltage reading and explain what this reading is indicating.
*0.02 Vce it is in the cut off regon. its not saturated only a little bit of ampage flow through.
In the plot given below what are the regions indicated by the arrows A & B?
How dose the transistor work in thise reagions? When it is in the saturated reagion the gate is fullyopen and we gain less heat and les Beta (B).
When it is in the Cut off reagion we gain more heat and more Beta. In the middle is the active region this is were the transistor works best.
What is the power dissipated by the transistor at Vce of 3 volts
45 V. if a car is running for a period of time ,as a quensiquence the car heats up .. as energy is bein used up the friction heats up the engin allowing it to get hot witch couses were and tere.
What is the Beta of this transistor at Vce 2,3,&4 volts.
Beta=Ic /Ib
B= 2/0.8
B=2.5//
Beta=Ic/Ib
B=4/0.2
B=20//
Beta=Ic/Ib
B=3/0.5
B=6//
Experiment 8.
Summary: Vary the base resistor and measure changes in voltage and current for Vce, Vbe, Ic,and Ib. Then plot a load line.
Set up the following circuit on a bread board. Use a 470R for Rc and a BC547 NPN transistor.
Set up the following circuit on a bread board. Use a 470R for Rc and a BC547 NPN transistor.
Pick five resistors between 2K2 and 1M for Rb. You want a range of resistors that allow you to see
Vce when the transistor is the saturated switch region and when it is in the active amplifier region. I
used 47K, 220K, 270K, 330K and 1M, but this can vary depending on your transistor. Some may
need to use 2K2. Put one resistor in place, and measure and record voltage drop across Vce and
Vbe. Also measure and record the current for Ic and Ib. Then change the Rb resistor and do all the
measurements and record the new readings. Do this for each of the resistor values above.
Record here:
Rb:2.2K--Vbe:0.79V--Vce:0.02V--Ib:2.05Ma--Ic:8.65Ma
Rb:47K--Vbe:0.68V--Vce:1.11V---Ib:0.01Ma---Ic:4.55Ma
Rb:220K-Vbe:0.69V-Vce:0.65V--Ib:0.02Ma---Ic:5.48Ma
Rb:270k-Vbe:0.78V--Vce:1.21V--Ib:0.06ma---Ic:4.55ma
Rb:330K-Vbe:0.67V-Vce:1.15V--Ib:0.01ma---Ic3.77ma
Your voltage drop measurements across Vce should vary from below 0.3 v (showing the transistoris in the saturated switch region) to above 2.0 v (showing the transistor is in the active amplifier region) If this is not the case, you may have to try a smaller or bigger resistor at Rb. Talk to your teacher to get a different size resistor, and redo your measurements.
Discuss what happened for Vce during this experiment. What change took place, and what caused the change?
- As i changed the resistor to a higher one the Vce kept rising becouse current flowing from collector to the emitor is less, therefor it results in higher volt drop.
Discuss what happened for Vbe during this experiment. What change took place if any, and what caused the change?
- As i changed the resistor the voltage droped from bace to the emitor becouse voltae drop is low from bace to emitor. (data sheet display 0.6V ..)
- The Ib drops as i changed the resistor becouse at the bace there is less current higher the resister got less currenf can flow the cuircuit .
- The Ic drops as the resistor changes, becouse there is less current flowing threw the collector & emitor ,becouse the resistenc is higher at the bace resistor.
Calculated the Beta (Hfe) of this thransistor using the above graph..
B= Ic/Ib
B8.65/2.05
B=4.21
