Yesterday i got 6 unknown resistors and found out the size with the color codes and a ohm meter.
1st resistor was Brown, black, orange and gold. 10,000 ohms +-5% 9500-10500 ohms range.
Measured at 9760 ohms.
2rd resistor was red, red , red and gold. 2200 ohms +-5% 2090-2310 ohms range.
Measured at 2093 ohms.
3rd resistor was red, violet, yellow and gold 270,000 ohms +-5%. Range is 256,000-283,500.
Measured at 250,700 ohms. This was out of the range possible damaged resistor.
4th resistor was yellow, violet, black and gold. 47 ohm +-5%. Range is 44.65-49.35 ohms.
Measured at 46.3 ohms.
5th resistor was orange, black, orange. 330,000 ohms.
Measured at 302,500 ohms.
Next i wired to resistors in series and recorded resistance.
Resistor 1= 2200 ohms
Resistor 2=220,000 ohms.
6th resistor was red, red ,yellow, gold. 220,000 +-5%. Range is 209,000 - 231,000 ohms
Measured at 203,200 oh
2200+220,000= 222,200 ohms.
Calculated value= 222,200 ohms
Measured value= 204,500 ohms.
Next the same two resistors in parallel.
2200 x 220,000
2200 + 220,000 = 2178.2
Calculated value= 2178.2 ohms
Measured value= 2130 ohms
The rule of electricity is that in parallel the total circuit resistance will always be lower than the smallest resistor.
Experiment 2
Voltage drop in forward bios direction
L.E.D= 1.7V
Diode= 0.51V
Voltage drop in Reverse bios direction
L.E.D= OL
Diode= OL
Identifying anode and cathode.
You can identify the cathode by the longest leg on an l.e.d or by the flat side of an l.e.d.
Diodes have a grey or black line for the cathode.
Diodes sketches.
DIAC
Diode (rectifier diode)
Light-emitting diode (LED)
Zener diode
Exercise: For Vs=5V, R= 1K, D= 1N4007 build the following circuit.
Calculated Current.
5/1000= 0.005A
Measured
0.004A.
That was the reading i expected because it was what i calculated.
Calculate Voltage drop across diode measure and calculate.
Calculated=0.6V
Measured=0.65V
Maximum current of a 1n4007 is 1A in ether direction.
Maximum voltage of a 1n4007 is 1000V in the reverse direction.
Replace diode with a l.e.d and measure current.
Calculated
5mA
Measured
35mA
I observed the l.e.d light up very dully. The current was slightly larger with the diode
Experiment 3
Components: 2 x resistors, 1 x 5V1 400mW Zener diode (ZD).
Exercise: Build the following circuit.
R=100 ohm
Rl-100 ohm
The value of VZ was 4.9V
When i varied volts between 10-15V these were the results.
10V=4.59
11V=4.87V
12v=4.9V
13v=5.01V
14v=5.05V
15v=5.10V
The voltage is staying at around 5v regardless of the supply voltage increasing. This is because
the zenor diode is acting as a regulator because it is in reversed bios. This is keeping the circuit at 5v. This circuit could be used as a 12v to 5v regulator.I then reversed the polarity of the diode and VZ now read 0.851V because it is acting like a normal diode when the zenor diode is in forward bias.
Experiment 4
Components: 1 x resistors, 1 x 5V1 400mW Zener diode, 1X Diode1N4007 .
Exercise:Build the following circuit.
Vs=10 & 15v, R=1K ohms
10V
Voltage drop V1= 5.06V
Voltage drop V2= 0.751V
Voltage drop V3= 5.81V
Voltage drop V4 = 0.0409
Calculated current= 0.01A
15V
Voltage drop V1= 5.16V
Voltage drop V2= 0.783V
Voltage drop V3= 5.96V
Voltage drop v4= 0.870V
Calculated current= 0.015A
The voltage is staying around the same because the zenor diode is positioned in reverse bios. Acting like a regulator dropping 5V.
The other diode is in forward bios.
Experiment 5
Capacitors take time to charge. It doesn’t happen instantly. The charge time is dependent on the
resistor in the circuit and the size of the capacitor. And it is expressed in the equation:
R x C x 5=T
Capacitors are often used for timing when events take place. And often the voltage only has to get
up to about 2/3 the applied voltage, and this happens at about 1/5 the time of their charging. So
this is why the 5 is built into the equation. The concept of “time constants” is used here, where
whatever the time it takes for a capacitor to build up to the full charge, it takes about 1/5 of that
time to build up close to 2/3 of the charge. So you can divide the charge time into 5 segments, and the first time segment is often the time you are interested in.
From Autotronics workbook.
Circuit One. 100uF capacitor. 1K resistor.
Circuit two. 100uF capacitor. 100 ohm resistor.
Circuit three. 100uF capacitor 470 ohm resistor.
Circuit four. 330uF capacitor. 1000 ohmThe bigger the resistance the more time the capacitor will take to charge because there is
less current. The smaller the resister the faster the capacitor will charge.
Bigger capacitors also take longer to charge.
Experiment 6
Bipolar transistors are constructed of a three-layer semiconductor “sandwich,” either PNP or
NPN. As such, transistors register as two diodes connected back-to-back when tested with a
multimeter’s “diode check” function as illustrated in Figure 9.1. Low voltage readings on the base
with the black negative (-) leads correspond to an N-type base in a PNP transistor. On the symbol,
the N-type material corresponds to the “non-pointing” end of the base-emitter junction, the base.
The P-type emitter corresponds to “pointing” end of the base emitter junction the emitter.
BIPOLAR JUNCTION TRANSISTORS
Here I’m assuming the use of a multimeter has a diode test function to check the PN junctions.
If your meter has a designated “diode check” function, and the meter will display the actual forward
voltage of the PN junction and not just whether or not it conducts current.
Meter readings will be exactly opposite, of course, for an NPN transistor, with both PN
junctions facing the other way.
Low voltage readings with the red (+) lead on the base is the “opposite” condition for the NPN
transistor. If a multimeter with a “diode check” function is used in this test, it will be found that
the emitter-base junction possesses a slightly HIGHER forward voltage drop than the collector
base junction. This forward voltage difference is due to the disparity in doping concentration
between the emitter and collector regions of the transistor; the emitter is a much more heavily
doped piece of semiconductor material than the collector, causing its junction with the base to
produce a higher forward voltage drop.
Knowing this, it becomes possible to determine which terminal is which on an unmarked transistor.
This is important because transistor packaging, unfortunately, is not standardised. All
bipolar transistors have three terminals, of course, but the positions of the three terminals on the
actual physical package are not arranged in any universal, standardised order.
Suppose a technician finds a bipolar transistor and proceeds to measure voltage drop with a
multimeter set in the “diode check” mode. Measuring between pairs of terminals and recording the values displayed by the meter.
From
Work Book Autotronics. How to identify a transistor
I measured 2 transistors.
I identified if they were PNP or NPN.
One was PNP.
One was NPN.
You identify the emitter because it has a slightly larger voltage in diode check .
The collector has the lower voltage.
Experiment 7
I connected my voltmeter between base and emitter.
My reading was 0.738V that was a good reading. That means the transistor is
saturated and working.
Next i connected the voltmeter between the collector and emitter.
My reading was 54mV. Both base to emitter and base to collector junctions are forward biased. This means there will be a 0.7 v drop from base to emitter and a 0.7v drop from base to collector. Since the base to emitter and base to collector voltages are almost the same, the collector to emitter voltage is almost zero if the junctions were matched, the saturated voltage would be 0V.
In region A the transistor is fully saturated and working.
Region B is the cut off where the transistor isn't saturated and isn't working because of lack of current.
Power dissipated by the transistor at Vce of 3v.
P=VI
13mA
3x0.013
=0.039W
Beta of this transistor at Vce of 2,3 and 4V.
B=Ib/Ic
Vce 4v= Beta= 25
Vce 2v= Beta= 25
Vce 3v= Beta=25
Experiment 8
I varied Rb from 2200 ohms to 330,000 ohms.
My results were
RB= 330,000 ohms
Vbe= 0.67V
Vce= 1.5V
Ib-1.3 micro amps
Ic=3.6mA
RB= 270,000 ohms
Vbe= 0.68v
Vce=1.2v
Ib= 164 micro amps
Ic=4.47mA
RB=2200 ohms
Vbe= 0.70V
Vce= 28mV
Ib= 2mA
Ic= 6.70mA
RB=10,000 ohms.
Vbe= 0.74V
Vce=483mV
Ib= 397 micro amps
Ic=6.4mA
Rb=220,000 ohms.
Vbe= 0.68V
Vce=0.70V
Ib= 202 micro amps
Ic=5.5mA
Discuss what changes happened for Vce.
Vce has more voltage when there is more resistance. There is more voltage lost because there
is more resistance.
Discuss what changes happened for Vbe.
Vbe was at 0.7V because the transistor only passes current when it exceeds 0.7V.
For collector current to flow VBE must be above 0.7V. This is when the transistor is considered fully saturated.
Discuss what changes happened for Ib
Current across the base was always very small. Usually in micro amps.
This current is so small it is not usually used in equations. RB is there to control the current that flows into the base. The bigger RB , the less current at the base.
Discuss what changes happened for Ic.
The bigger the resistor the less current is across the collector. Because the resistor blocks the current across IB.
