APPLIED ELECTRONICS Outcome 2Outcome 2 - Design and construct electronic systems, based on operational amplifiers, to meet given specificationsWhen you have completed this unit you should be able to: State the characteristics of an ideal Operational Amplifier Identify the various op. amp. Configurations Carry out calculations involving op. amps Select a suitable op. amp. circuit for a given purpose Design op. amp. circuits for a given purpose.
APPLIED ELECTRONICS Outcome 2The operational amplifier (op. amp) This ic was designed to perform mathematical operations and was originally used in analogue computers. The op. amp. can be used to add, subtract, multiply, divide, integrate and differentiate electrical voltages. It can amplify both d.c. and a.c. signals. POSITIVE SUPPLY LINE (+Vcc) INVERTING INPUT OUTPUT NON-INVERTING INPUT NEGATIVE SUPPLY LINE (-Vcc)
APPLIED ELECTRONICS Outcome 2An "ideal" amplifier should have the following qualities An infinite input resistance (typically 1M or more) - so that very little current is drawn from the source Zero output resistance (typically 100 or less) - so that variations in load have very little effect on the amplifier output An extremely high gain (typically 100,000) No output when the input is zero (in practice this isseldom achieved, however manufacturers provide an "offset - null"to compensate for this)
APPLIED ELECTRONICS Outcome 2 It can be seen from the diagram that the op. amp. has two inputs. The op. amp. is designed as a differential amplifier i.e. it amplifies the difference between the two input voltages. The two inputs are indicated by a "-" and "+". POSITIVE SUPPLY LINE (+Vcc) INVERTING INPUT OUTPUT NON-INVERTING INPUT NEGATIVE SUPPLY LINE (-Vcc)
APPLIED ELECTRONICS Outcome 2 N U LL NO T 1 8 D E F L E C T IO N CO NNECTED IN V E R T IN G P O S IT IV E S U P P L Y IN P U T 2 7 L IN E (+ V c c ) N O N -IN V E R T IN G 3 6 OUTPUT IN P U TN E G A T IV E S U P P L Y 4 5 N U LL L IN E (-V c c ) D E F L E C T IO N Op. amp. ics come in two forms, the most popular of which is the dil (dual - in - line) package. The pin diagram is shown above The other form of IC is Surface Mount. This is normally used in automated assembly processes.
APPLIED ELECTRONICS Outcome 2 GAIN The op. amp. was designed as a voltage amplifier. The voltage gain of any amplifier is defined as Voltage output Voltage gain = Voltage input Vo AV = Vi For a differential amplifier, the voltage input is the difference between the two inputs. Vi = ( V(at non - inverting input) - V (at inverting input) )
APPLIED ELECTRONICS Outcome 2Pupil Assignment 11. If V (at non - inverting input) = 3.10 V and V (at inverting input) = 3.11 V. Calculate the input voltage and hence the output voltage if the gain is known to be 100.2. The gain of an op. amp. is known to be 100,000. If the output voltage is 10 V, calculate the input voltage.3. The gain of an op. amp. is known to be 200,000. If V(at non - inverting input) = 2.5 V and V (at inverting input) = 2.2 V, calculate the output voltage.
APPLIED ELECTRONICS Outcome 2The answer to Q3 is obviously unrealistic since the outputvoltage from an op. amp. cannot be greater than the supplyvoltage.As the output of the op. amp. increases, saturation starts tooccur and a "clipping" effect will be noticed. This normallyoccurs when the output reaches 85% of VCCAny further increase in the input will cause no furtherincrease in the output since the op. amp. has reachedsaturation.The inherent voltage gain of an op. amp. (i.e. when noexternal components are connected) is designed to be verylarge (200,000 in some cases). This is sometimes called
APPLIED ELECTRONICS Outcome 2Controlling Gain In order to reduce the gain, a smallpart of the output signal isfed back to theinverting inputthrough a feedbackresistor, Rf. Since this signal is going to the inverting input, it is a form of negative feedback and has the effect of reducing the overall gain of the circuit. The closed loop voltage gain, AV, of the circuit will depend on the circuit configuration.
APPLIED ELECTRONICS Outcome 2The inverting amplifier configuration Rf The signal is connected to the inverting input R 1 through an input resistor (R1). V in The non - inverting input V out is connected to ground 0V either directly or through a biasing resistor Rb.
APPLIED ELECTRONICS Outcome 2The inverting amplifier configuration RfWorked exampleAn op. amp. is used in a circuitas shown with R1 = 15 k and Rf R 1= 470 k.Calculate the gain of the circuit V in V outand determine the outputvoltage when an input signal of 0V0.2 v is applied.
APPLIED ELECTRONICS Outcome 2The inverting amplifier configurationStep 1 Rf 470kCalculate the gain 15k Rf 470k R 1Av = − =− = −3133 . R1 15k V in V outStep 2 0VCalculate the output voltageVout = Av x Vin = - 31.33 x 0.2 = - 6.266 V
APPLIED ELECTRONICS Outcome 2The inverting amplifier characteristics Closed loop voltage gain (negative sign indicates inversion) Rf Rf Av = − R1 R 1 Input resistance of V in the circuit = R1 V out 0V For an inverting amplifier, the sign of the output voltage is the opposite of the input voltage. In order to obtain the same sign, the output signal could then be fed through another inverter (with Rf = R1, so that the gain = -1).
APPLIED ELECTRONICS Outcome 2Pupil Assignment 2 A thermocouple known to produce an output of 0.040 volts per oC is connected to an op. amp. Circuit as shown below. 1M THERMO-COUPLE 10k Vout 0V Calculate the gain of the circuit Determine the output voltage if the thermocouple is heated to a temperature of 1000 oC.
APPLIED ELECTRONICS Outcome 2 The non - inverting amplifier configuration The signal is connected directly to the non - inverting input. Rf and R1 form a voltage divider circuit feeding back some of the output signal to the inverting input. The figures below show two different ways of drawing the same circuit. Rf Rf Vin VoutVin R1 Vout R1 0V AA 0V
APPLIED ELECTRONICS Outcome 2Characteristics of the non - inverting amplifier Rf Closed Loop Voltage Gain Rf AV = 1 + R1 (no inversion, gain is positive) Vin R1 Vout Input resistance of the circuit = input resistance of the op. 0V amp. (very high)Note: because of the high input resistance, this circuit is useful when inputtransducers do not provide large currents.
APPLIED ELECTRONICS Outcome 2Pupil Assignment 3To build a simple light meter, a light dependent resistor (LDR) is connectedinto a circuit as shown. +8VIn bright sunlight, the LDR has a 50k resistance of 1 k. In shade, its resistance increases to 15 k. LDRDetermine the voltages that would appear on the voltmeter in both light conditions. 15k 100kHow could the circuit be altered to 0V indicate changes in temperature?
APPLIED ELECTRONICS Outcome 2The voltage follower This is a special case of the non-inverting amplifier in which 100% negative feedback is obtained by connecting the output directly to the inverting input. Since Rf = 0, the gain of this circuit is 1 i.e. The output voltage = input voltage. The practical application of this circuit is that it has a very high input resistance and a very low output resistance. It is therefore used in matching a source that can only produce a low current to a load which has a low resistance
Impedance Matching Practical Build the circuit shown on the left. You should find that the signal from the voltage divider collapses when the lamp is added to the circuit. Hence the lamp fails to light!
Impedance Matching Practical Now build the circuit shown. The lamp should now light. However, its brightness will be no where near its optimum. The activity shows the buffering effect of a voltage follower. To illuminate the bulb fully would require a transistor amplifier in the circuit.
APPLIED ELECTRONICS Outcome 2Circuit Simulation using Croc Clips Set the input voltage to 2 Volts, 0.25 Hz. Set the oscilloscope to a maximum voltage of 10 V and a minimum voltage of - 10 V Start the trace on the oscilloscope and compare the input and output voltages. Now increase the size of the feedback resistor to 50 k and repeat the exercise. This time you should observe “clipping” of the output signal. 20k 10k 0.25Hz
APPLIED ELECTRONICS Outcome 2Circuit Simulation using Croc Clips Set the input voltage to 2 Volts, 0.25 Hz. Set the oscilloscope to a maximum voltage of 10 V and a minimum voltage of - 10 V Start the trace on the oscilloscope and compare the input and output voltages. Now increase the size of the feedback resistor to 50 k and repeat the exercise. This time you should observe “clipping” of the output signal. 20k Op Amp rails +/- 10k 9v 0.25Hz
APPLIED ELECTRONICS Outcome 2 Circuit Simulation using Croc Clips 1k 40° 10k -t O C -40° 9V 1k 10k 9V 100k VrSet the “temperature” to 0oC, adjust the variable resistor (Vr) until the voltmeterreads 0.00.Increase the “temperature” to 40 oC, adjust the feedback resistor in the invertingamplifier until the voltmeter reads 0.40The electronic thermometer is now calibrated to read 0.00 at 0oC and 0.40 at40oC.Investigate voltage readings at various other “temperatures” and suggest why thiswould not make a good thermometer
APPLIED ELECTRONICS Outcome 2 The summing amplifier Here, two (or more) signals are connected to the inverting input via their own resistors. The op. amp. effectively amplifies each input in isolation of the others and then sums the outputs. Notes: Rf Any number of inputs can be R 1 added in this way. Rf affects the gain of every R 2 input.V 1 If all the resistors are the V 2 V o u t same size, then the gain for each input will be -1 and 0V S D Vout = - ( V1 + V2 + V3 + ......)
APPLIED ELECTRONICS Outcome 2 Characteristics of the Summing Amplifier Each input signal is amplified by the Rfappropriate amount (seeinverting mode) R 1 R 2 Rf RfVout = ( − × V1 ) + ( − × V2 ) V 1 V V out R1 R2 2 0V SD V1 V2Vout = −Rf ( + +.......) R1 R2
APPLIED ELECTRONICS Outcome 2Digital-to-analogue converter Digital devices produce ON/OFF signals. Processing takes place using the binary number system (as opposed to the decimal system). Construct the circuit shown 10k S3 S2 S1 R1 10k 10k 10k 1V R2 10k R3 10k Continued over
APPLIED ELECTRONICS Outcome 2 10k Since all inputs are S3 S2 S1 R1 10k amplified by the same amount (same 10k 1V R2 10k value of input resistors) the output 10k R3 voltage = input 10k voltages e.g. S1, ON (connected to 1V) and S2, ON (connected to 1V) , the output voltageNow change the circuit so that R2 = 5k and R3 = 2.5k should = (1 + 1) = 2VConstruct a table to show the state of the input switchesand the output voltage.
APPLIED ELECTRONICS Outcome 2Pupil Assignment 4Mixers allow different signals to beamplified by different amounts beforebeing fed to the main amplifier. Signalsmight come from microphones, guitarpick-ups, vocals, pre-recorded soundtracks etc. 10kSimulate the circuit shown S 1 S 2 S 3opposite.Adjust the frequencies of thesignals as shown and adjust theoscilloscope to give a maximum 0 .2 5 H z O .5 H z 0 .7 5 H zvoltage of 10 V and a minimumof -10 V.
APPLIED ELECTRONICS Outcome 2 Pupil Assignment 4 continuedPutting each switch onindividually will allow you to 10k“see” each of the input S 1 S 2 S 3signals in turn.Putting more than oneswitch on at a time will showyou the sum of the inputsignals. 0 .2 5 H z O .5 H z 0 .7 5 H zAdjusting the size of theinput variable resistors altersthe amplification of thatparticular input signal.
APPLIED ELECTRONICS Outcome 2Pupil Assignment 5 10k 4V 10k 5kDetermine the output INPUTS 2Vvoltage for the circuit shown 2kin figure 12 1V V out 0V AA
APPLIED ELECTRONICS Outcome 2Pupil Assignment 6 A personal stereo has both tape and radio inputs which produce output signals of 50 mV and 10 mV respectively. The amplifying system consists of a main amplifier and uses an op. amp. as a pre - amplifier. Design a possible pre - amplifier circuit so that an output of 1 volt is produced when either the tape or radio inputs are used.
APPLIED ELECTRONICS Outcome 2 Pupil Assignment (Extension)Design a operational amplifier circuit which will give a maximum output of4 volts. The amplifier should be capable of outputting 7 equal stepsbetween 0V and 4V. The input to each bit is 3V. Solution: Vout = 3(-Rf/R1)+(-Rf/R2)+(-Rf/R3) -4 = -3Rf/10 – 3Rf/5 – 3Rf/2.5 -4 = -3Rf – 6Rf – 12Rf/10 40 = 21Rf Rf= 40/21 = 1.9K
APPLIED ELECTRONICS Outcome 2 The difference amplifier configurationHere both inputs are used. RfThe op. amp. amplifies the R1difference between the twoinput signals R2To ensure that each input is V1 Voutamplified by the same amount, V2 R3the circuit is designed so that the 0Vratio: Rf R3 = R1 R2
APPLIED ELECTRONICS Outcome 2The difference amplifierconfiguration Rf Rf AV = R1 R1 R2 V1 Rf Vout Vout = × (V2 − V1 ) V2 R3 R1 0VNote: If R1 = Rf then AV = 1 and Vout = (V2 - V1) , the circuit works as a "subtracter". The output will be zero if both inputs are the same. This circuit is used when comparing the difference between two input signals.
APPLIED ELECTRONICS Outcome 2 Pupil Assignment 7 Two strain gauges are connected to a difference amplifier as shown. RA = RB = 1 k, when not understrain, Rg1 = Rg2 = 200 Ohm Calculate the voltage at X and Y when both gauges are not under strain and 6V hence determine the output voltage of Ra Rb 42k the amplifier. 4k2 Y As the strain of Rg2 increases, its Xresistance increases from 200 to 210, 390R 3.9Kdetermine the new output voltage. Rg Rg2 Vout 39k 1 What would you expect to happen to 0V AA the voltage divider output voltage ifboth gauges were put under the sameamount of strain?
APPLIED ELECTRONICS Outcome 2The comparator Configuration This is a special case of the difference amplifier in which there is nofeedback. The gain of the circuit is therefore Ao and any small difference in thetwo input signals is amplified to such an extent that the op. amp.saturates (either positively or negatively). V 1 V out V 2 0V AA
APPLIED ELECTRONICS Outcome 2The comparator Configuration AV = Ao Vout = Ao x (V2 - V1) If V2 > V1, Vout is positive V 1 V out If V2 < V1, Vout is negative V 2 This is commonly used in 0Vcontrol circuits in which loads are AAmerely switched on and off
APPLIED ELECTRONICS Outcome 2Temperature Sensing System 6V Vr is adjusted until V1 is just greater than V2, the V r output will therefore be negative and the led will V 1 be off. V 2 As the temperature falls, the resistance of the ntc -t thermistor rises and therefore V2 starts to rise. Eventually, V2 > LED V1, the output goes positive and the led 0V lights.
APPLIED ELECTRONICS Outcome 2Circuit Simulation Simulate the following circuit Alter the temperature and observe the circuit operation Describe how the circuit works Led 2 LED 1
APPLIED ELECTRONICS Outcome 2 Driving External Loads The maximum output current that can be drawn from an op.amp. is usually low (typically 5 mA). If larger currents are required, the output could be connected to a transducer driver either a bipolar transistor or MOSFET (and relay circuit if required).Describe the operation of the circuit shownand state the purpose of the variableresistor Vr and the fixed resistor Rb. RbFor clarity, the d.c. power supply has notbeen shown) Vr
APPLIED ELECTRONICS Outcome 2 Control SystemsIn a control or servomechanism system a feedback loop is included in the circuit.This monitors the output and necessary changes are made to ensure that thelevel of the output remains at a constant level. S E N S IN G TRANSDUCER IN P U T O UTPUT S E T T IN G CO NTRO L SYSTEMThe difference between the input setting and the actual output as monitored by thetransducer will produce an error. This error is then used to alter the output of the controlsystem.e.g. The temperature control of a freezer is set at a given value. A transducer thenmonitors the temperature and switches the freezer pump on and off accordingly.
APPLIED ELECTRONICS Outcome 2Control Systems S E N S IN G TRANSDUCER IN P U T O UTPUT S E T T IN G CO NTRO L SYSTEMIn its simplest form, a feedback (or closed-loop) system provides an on/offoutput in which a mechanical or electronic relay, switches the power circuit onor off. This on/off operation will cause the output to "hunt" above and belowthe required level.In some cases, an on/off system may be all that is required. CAN YOU SUGGEST SUCH A SITUATION?
APPLIED ELECTRONICS Outcome 2 Control Systems OPEN LOOP PROBLEMS In a non-feedback system (sometimes known as an open-loop system), the inputs are adjusted to give the expected output and then left. Changes in conditions (load, environment, wear & tear etc.) may result in the output varying from the level set by the inputs. These changes are not taken into account by the open-loop system. For example, the speed of an electric motor may be set by an input variable resistor, load on the motor however will cause it to slow down and the output speed will be less than expected for the given input conditions. GIVE AN EXAMPLE OF AN OPEN LOOP SYSTEM
APPLIED ELECTRONICS Outcome 2OPEN LOOP An open loop control system represents the simplest and cheapest form of control. However, although open loop control has many application, the basic weakness in this type of control lies in the lack of capability to adjust to suit the changing output requirements.
APPLIED ELECTRONICS Outcome 2CLOSED LOOP Closed loop control systems are capable of making decisions and adjusting their performance to suit changing output conditions. A personal cassette player is capable of detecting the end of the tape and switching the motor off, hence protecting the tape from snapping (or the motor burning out). END O F TA P E CO NTRO L O UTPUT D R IV E R
Programmable Systems Outcome 2 Three main types of Electronic Control, 1) On/Off Open Loop Control 2) On/Off Closed Loop Control 3) Proportional Closed Loop ControlWe will now analyse each type in turn.
APPLIED ELECTRONICS Outcome 26V ON/OFF OPEN LOOP CONTROL V r V 1 Describe the operation of the circuit V 2 -t LED 0V
APPLIED ELECTRONICS Outcome 2 T E M P E R A T U R E F E E D B A C K S IG N A L +V CCR R T ON/OFF CLOSED LOOP CONTROLR V Describe the operation of the circuit R VR O V -V CC
APPLIED ELECTRONICS Outcome 2Control Systems Proportional ControlA better form of feedback loop is where the output is proportional to thedifference between the preset level and the feedback signal.This results in smoother control, for example, in an electrical heater where theoutput power of the heater can be varied according to the difference betweenthe preset temperature and the actual temperature.If the temperature difference is large, the heater might be working at full power,as the temperature of the room increases, the temperature difference betweenthe preset value and the actual temperature will decrease and therefore theoutput power of the heater will decrease. The Difference Amplifier is the basis of proportional control
APPLIED ELECTRONICS Outcome 2Control Systems Proportional Control Rf R1 R2 V1 Vout V2 R3 0VAdd the necessary components to the above circuit to achieve a proportionalclosed loop control circuit.
APPLIED ELECTRONICS Outcome 2Strain Gauges and Measurement of Load CLAM PStrain gauges can be used to investigatethe load on particular members of aconstruction. S T R A IN G AUG EThe resistance of a strain gauge depends M E TA Lon whether it is under tension or S T R IPcompression. It will also however depend ontemperature. LO ADThe diagram shows a single strain gaugebonded to the upper face of a metal strip.
APPLIED ELECTRONICS Outcome 2Strain Gauges and Measuring Load +15V 1k 560R As loads are placed 1M on the strip, it bends and stretches the 10kstrain gauge which in 47kturn changes its resistance. This change in 10k Voutresistance can be 1k Rg 1Mamplified using a differentialop. amp. 0V -15V
APPLIED ELECTRONICS Outcome 2Strain Gauges and Measuring Load Since the resistance of the gauge also depends on temperature, any temperature change will be "recorded" as a change in load. In order to overcome this, it is normal to use two strain gauges in a voltage divider circuit. +15V 1k Rg 1M 10k 47k 10k Vout 1k Rg 1M 0V -15V
APPLIED ELECTRONICS Outcome 2Strain Gauges and Measuring Load Assuming both gauges remain at the same temperature, they will both change resistance by the same amount and therefore the circuit will remain in balance. +15V 1k Rg 1M 10k 47k 10k Vout 1k Rg 1M 0V -15V
APPLIED ELECTRONICS Outcome 2 100k Unit Assignment (1) 50kThe input to the circuit is monitored by asingle beam oscilloscope, the graphicshows the screen display and the Vincontrol settings. Vout 0V Determine the frequency of the pulses at the input. Assuming the oscilloscope controls are not adjusted, redraw the trace you would expect to see if the oscilloscope was connected to the output of the circuit. 1.0 2.0 50 The same components are now used to wire the0.5 5.0 1.0 100 op. amp. in the non-inverting configuration. Re- y - GAIN 0.1 500 draw the new circuit diagram and the trace you would expect at the output this time. TIMEBASE Volts/cm ms/cm
APPLIED ELECTRONICS Outcome 2Unit Assignment (2) When a tacho generator is rotated, it produces a voltage Proportional to its angular velocity. The tacho is to be used to monitor the speed of coolant flowing along a pipe. If the speed is below a recommended level then an alarm should come on, if the speed is above this level then a green light should come on. Draw a circuit diagram of the system that could be used and explain how the system works.
APPLIED ELECTRONICS Outcome 2Unit Assignment (3) Part of a disco audio system consists of the microphone connected to a circuit containing op. amps. as shown below. The circuit should provide a visual indication of sound level. 6V Name the Op Amp configurations used It was found that the circuit did not operate as expected. Simulate theMICROPHONE 1M circuit and suggest a solution with 1k justification.
APPLIED ELECTRONICS Outcome 2Unit Assignment (4) A robot is designed to follow a white line painted on the floor. The robot moves by means of suitably geared motors which can be switched on and off independently. The left sensor controls the right motor (and vice versa). When a sensor detects a white line, the motor is switched on. Suggest a suitable detector that could be used to detect the white line. Explain how the robot follows the white line. W H IT E L IN E Design a suitable circuit that could be used to control one of the motors.
APPLIED ELECTRONICS Outcome 2SEB & SQA Past Paper exam Questions 1993, Paper 1, question 1 Name the configuration of the amplifier shown below Calculate the gain of the amplifier, (i) If the input signal Vi is 0.5 V, what is the value of the output signal Vo? (ii) Explain your answer. +15V 100k 2k V1 V0 0V -15V
APPLIED ELECTRONICS Outcome 21997, Paper 1, question 6 The circuit below shows an operational amplifier circuit, which includes an ORP12 light dependent resistor as an input sensor. When the light level on the LDR is 50 lux, determine: the resistance of the LDR; the voltage gain of the operational amplifier; the current flowing through the load resistor RL, stating clearly the direction in which it is flowing. 6k 1k +Vcc A ORP12 4V -Vcc RL= 300R B
APPLIED ELECTRONICS Outcome 21994, Paper 1, question 10A technological experiment involves recording the total effects of light andtemperature. It utilises an operational amplifier configured as shown below. +9V 10k 3k5 20k 20k Vo 10k t 0V -9V Continued
APPLIED ELECTRONICS Outcome 21994, Paper 1, question 10 +9VName the configuration 10k of the amplifier used 3k5 in the experiment.Explain clearly how the 20k system operates.Determine the gain of the amplifier.Comment on the 20k suitability of the Vo 10k value of the gain in t this particular circuit. 0V -9V Continued
APPLIED ELECTRONICS Outcome 2The following graph shows the actual temperature and light readingsrecorded during the experiment between the hours of 1400 and 2000 onone particular day.The characteristics of the light dependent resistor and the thermistor usedin the circuit are shown below L IG H TDetermine the output voltage (L U X ) T E M P (o C ) (Vo) from the circuit at 1700 hours. 1000 50For later processing, this value 800 40 of (Vo) must be positive.Name an additional device 600 30 which can be added to the 20 TEM PER ATU R E 400 circuit to produce a positive value for Vo. 200 10 L IG H TDraw the circuit symbol for this additional device and 1400 1500 1600 1700 1800 1900 2000 T IM E O F D A Y indicate the value of any components used.
APPLIED ELECTRONICS Outcome 2 1997, Paper 1, 12V question 9 t 400RA deep-fat fryer Cincorporates acooking oiltemperature 300Rindicator. An arrayof LEDs is shown on Bthe control paneland, as the 200Rtemperature of thecooking oil Aincreases, the LEDslight in a laddersequence. 100R 15k 0V -12V
APPLIED ELECTRONICS Outcome 21997, Paper 1,question 9 In which amplifier mode are the 741 Ics being used ? Explain in detail why the LEDs light up in sequence as the temperature of the oil increases. The function of the components in the circuit should be included in your explanation. At what temperature will LED “C” light ? If the current through each LED is to be limited to 200 mA, determine what value of resistor should be connected in series with each LED. (Ignore any voltage drop across the LEDs.)
APPLIED ELECTRONICS Outcome 21998, Paper 1, question 7 A camera manufacturer is evaluating a design for a light level indicator, details of which are shown below. For this circuit, determine the range of values of the input voltage Vin over which the LED will glow to indicate that a photograph may be taken. Show all calculations +5V 6k 4k +Vcc +Vcc -Vcc -Vcc Vin 4k 16k 0V
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