Laboratory Manual for
LINEAR ELECTRONICS-I
B. E.
SEM. III (EC)
January 2010
Faculty of TechnologyDharmsinh Desai University
Nadiad
TABLE OF CONTENTS
Sr. No. Title Page No.
1. Characteristics of PN Junction Diode. 4
2. Characteristics of Zener Diode. 6
3. Study of Clipper Circuit. 8
4. Study of Clamper Circuit. 9
5. Four-Diode Bridge Sampling Gate circuit. 11
6. Input and Output Characteristics of a CE Configuration. 13
7. Input and Output Characteristics of a CC Configuration. 17
8. Frequency Response of RC Coupled CE Amplifier. 21
9. Drain and Transfer Characteristics of FET. 23
10. Square Wave Response of CE Amplifier. 25
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LAB MANUAL
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LAB 1
Characteristics of PN Junction Diode.
AIM: To verify the characteristics of PN Junction diode under forward bias & reverse bias.
APPARATUS: Bread board, D.C. Power Supply, Digital Multimeter, Diode IN4001, Resistor 1KOhm.
THEORY:
The diode has a very important property, it permits only unidirectional conduction. It conducts well in the forward direction and poorly in the reverse direction. An ideal diode acts as a closed switch, when forward biased and as an open switch when reverse biased. In practical case the diode conducts when voltage across the diode is more than the barrier voltage i.e. 0.7 V for Silicon Diode and 0.3 V for Germanium diode and a current of few milli Ampere flows through the Diode.
In reverse bias a small current of a few microamperes flows through the diode. Care should be taken that the reverse voltage applied is not greater than the specified breakdown voltage otherwise the diode will get permanently damaged.
CIRCUIT DIAGRAM:
Fig 1. Forward Biased Diode
Fig2. Reverse Biased Diode
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PROCEDURE:1. For Forward Characteristics Connect the circuit as shown in Figure.12. Vary Vs in steps of 0.2 V up to 1.0 V.3. After 1 V go in steps of 1 V up to 10 V4. Measure the voltage drop across the diode (Vd) and Resistance (Vr) for each value
of Vs.5. Calculate the current through the diode by Vr/R. where Vr is the drop across the
resistance and R is the value of the resistance. 6. Note down the readings in the observation table.7. For Reverse Characteristics connect the circuit as shown in fig 2.8. Adjust the Reverse voltage to 20 V, Note down the reverse current.9. Plot the graph of Vs vs I
OBSERVATION TABLE:
Vs(Volts) Vd(Volts) I= VR/R(mA)
CONCLUSION:
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LAB 2
Characteristics of Zener Diode.
AIM: To verify the characteristics of Zener diode under reverse bias.
APPARATUS: Bread board, D.C. Power Supply, Digital Multimeters, Zener Diode, Resistor
THEORY:
Zener diodes are designed to operate in the reverse biased breakdown region without damage. By varying the doping levels it is possible to produce zener diodes with breakdown from about 2 V to 200V. In forward bias it acts a normal P –N junction diode. In the case of reverse bias large amount of current flows after the input voltage has increased its breakdown voltage.This is because of two effects (1) Zener Effect and (2) Avalanche effect. If the applied reverse voltage exceeds the breakdown voltage, a zener diode acts like a constant voltage source. The most common application of zener diode is voltage regulator.
CIRCUIT DIAGRAM:
Fig: Reverse Biased Zener Diode
PROCEDURE:1. Connect the circuit as shown in Figure for studying the reverse characteristics. 2. Increase Vs in steps of 2 V up to 20 V insteps of 2 V.3. Measure the current through the diode IZ and the voltage drop across the diode VZ.
4. Connect the circuit so that the Zener Diode is forward biased. Note down the readings of Vz and Iz by increasing supply voltage from 0. 1 to 10 V. from 0.1 to 1 V increase in steps of 0.2 V
5. Plot the V-I characteristics for forward and reverse bias and note down your conclusion.
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OBSERVATION TABLE:
Vs(Volts) VZ(Volts) IZ
CONCLUSION:
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LAB 3
Study of Clipper Circuit
AIM: To Observe the clipping waveform in different clipping configurations using diode.
APPARATUS: Bread Board, CRO, DC power supply, millimeters, resistors, diode.
THEORY:
Once the input voltage exceeds V1 the diode D1 starts conducting but diode D2 does not, since it is in reverse bias. As a result the output voltage does not go above V1. On the negative swing of input, the diode D2 conducts, but D1 does not, hence it clips the input voltage at negative side, so that the output does not exceed above V2.The clipping levels are adjusted by the voltages V1 and V2. The source resistance Rs should be much larger than the forward resistance of diode but should be smaller than the reverse resistance of diode.
CIRCUIT DIAGRAM:
Fig: Two voltage level clipper
PROCEDURE:1. Connect the circuit as shown in fig.2. Apply sine wave of 1 kHz signal from function generator.3. Observe the output waveform on CRO.4. Draw the input and output waveform on graph with scale.
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CONCLUSION:
LAB 4
Study of Clamper Circuit
AIM: To study the negative clamper circuit.
APPARATUS: Bread board, Diode IN4001, Resistor 10 K, Capacitor 1F, Signal generator (0-100kHz), CRO.
THEORY:
A simple RC coupling circuit blocks direct current but lets the alternating current pass unchanged. So to restore the DC component along with AC, we need a clamper circuit as shown in figure above. Under the ideal conditions, during positive cycle of the sinusoidal signal having peak amplitude Vm, the diode will conduct and hence the capacitor gets charged to the peak value of the input signal (i.e. Vm).When the input voltage becomes positive again, then the output voltage will be input voltage minus the voltage across the charged capacitor and hence in our case maximum output voltage during positive cycle will be (Vm – Vm) = 0. This means the diode will not conduct after the capacitor is fully charged. The charged capacitor has the same effect as a battery of voltage Vm in series with input.
Using the principle of superposition, the output voltage is the ac input voltage minus a dc value of Vm. The shape of the output voltage and the maximum negative output voltage depends on the time constant R1C1.If input frequency f = 10 kHz, then T = 1/f = 10-2 sec = R1C1
Assume C1 = 1µF then R1 = 10 KΩ
CIRCUIT DIAGRAM:
Fig: Negative clamper circuit
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PROCEDURE:1. Connect the circuit as per the circuit diagram on the bread board.2. Set the input signal voltage (say 5V, 10 kHz sine wave) using the signal generator.3. Observe the output waveform using CRO (DC-mode).4. Sketch the observed waveform with reference to the input waveform.5. Modify the input frequency (below and above 10 kHz) and see the effect on the
output waveform.6. Now connect the diode in reverse manner and repeat all steps from 1) to 5).
CONCLUSION:
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LAB 5
Four-Diode Bridge Sampling Gate circuit
AIM: To observe the output of a four-diode bridge sampling gate circuit.
APPARATUS: Bread board, two Signal generators, C.R.O., Diodes, Resistors.
THEORY:
An ideal sampling gate is a transmission circuit in which the output is an exact reproduction of an input waveform during a selected time interval and is zero otherwise. The time interval for transmission is selected by an externally impressed signal, called control or gating signal and is usually square wave with different duty cycle. These sampling gates are also referred to as transmission gates or time-selection circuits. A four diode sampling gate is shown in figure. In this circuit the sinusoidal signal is applied to the point where anode of D4 and cathode of D3 are joined and the control voltage is applied through the resistance R1 = R2. In most practical systems the period of control signal should be equal or be an integral multiple of input signal.
During the positive cycle of control voltage, all the four diodes conduct and hence the output voltage will be same as input signal. During the negative cycle of the control voltage, all the four diodes are non-conducting and hence the current through R3 is zero, so output voltage is zero. To ensure that all four diodes conduct during positive cycle of control voltage, the minimum value of control voltage should be equal to [2 + (R1/R3)] time signal voltage.
CIRCUIT DIAGRAM:
Fig: Sampling gate circuit
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PROCEDURE:1. Connect the circuit on bread board as shown in circuit diagram.2. Apply the sinusoidal signal of 1 kHz (say 3 V peak) to the terminals as shown in
figure.3. Apply the square wave of 1 kHz (having positive amplitude of three times peak
sinusoidal voltage) with proper duty cycle (say 70%) to the resistors R1 and R2 as shown in figure.
4. Observe the output waveform on C.R.O.5. Vary the duty cycle of the control voltage and see the effect on the output wavefrom.
OBSERVATION:
Plot the output waveform with reference to the input sinusoidal signal and control voltage.
CONCLUSION:
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LAB 6
Input and Output Characteristics of a CE Configuration
AIM: Plot the input and output characteristics of a CE configuration and to determine its h-parameters.
APPARATUS: Bread board,resistors, capacitors, transistor(BC 547), DC power supply(0-15 V), Signal generator(0-1Mhz), DMM, CRO.
THEORY:
Input characteristics:Input current IB is plotted as a function of VBE and VCE. The VCE is held constant and
then IB is plotted versus VBE. The model graph shows that the input characteristics are those of a forward biased PN junction. For a given value of VBE, IB reduced when higher VCE levels are employed. This is because higher VCE produces greater depletion region penetration into the base reducing the effective base region.Output characteristics:
IC is plotted as a function of IB and VCE. First VCE is held constant and IC is plotted versus IB. As IB is increased, IC increases linearly up to a certain level than IC gets saturated and it does not increase with increase in IB. The output chacteristics curves are not exactly horizontal because as VCE increases α increases and as a result β increases. So, for a constant value of IB as VCE increases IC increases slightly.
CIRCUIT DIAGRAM:
Fig: CE configuration
THEORY:
Input characteristics:Input current IB is plotted as a function of VBE and VCE. The VCE is held constant and
then IB is plotted versus VBE. The model graph shows that the input characteristics are those of a forward biased PN junction. For a given value of VBE, IB reduced when higher VCE levels
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are employed. This is because higher VCE produces greater depletion region penetration into the base reducing the effective base region.Output characteristics:
IC is plotted as a function of IB and VCE. First VCE is held constant and IC is plotted versus IB. As IB is increased, IC increases linearly up to a certain level than IC gets saturated and it does not increase with increase in IB. The output chacteristics curves are not exactly horizontal because as VCE increases α increases and as a result β increases. So, for a constant value of IB as VCE increases IC increases slightly.
PROCEDURE:
Input characteristics:1. Connect the circuit as per the circuit diagram.2. Set VCE at 5V, vary VBB in steps of 1V to 20V and note down the corresponding IB and
VBE. Repeat the above procedure for 10V, 15V etc.3. Plot the graph: VBE vs IB for a constant VCE.4. Find the h-parameters:
a) hie = input impedance of a transistor = ΔVBE/ΔIB|VCE constant. = (VBE2-VBE1)/(IB2-IB1) | VCE constant.
b) hre: reverse voltage gain hre = ΔVBE/ΔVCE|IB constant = (VBE2-VBE1)/(VCE2-VCE1) |IB
constant.
Output characteristics:1. Connect the circuit as per the circuit diagram.2. Set IB at 20 µA, vary VCC in steps of 1V and note down the corresponding IC and VCE.
Repeat the above procedure for 40µA, 60µA, 80µA.3. Plot the graph: VCE vs IC for a constant IB.4. Find the h-parameters:
a) hfe = forward current gain hfe = ΔIC/ΔIB| VCE constant = (IC2-IC1)/(IB2-IB1)| VCE
constantb) hoe: output admittance hoe = ΔIC/ΔVCE|IB constant = (IC2-IC1)/(VCE2-VCE1) |IB
constant.
OBSERVATION TABLE:
Input characteristics:
VCE(Volt) VBE(volt) IB(µA)
VCE1
VCE2
VCE3
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Output characteristics:
IB(µA) VCE(volt) IC(mA)
IB1
IB2
IB3
RESULT:
Parameters Practical readings
hfe
hie
hre
hoe
CONCLUSION:
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MODEL GRAPH:
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LAB 7
Input and Output Characteristics of a CC Configuration
AIM: To Plot the input and output characteristics of a CC configuration and to determine its h-parameters.
APPARATUS: Bread board, resistors, capacitors, transistor (BC 547), DC power supply (0-15 V), Signal generator(0-1Mhz), DMM, CRO.
THEORY:
Input characteristics:The common collector input characteristics as shown in model graph, are quite
different from either common-base or common emitter input characteristics. The difference is due to the fact that input voltage (VBC) is largely determined by the VEC level.
VEC = VEB + VBC Or VEB = VEC – VBC
Increasing the level of (input voltage ) VBC with VEC held constant , reduces the level of VEB, and thus reduces IB.Output characteristics:
To obtain the output characteristics IE is plotted vs. VCE for several fixed values of IB. Because IC is approximately equal to IE, the common-collector output and current gain characteristics are practically identical to those of the common-emitter.
CIRCUIT DIAGRAM:
Fig: CC configuration
PROCEDURE:
Input characteristics:1. Connect the circuit as per the circuit diagram.2. Set VEC at -5V(say), vary VBB in steps of -1V to -20V and note down the
corresponding IB and VBC. Repeat the above procedure for VEC -10V, -15V etc.
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3. Plot the graph: VBC vs IB for a constant VEC.4. Find the h-parameters:
a) hic = input impedence of transistor = ΔVBC/ΔIB|VEC constant. = (VBC2-VBC1)/(IB2-IB1) | VEC constant.
b) hrc: reverse voltage gain hrc = ΔVBC/ΔVEC|IB constant = (VBC2-VBC1)/(VCE2-VCE1) |IB
constant.Output characteristics:
1. Connect the circuit as per the circuit diagram.2. Set IB at 20 µA, vary VEE in steps of 1V and note down the corresponding IC and VCE.
Repeat the above procedure for 40µA, 60µA, 80µA.3. Plot the graph: VEC vs. IC for a constant IB.4. Find the h-parameters:5. Find the h-parameters:
a) hfc = forward current gain hfc = ΔIE/ΔIB| VEC constant = (IE2-IE1)/(IB2-IB1)| VCE
constantb) hoe: output admittance hoe = ΔIE/ΔVEC|IB constant = (IE2-IE1)/(VCE2-VCE1) |IB
constant.
OBSERVATION TABLE:
Input characteristics:
VEC(Volt) VBC(volt) IB(µA)
VEC1
VEC2
VEC3
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Output characteristics:
IB(µA) VEC(volt) IC(mA)
IB1
IB2
IB3
RESULT:
Parameters Practical readings
hfe
hie
hre
hoe
CONCLUSION:
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MODEL GRAPH:
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LAB 8
Frequency Response of RC Coupled CE Amplifier
AIM: To plot the frequency response of RC coupled CE amplifier and to determine the cut- off frequencies and bandwidth.
APPARATUS: Bread board, resistors, capacitors, transistor (2N2222A / BC547), D.C. poser supply( 0-15 V),Signal Generator (0-1MHz), C.R.O.
THEORY:
One of the most important functions of electronic circuit is amplification. Almost all electronic systems use amplifier. Generally people think that amplifier amplifies the signal of all frequencies by the same amount. But real fact is that at some low and high frequencies the gain provided by the amplifiers is less than the mid-band gain (maximum gain in between lower cutoff frequency and higher cut-off frequency). At low frequency the gain is reduced due to the coupling and bypass capacitors which are present in series state as we look from input to output (in worst condition act as open circuit at very low frequencies).At high frequency the gain is reduced due to the internal parasitic capacitance which are virtually present in parallel as we look from input to output (in worst condition act as short circuit at very high frequency). The frequencies, at lower side and higher side,
where the gain of the amplifier is 0.707 times of maximum mid-band gain is are known as lower cut-off frequency (fL) and higher cut-off frequency respectively (fH). The difference between fH and fL is known as bandwidth of the amplifier.
CIRCUIT DIAGRAM:
Fig: RC coupled single stage CE Amplifier.
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PROCEDURE:1. Connect the circuit on the bread board as shown in the figure2. Set input signal ac voltage to 20 mV (peak to peak) using the signal generator.3. Keeping the input voltage constant, vary the frequency from 0 Hz to 1 MHz in regular
steps and note down the corresponding output voltage.4. Plot the frequency response: Gain (dB) vs. Frequency (Hz) on semilog graph
paper.5. Calculate the bandwidth from the graph.
OBSERVATION TABLE:
vi (p-p) = 20mV
Frequency(Hz) vo(p-p) (V) Gain = vo/vi Gain(dB) = 20 log(vo/vi)
CONCLUSION:
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LAB 9
Drain and Transfer Characteristics of FET
AIM: To plot the drain for the given JFET and determine the drain resistance.
APPARATUS: FET (BFW10) , bread board, power supply, volt meter, variable resister, resistors.
THEORY:
Like BJT, FET is also three terminal semiconductor device and basically as an amplifier but unlike BJT, a FET is a unipolar device (i.e. the current is carried by only one type of charge carrier either electrons and holes). FET has very high input impedance compared to BJT. JFET can be a n channel FET and p channel FET. The important parameter of JFET are defined below :-Mutual Conductance g(m ) = Id / Vgs Vgs is const.Amplification factor µ = Vds / Vgs -- Id is constantrd = gm * µ
CIRCUIT DIAGRAM:
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Fig: CS configuration
PROCEDURE: 1. Note down the type number of JFET used.2. Connect the circuit as shown in fig.3. Fix VGS say zero, increase drain voltage VDS from zero and note down
corresponding values of ID.4. Repeat the same procedure for different value of VGS
5. Plot the graph of ID VGS for constant VGS.6. Calculate mutual conductance gm = ID /VGS for constant VDS from graph.7. Calculate Amplification factor µ= VDS /VGS for constant ID from graph.8. Finally, rd =µ/gm.
OBSERVATION TABLE:
VGS = -------
Sr. No. VDS (Volts) ID (mA)
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CHARACTERISTIC:
CONCLUSION: Make your own conclusion.
LAB 10
Square Wave Response of CE Amplifier
AIM: To plot the of square wave response of CE amplifier and determine the rise time and fall time.
APPARATUS: Transistor, Resistors, Capacitors, DC Power Supply, Function Generator, CRO, Bread Board, Connecting Wires.
THEORY:
A sense for the frequency response of an amplifier can be determined by applying a square wave signal to the amplifier and observing the output response. The shape of the output waveform will revel whether the high frequency or low frequencies are properly amplified.
If the response of an amplifier to an applied square wave is an undistorted replica of the input, the frequency response of the amplifier is obviously sufficient for the applied input. If the output waveform is distorted at the horizontal edge of the output waveform then the low
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frequencies are not being amplified properly and lower cut-off frequency is investigated and if the output waveform is distorted at vertical edges of the output waveform then the high frequencies are not being amplified properly.
The high frequency can be determined from the output waveform by carefully measuring the rise time defined between 10% and 90% of the peak value.
CIRCUIT DIAGRAM:
Fig: CE amplifier
PROCEDURE:1. Implement the components on the bread board as per circuit diagram.2. Apply square wave input of 50KHz. from the function generator.3. Observe the output on the CRO.4. Measure the rise time of output square waveform.5. Also measure the amplitude at the trailing edge and leading edge of the out put
square waveform and calculate percentage of tilt.
OBSERVATION TABLE:
Plot the square wave response and determine the rise time and tilt from the graph.
CONCLUSION: Make your own conclusion.
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Datasheet of BJT BC 547
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BC546/547/548/549/550
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Symbol Parameter Value Units VCBO Collector-Base Voltage : BC546 : BC547/550 :
BC548/549 80 50 30 V V V
VCEO Collector-Emitter Voltage : BC546 : BC547/550 : BC548/549
65 45 30 V V V
VEBO Emitter-Base Voltage : BC546/547 : BC548/549/550 6 5 V V
IC Collector Current (DC) 100 mA
PC Collector Power Dissipation 500 mW
TJ Junction Temperature 150 °C
TSTG Storage Temperature -65 ~ 150 °C
Symbol Parameter Test Condition Min. Typ. Max. Unit
s
ICBO Collector Cut-off Current VCB=30V, IE=0 15 nA
hFE DC Current Gain VCE=5V, IC=2mA 110 800
VCE
(sat) Collector-Emitter Saturation Voltage IC=10mA, IB=0.5mA 90 250 mV
IC=100mA, IB=5mA 200 600 mV
VBE
(sat) Base-Emitter Saturation Voltage IC=10mA, IB=0.5mA 700 mV
IC=100mA, IB=5mA 900 mV
VBE (on) Base-Emitter On Voltage VCE=5V, IC=2mA VCE=5V, IC=10mA
580 660 700 720
mV mV
fT Current Gain Bandwidth Product
VCE=5V, IC=10mA, f=100MHz
300 MHz
Cob Output Capacitance VCB=10V, IE=0, f=1MHz 3.5 6 pF
Cib Input Capacitance VEB=0.5V, IC=0, f=1MHz 9 pF
NF Noise Figure : BC546/547/548 VCE=5V, IC=200µA 2 10 dB : BC549/550 : BC549 f=1KHz, RG=2KΩ VCE=5V, 1.2 4 4 dB
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hFE Classification
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• General purpose switching and amplification.
DESCRIPTION
NPN transistor in a TO-92; SOT54 plastic package. PNP complements: BC556 and BC557.
PIN DESCRIPTION
1 emitter
2 base
3 collector
• Low current (max. 100 mA)
• Low voltage (max. 65 V).
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FEATURES PINNING
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APPLICATIONS
LIMITING VALUES
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SYMBOL
PARAMETER CONDITIONS MIN. MAX. UNIT
VCBO collector-base voltage open emitter
BC546 − 80 V
BC547 − 50 V
VCEO collector-emitter voltage open base
BC546 − 65 V
BC547 − 45 V
VEBO emitter-base voltage open collector
BC546 − 6 V
BC547 − 6 V
IC collector current (DC) − 100 mA
ICM peak collector current − 200 mA
IBM peak base current − 200 mA
Ptot total power dissipation Tamb ≤ 25 °C; note 1 − 500 mW
Tstg storage temperature −65 +150 °C
Tj junction temperature − 150 °C
Tamb operating ambient temperature −65 +150 °C
SYMBOL PARAMETER CONDITIONS VALUE UNIT
Rth j-a thermal resistance from junction to ambient
note 1 0.25 K/mW
SYMBOL PARAMETER CONDITIONS MIN. TYP. MAX.
UNIT
ICBO collector cut-off current IE = 0; VCB =30V − − 15 nA IE = 0; VCB = 30 V; Tj = 150 °C
− − 5 µA
IEBO emitter cut-off current IC = 0; VEB =5V − − 100 nA hFE
DC current gain BC546A IC =10 µA; VCE =5V; see Figs 2, 3 and 4
− 90 − BC546B; BC547B − 150 −
BC547C − 270 −
DC current gain IC = 2 mA; VCE =5V;
BC546A see Figs 2, 3 and 4 110 180 220
BC546B; BC547B 200 290 450
BC547C 420 520 800
BC547 110 − 800
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In accordance with the Absolute Maximum Rating System (IEC 134).
Note
1. Transistor mounted on an FR4 printed-circuit board.
THERMAL CHARACTERISTICS
1. Transistor mounted on an FR4 printed-circuit board.
CHARACTERISTICS
Tj=25 °C unless otherwise specified.
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BC546 110 − 450
VCEsat collector-emitter saturation IC = 10 mA; IB = 0.5 mA − 90 250 mV voltage IC = 100 mA; IB =5mA − 200 600 mV
VBEsat base-emitter saturation voltage
IC = 10 mA; IB = 0.5 mA; note 1
− 700 − mV
IC = 100 mA; IB = 5 mA; note 1
− 900 − mV
VBE base-emitter voltage IC = 2 mA; VCE = 5 V; note 2 580 660 700 mV
IC = 10 mA; VCE =5V − − 770 mV
Cc collector capacitance IE =ie = 0; VCB = 10 V; f = 1 MHz
− 1.5 − pF
Ce emitter capacitance IC =ic = 0; VEB = 0.5 V; f = 1 MHz
− 11 − pF
fT transition frequency IC = 10mA; VCE = 5 V; f = 100 MHz
100 − − MHz
F noise figure IC = 200 µA; VCE =5V; RS
=2kΩ; f = 1 kHz; B = 200 Hz − 2 10 dB
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Notes
1. VBEsat decreases by about 1.7 mV/K with increasing temperature. 2. VBE decreases by about 2 mV/K with increasing temperature.
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103
IC (mA) BC546B; BC547B.
Fig.3 DC current gain; typical values.
hFE
200
100
50
0 10−2 10−1 1 10 102 103
IC (mA)
BC546A.
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Fig.2 DC current gain; typical values.
100
200
300
0 10−2 10−1 1 10 102
hFE
400
200
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