There are two different inputs which are called inverting(-) and non-inverting(+) input. If the non inverting input is higher than inverting input the output will follow the V+ of voltage rail. On the other hand, if the inverting input is higher the output will go the V- of voltage rail. Therefore, I can make various circuits with different tendency of output values from different input voltages. reference by http://moodle.unitec.ac.nz/file.php/1309/Theory/Lesson_6_Opamps/Lesson6_Operational_Amplifiers.pdf
- Non inverting OP-AMP Vout= ((Rf/Rs)+1) x Vin
The non-inverting OP-AMP will need a path for DC to ground. If the signal source might not give this, or if that source requires a given load impedance, the circuit will require another resistor from input to ground. In either case, the ideal value for the feedback resistors (to give minimum offset voltage) will be such that the two resistances in parallel roughly equal the resistance to ground at the non-inverting input pin. - Inverting OP-AMP Vout= ((Rf/Rs) x -Vin The output voltage changes in an opposite direction to the input voltage
- Applications
· audio- and video-frequency pre amplifiers and buffers
· differential amplifiers · differentiators and integrators
· filters
· precision rectifiers
· precision peak detectors
· voltage and current requlators · analog calculators
· analog to digital converters
· digital to analog convrters
· voltage clamps
· oscillators and waveform generators
2. Mosfets(Metal Oxide Semiconductor Field Effect Transistor) A mosfet operates device with its switching function which is similar to transistors with simiconductor material. The transistor's switching is controlled by the current, however, mosfets are voltage controlled device. - Comparison
- Types Enhancement mode : The transistor requires a Gate-Source voltage, to switch the device "ON". The enhancement mode MOSFET is equivalent to a "Normally Open" switch. N-channel type
There are three different terminals in a N-channel type. When the certain voltage gains in a gate, the drain will connect the source(-). Electron flows from negative to positive. This type is similar to PNP transistor.
P-channel type
The diode direction is only changed. When the gate voltage is at 0v, this mosfet turns on so the between Drain and source will be connected. This type is similar to NPN transistor.
Depletion mode : The transistor requires the Gate-Source voltage, to switch the device "OFF". The depletion mode MOSFET is equivalent to a "Normally Closed" switch.
When the resistors are connected in series the Vout (available voltage) will be different values due to different resistor values. Therefore, I can make the various circuits from the output voltages, such as an O2 sensor indicator. Formula Vout1 =(R2+R3)/RT x Vin, Vout2 =R3/RT x Vin If I know the Vout1 and Vout2 the R1, R2 and R3 of the resistor values would be determined. Chose any one resistor value. Voltage drop of R1=Vin-Vout1, I=V/R, (calculate the amp value) Voltage drop of R2=Vout1-Vout2 R2=Vd2/I Voltage drop of R3=R3-0 R3=Vd3/I
2. Calculations R5? Power supply voltage is at 12v, Vd of crossing diode D2=0.6v, Vd of zener diode D1=9.1v Vd of R5 = 12-0.6-9.1= 2.3v I=5.6mA=0.0056A R=V/I =2.3/0.0056 =411Ω I have chosen a 380Ω resistor which is similar to 411Ω resistor. R8 & R7? R6=10KΩ Voltage drop in crossing R6 is at 8.47v
The voltage are at 9.1v and 0.63v before the R6 and after the R6 each.
So, Vd=9.1-0.63 = 8.47v
Using Ohm's law I=V/R =8.47/10,000 =0.000847A R8 consumes the voltage which voltage drop is at 0.4v Vd= 0.63 - 0.23 =0.4v R=V/I =0.4/0.000847 =472Ω R7 consumes the voltage which voltage drop is at 0.23 Vd= 0.23 - 0 =0.23Ω R=V/I =0.23/0.000847 =271.5Ω Total resistance RT=R6+R8+R7 =10,743Ω I=9.1/10,743 =0.000847A R2, R3 & R4? I= 9.5mA =0.0095A The voltage drop in R2 is at 9.6v, Vd=12 - 0.6 - 1.8 =9.6v R=V/I =9.6/0.0095 =1010.5Ω The voltage drop in R3 is at 9v, Vd=12 - 0.6 - 0.6 - 1.8 =9v R=V/I =9.6/0.0095 =947Ω The voltage drop in R4 is at 9.6v, Vd=12 - 0.6 - 1.8 =9.6v R=V/I =9.6/0.0095 =1010.5Ω 3. A Technical Explanation
There are three different resistors(R6, R8 & R7) in this circuit and these resistors are connected in series which produces different voltage drop and available voltages and these voltages operate the Op-Amp.
A green LED turns on. If the sensor input (12 IN+) is lower than 0.23v and the 13IN- is at 0.23v the current flows through the green LED and runs from 14OUT to 11V- of rail voltage. The other two LEDs will turn off due to lower input voltage.
A yellow LED turns on. If the sensor signal is between 0.23v and 0.63v the 9IN- is at 0.23v to 0.63v. The current flows through the yellow LED and runs from 8OUT to 11V- of rail voltage.The other two LEDs will turn off due to lower and higher input voltages.
A red LED turns on. If the sensor signal is higher than 0.63v the 5IN+ is at 0.63v. The current flows through the red LED and runs from 7OUT to 11V- of rail voltage. The other two LEDs will turn off due to higher input voltage.
4. Test Procedure - Building a circuit in a board : I have chosen correct components with calculated resistors. There are four Op-Amps in a LM324.
When the signal voltage is lower than 0.23v, the green LED turns on and other two LEDs turn off. When the O2 sensor signals are between 0.23v and 0.63v, the yellow LED turns on and other two LEDs turn off. Finally, When the signal voltage is higher than 0.63v, the red LED turns on and other two LEDs turn off. Therefore, this circuit was working properly without any problems.
- Voltage drop When I tested the voltage drop in each resistor, the values were similar between calculated and actual measurement.
- Making an actual board with soldering This circuit was working properly without any problems.
5. Problems I have made a disconnection of the circuit. So I could find various faults and repair that problems. The points of the broken wiring
Point1 : The yellow LED stays turning off. The pin2 will get 0v and the pin3 of signal voltages is always higher than pin2. So, the current flows through the pin4 of voltage rail and D3 diode then the yellow LED always turns off. Point2 : The yellow LED keep turning on over 0.63v. The current will not flow through D3 diode and after the yellow LED. When the red LED turns on at over 0.63v, above current flowing will turn the yellow LED off. From the disconnection, the yellow LED keep turning on at over 0.63v. Point3 : The yellow LED keep turning on always. The pin10 will get 0v and a pin9(signal voltage) is always higher than the pin10. Therefore, the current flows through the yellow LED and pin11(earth) of voltage rail and the yellow LED always turns on. Point4 : The yellow LED keep turning on always. The pin10 will get 0v and a pin9(signal voltage) is always higher than the pin10. Therefore, the current flows through the yellow LED and pin11(earth) of voltage rail and the yellow LED always turns on. Point5 : The green LED keep turning on always. The pin13 will always receive 9.1v and the O2sensor signals are lower than 9.1v. So, the current flows through the green LED and pin11(earth) of voltage rail and the green LED always turns on. Point6 : The all LEDs and Op-Amps are turned off. The supply voltage is cut for the operation of all Op-amps and LED. So these components will not work.
6. Reflection It was a good opportunity that I have understood the operation of Op-Amps from building an O2 sensor display unit. In addition, I could make various voltages from the voltage divider with using different resistors. These voltages are used input of Op-Amps and two differnt inputs determine the flowing output voltage and it controls the switching LEDs on and off. When I took the display on an engine, it works correctly. When the engine rpm suddenly goes up, the red LED turns on due to rich mixture. When the engine is idling, the LED lights are changed green, yellow and red in order which means that the O2sensor monitors the state of exhaust gases then the ECU controls fuel injection for proper engine running.
2. Calculations R14, R15 ? Supply voltage 12v LED 1.8V & 20mA The LED consumes 1.8v for its operation and the current of 20mA flows through anode to cathode. Transistor's voltage drop between collector and emitter 0.2v when the gate is opened. The voltage drops in resistor R14 and R15 are at 10v. (Vt= Vr + Vl + Vce, Vr=Vt - Vl - Vce) Vr=12v - 1.8v - 0.2v=10v The calculations are shown the ohm's law. V=I X R, R=V/I R= 10/0.020 =500Ω Therefore, I chose the resistors value of 500Ω which is near 470Ω for R14 and R15. R13, R16 ? PWM0, PWM1 5v (Injector's signals) Vbe= 0.6v (voltage drop between base and emitter in a transistor BC547) The voltage drops in resistor R13 and R16 are at 4.4v. (Vt= Vr + Vbe, Vr=Vt - Vbe) Vr=5v - 0.6v =4.4v Base to Emitter saturation amps Ib=5mA, Ic=100mA from the BC547 data sheet. V=I X R, R=V/I
R= 4.4/0.005 =880Ω Therefore, I chose the resistors value of 880Ω which is near 820Ω for R13 and R16. 3. A Technical Explanation
This circuit has two individual circuits which is called parallel circuit using 12v supply voltage. When the PWM0 and PWM1 of input signals at 5v are received in this circuit, the small current flow through base(2) to emitter(3) with operating voltage at around 0.6v. And then, the gate between collector and emitter is opened and higher current flows through collector to emitter. If there are no signals in the PWM0 and PWM1 the circuit of the 12v supply voltage will have open circuit because the gate is closed and the current will not run through base to emitter. As a result, the LED turns on when the input signals(PWM0 & PWM1) of injectors are applied. When the duty cycle of the input signals increases, the two LED flash quickly or turn on continuously. 4. Test Procedure
- From the calculated resistor values, I have made the circuit and tested its working.
R14, R15 = 470Ω, R13, R16= 820Ω
When the duty cycle of 5v signals is increased, the two green LED flash rapidly or turn on constantly.
- I have also created the simulation circuit from the lochmaster program before actual soldering on a board. The board, resistors, LEDs and transistors are all the same shape with real components. This simulation board is checked by my tutor and can be right.
- Then, I have made a actual injector circuit in a soldering board, such as above picture.
- Testing Available voltage and Voltage Drop
The voltage drop of LED is slightly higher than ideal voltage at 1.8v. However,all the voltage drops in the each component are similar to calculated values.
- Testing actual Amp (Ib,Ic)
Ib=4.3mA, Ic=20.03mA
5. Problems When I made a circuit, such as temporary circuit, the one LED do not flash. So I have checked the circuit flowing and eventually, I have found the problem which was anode and cathode direction of the LED. Then, I have changed the direction and the fail LED was flashing.
The duty cycle value of frequency device is at around 2v and I could not read at 5v and get wrong voltage drop and available voltage. Therefore, I have changed the signal voltage with another power supply at 5v. And then, I could read correct voltage drop and available voltage.
6. Reflection
It was a good opportunity that have understood the electric circuit of the operation with building simple injector circuit. This circuit shows transistor's switching operation with the relationship of Base, Collector and Emitter. If the wrong resistors were chosen the circuit would not run correctly and higher resistors would make dark brightness of LED.
If this circuit is installed in a vehicle the signals of injectors indicate the fuel injection. In addition, when the engine RPM is increased, the duty cycle of signals also goes up therefore LEDs flash rapidly and turn on continuously.
Ohm's lawThe current flow through a Resistance there must be a voltage across that resistance. Ohm's law shows the relationship between the voltage current and resistance.
V=I X R, I=V/R, R=V/I - When a larger light bulb(consumer) uses in a circuit the current increases more than using a small light bulb and if the larger current flows through the circuit the resistance decreases.
Kirchoff's law The sum of all the voltage drop around the loop equal zero.This idea by Kirchoff is known as the Conservation of Energy.
- The current flow first through one light bulb(bulb1), then the other(bulb2). The main voltage drop is light bulb1 and light bulb2, at 6.34V and 6.32V each. As a result, the supply voltage is shared by two light bulbs. When the voltage is flowing in series circuit the consumers share the voltage. - When the amp measures at different parts of the circuit the amp values are all the same at 0.22A. The amp in the individual circuit is higher than the series circuit. Because, the resistance is lower in individual circuit with a light bulb. If the lower resistance is produced in a circuit the amp increases from the ohm's law. - The watts of two bulbs is decreased because the voltage is shared by two bulbs. Therefore, the brightness of two light bulbs is darker than a bulb of individual circuit. - When the three bulbs connect in series circuit the supply voltage is shared by three bulbs.The amp is reduced due to one more resistance is produce by the third bulb. The total resistance also increases in three bulbs series circuit and the brightness is darker than two bulbs with lower watts. - Voltage drop: I can check the voltage consumption each parts in detail. - Available voltage: The total consumption and remaining voltage are checked from this test.
Parallel circuit
- The available voltage and the voltage drop are similar in light bulb1 and bulb2. The voltage is not shared in parallel circuit which keeps the individual circuit from each light bulbs. - The current flows each bulb and the total amp in parallel circuit is higher than individual circuit the total amp value is the sum of the amp of each bulb. - Total resistance of Parallel circuit the total resistance value is smaller than the smallest resistance in parallel circuit. - The watts in parallel circuit is higher than series circuit. Because, the using voltage in each bulb is near supply voltage and the amp is also higher than series circuit. As a result, from this parallel circuit light bulbs consume more energy and show two individual circuit.
Available Voltage
The current flows from positive to negative and this circuit is close circuit when the switch is closed. Available voltage is around 12V between negative terminal before the light bulb and then there are no available voltage after the light bulb because the supply voltage is consumed by the light bulb. The used and remaining voltage are checked by the available voltage at the different points.
Voltage Drop
The voltage drop is measured across the different components of the circuit with using a voltmeter. There are small voltage drop at 0.01V to 0.02V, such as switch, terminal and wire because the resistance is produced from these components and the main voltage drop is from the input to the bulb to the output of the bulb. The light bulb uses supply voltage with a main consumer in this individual circuit.
THEORY A transistor is one of electronic component and semiconductor device that can basically amplify the current. The transistor is like the relay and it has a switching operation. There are three different terminals, such as base, collector and emitter. When the current flow from base to collector with enough voltage at around 0.6v for its operation, the gate will open between collector and emitter and the higher current runs with the lower voltage. In addition, there are two different types of the transistor which are called NPN and PNP type.
EXPERIMENT No. 6 Meter check of a transistor 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.
9.1: PNP transistor meter check: (a) forward B-E, B-C, voltage is low; (b) reverse B-E, B-C, voltage is OL.
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, the technician obtains the data in Figure 9.2.
Figure 9.2: 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 9.3.
Figure 9.3: BJT terminals identified by meter.
Please 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 Symbol and semiconductor construction shown below.
Fig 10
Identify the legs of your transistor with a multimeter. For identifying and testing purposes, refer to the representation shown above.
EXPERIMENT No. 7 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 base and emitter.
Note the voltage reading and explain what this reading is indicating.
0.8v The current flows from Base to Emitter with transistor's operation at around 0.6v and the lower current goes through Base to Emitter.
Connect the multimeter between collector and emitter.
Note the voltage reading and explain what this reading is indicating.
0.06v The gate is opened between collector and emitter so the higher current flows from collector to emitter with lower voltage drop of 0.06v.
In the plot given below what are the regions indicated by the arrows A & B?
How does a transistor work in these regions? Explain in detail: A(saturated) : The supply voltage will appear across the load collector current will flow between collector and emitter. B(cut-off) : Base current is too low or zero. So collector current will not flow from collector to emitter. The transistor operates correctly in the active region with an amplifier and a switching function. As a result, higher current flows between collector and emitter. reference by http://moodle.unitec.ac.nz/file.php/1309/Theory/Lesson_4_BJT/Lesson_4_BJT.pdf
What is the power dissipated by the transistor at Vce of 3 volts?
P(W)=I X V Ic=13mA=0.013A (the Vce of 3v form the above fig13) P=0.013 X 3 =0.039W
What is the Beta of this transistor at Vce 2,3 & 4 volts?
Beta=gain
β= Ic/Ib
Vce=2v, Ic=20mA, Ib=0.8mA
β= 20/0.8= 25
Vce=3v, Ic=13mA, Ib=0.5mA
β= 13/0.5= 26
Vce=4v, Ic=5mA, Ib=0.2mA
β= 5/0.2= 25 EXPERIMENT No. 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.
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 47KVbe 0.7v Vce 0.09v Ib 0.11mAIc 6.17mA
Rb 220KVbe 0.68v Vce 0.16vIb 0.03mA Ic 6.02mA
Rb 270K Vbe 0.68v Vce 0.20vIb 15.9μA Ic 5.95mA
Rb 330K Vbe 0.68v Vce 0.32v Ib 13.1μA Ic 5.71mA
Rb 1MVbe 0.65v Vce 2.16v Ib 4.2μA Ic 2.02mA
Your voltage drop measurements across Vce should vary from below 0.3 v (showing the transistor is 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? When I changed five different resistors(Rb) from lower values to higher values, the Vce(voltage between collector and emitter) values also changed. When the Rb increased, the Vce also goes up.
Discuss what happened for Vbe during this experiment. What change took place if any, and what caused the change? The Vbe dropped slightly when the resistor values are increased. However, the Vbe has constant values from base to emitter for transistor's operation at around 0.6v.
Discuss what happened for Ib during this experiment. What change took place, and what caused the change? When I changed five different resistors(Rb) from lower values to higher values, the Ib values also changed. When the Rb increased, the Ib decreased. The current Ib is lower than Ic values. The small current runs through the base for an operating transistor.
Discuss what happened for Ic during this experiment. What change took place, and what caused the change? When I changed five different resistors(Rb) from lower values to higher values, the Ic values also changed.When the Rb increased, the Ic dropped. The current Ic is higher than Ib values. This means that the gate between collector and emitter is opened so the higher current flows from collector to emitter.
Plot the points for Ic and Vce on the graph below to create a load line. Plan the values for so you use up the graph space. Use Ic as your vertical value, and Vce as your horizontal value.
Using Vbe on the Vce scale, plot the values of Ib so the finished graph looks similar to fig 13.
Calculate the Beta (Hfe) of this transistor using the above graph.
β= Ic/Ib
Rb=47K, Ic=6.17mA, Ib=0.11mA β= 6.17/0.11= 56.1 Rb=220K, Ic=6.02mA, Ib=0.03mA β= 6.02/0.03= 200.7 Rb=270K, Ic=5.95mA, Ib=0.0159mA β= 5.95/0.0159= 374 Rb=330K, Ic=5.71mA, Ib=0.0131mA β= 5.71/0.0131= 435.9 Rb=1M, Ic=2.02mA, Ib=0.0042mA β= 2.02/0.0042= 480.9 Explain what the load line graph is telling you. Discuss the regions of the graph where the transistor is Saturated, Cut-off, or in the Active area.
Vce is significantly large at 2.16v when the Rb is 1MΩ from the load line graph and figure. Ib will be nearly zero and no collector current will flow, this means that the transistor is cut-off. The other end of the load line Ib increases with maximum current at 0.11mA, then Vce will be nearly zero or close to it, this state is called transistor's saturation. The transistor is working in saturation with the highest current of base bias. In both cut-off and saturation, minimum power is consumed in the transistor. The transistor can be operated correctly in the active area with different values of Ib and transistor's amplification characteristic.
In addition, β(gain) value should be smaller for the chosen transistor. Because if the β is too high the heat will produce in a transistor and it will be damaged.
THEORY Capacitors are used to store an electrical charge and are used in timer circuits. If the supply voltage is too high or the circuit has sudden voltage changes, the capacitors prevent voltage spikes. A large capacitance means that more charge can be stored. Capacitance is measured in farads, symbol F.
EXPERIMENT No. 5 The capacitor stores electric charge.
A capacitor consists of two metal plates very close together, separated by an insulator. When connected to a battery or power source electrons flow into the negative plates and charge up the capacitor. The charge remains there when the battery is removed. The charge stored depends on the “size” or capacitance of the capacitor, which is measured on Farads (F). Types of capacitor:
Non-electrolytic capacitor
· Fairly small capacitance - normally about10pF to 1mF
· No polarity requirements - they can be inserted either way into a circuit.
· Can take a fairly high voltage. Variable capacitor
· Adjustable capacitor by turning a knob - similar to variable resistors.
· The maximum capacitance available is about 200pF.
· Used in radios.
Electrolytic capacitor
· Large capacitances - 1mF to 50000mF
· Warning: These must be corrected the right way round (polarity) or they can explode - the white terminal on the diagram above signifies positive.
· Black stripe with “-“ shows which terminal is the negative (usually the short one)
· Low voltage rating – from 25 ~ 100V DC
· They have a significant leakage current - this means that they will lose the charge stored over time.
Tantalum capacitor
· These have the same properties as the Electrolytic capacitor, but they are physically smaller & have lower leakage. As a result, though, they are more expensive. Identifying Capacitor “Size”If the Farad “size” is not printed on the capacitor, you may find an EIA code listed. Use the table below to figure out the capacitance.
* Values with asterisk are not usually expressed in this form RC Time Delay or “Charging Time”
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. This is the time it takes to charge up to the applied voltage.
For example, 1,000,000 Ω x 0.000001 F x 5 = 5 seconds to charge to applied voltage. This can also be expressed as 1 MΩ x 1 μF x 5 = 5 seconds.
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.
Practice watching the capacitors charge up in the exercise below.
Fig 8-Capacitor Charging Circuit
Components: 1 x resistor, 1 x capacitor. 1 x pushbutton N/O switch.
Exercise: First, calculate how much time it would take to charge up the capacitor. Then, connect the circuit as shown above. Measure the time taken by the capacitor to reach the applied voltage on an oscilloscope. Fill in the chart below. Also draw the observed waveforms in the graphs below, filling the details on each one.
Note: you will need to adjust the time base to enable you to observe the pattern.
3. 0.0001F X 470Ω X 5=0.235S(235ms) 4. 0.00033F X 1000Ω X 5=1.65S
Label the axis of each graph:
Circuit 1:
Capacitance 100uFResistance 1KΩ
1 : The supply voltage is not engaged in a capacitor. voltage=0v
2 : The supply voltage is charging through the capacitor. The voltage is significantly charged with higher speed at the beginning of charging time. However, when the capacitor is nearly fully charged, the changing voltage is small. The charging time reached at around 500ms.
3 : The supply voltage is fully charged in a capacitor at 12v. Circuit 2:
Capacitance 100uFResistance 100Ω
1 : The supply voltage is not engaged in a capacitor. voltage=0v
2 : The supply voltage is charging through the capacitor. The voltage is significantly charged with higher speed at the beginning of charging time. However, when the capacitor is nearly fully charged, the changing voltage is small. The charging time reached at around 50ms.
3 : The supply voltage is fully charged in a capacitor at 12v.
Circuit 3:
Capacitance 100uF Resistance 470Ω
1 : The supply voltage is not engaged in a capacitor. voltage=0v
2 : The supply voltage is charging through the capacitor. The voltage is significantly charged with higher speed at the beginning of charging time. However, when the capacitor is nearly fully charged, the changing voltage is small. The charging time reached at around 200ms.
3 : The supply voltage is fully charged in a capacitor at 12v.
Circuit 4:
Capacitance 330uF Resistance 1KΩ
1 : The supply voltage is not engaged in a capacitor. voltage=0v
2 : The supply voltage is charging through the capacitor. The voltage is significantly charged with higher speed at the beginning of charging time. However, when the capacitor is nearly fully charged, the changing voltage is small. The charging time reached at around 1.5s.
3 : The supply voltage is fully charged in a capacitor at 12v.
How does changes in the resistor affect the charging time?
When the resistor values are higher, the charging time of a capacitor increases under same capacitance values.
How does changes in the capacitor affect the charging time?
When the capacitance is higher with the same resistors, the charging time of the capacitors also goes up.
