should be performed at least 5 exercises for each topic
NOTE: If not stated explicitly component values \u200b\u200bare indicated by the value of their subscript.
Continue
Fig.1: Knowing that Vg = 120V find the current (Rtot = (r3 + r4) / / r2 + r3 Itot and I find the current divider are the other or use Millman )
Fig.2: Knowing that Vg = 220V find the current (use Millman)
Fig.3: Knowing that Vg = 220V find the current (r = (r5 + r6) / / r2 and then use Millman)
4: Knowing that Vg = 120V 2A and find the current Ig = (Millman)
5: Knowing that Vg = 120V find the current (r = r2 + r5 / / (r3 + r4) and Millman)
Fig.6: Knowing that VG1 = 120V, 130V and VG4 = Ig = 3A find Current
7: Knowing Vg = Ig = 120V and 1A find the current
Fig.8: Knowing that VG4 = VG1 = 120V and 100V find the current
9: Knowing and Vg2 VG1 = 100V = 50V find the current
Fig.10: Knowing and Ig = VG1 = 100V 4A find the current
AC
NB component values \u200b\u200band frequency can be chosen at will.
For all eleven circuits, Vi = 120V and the currents are determined in the various branches.
NB The procedure must be sgeuente:
1) Sketch the physical circuit
2) Calculate the reactance
3) Draw the circuit symbol with the impedance (complex numbers)
4) Determine the voltage and current (complex numbers)
Transfer functions
Fdt1: Calculate literally the parallel between a capacitor and a resistance
Fdt2: Calculate literally series between a capacitor and a resistance
Fdt3: Calculate the expression literally: z1 / / z2 + z3 = 1/Cs where z1, z2 and z3 = R = R + Ls
Fdt4: Calculate the expression literally: z1 + z2 / / z3 = 1/Cs where z1, z2 and z3 = R + Ls = R
Fdt5: Calculate the expression literally: (z1 + z2) / / z3 = 1/Cs where z1, z2 and z3 = R + Ls = R
Fdt6: Calculate the expression literally: z1 / (z1 + z2) where z1 and z2 = 1/Cs = R + Ls
Fdt7: Calculate literally the expression 1 / (+ 1/z1 1/z2) where z1 and z2 = 1/Cs = R + Ls
Fdt8: Calculate literally the expression: (1/z2) / (+ 1/z1 1/z2) where z1 and z2 = 1/Cs = R + Ls
NB
To determine the frequency response through the FDT is necessary: \u200b\u200b1
) Draw the physical circuit
2) Draw the circuit symbol in s (complex frequency)
3) Calculate Vo date Vg
4) Calculate A (s) = Vo / Vg
5) By calculating sine wave A ( jw) in magnitude and phase
6) Replace the numeric values \u200b\u200b(R, L and C)
7) plotted by varying the value of w from zero onwards
Fdt9: For all eleven circuits figure should be determined in the function Transfer Vu / Vi in literal form
Bode amplitude
Bode1:
Draw the function A (s) = [s +1000.1 s ^ 2 +100] / [(s + 1) (s +10000)]
Draw the function A (s) = [1000s +100] / [(s +1) (s +10000)]
Draw the function A (s) = s [1000s + 100] / [(s +1) (s +10000)]
Draw the function A (s) = s [1000s +1] / [(50s +1) (5s +10000)]
the A Draw (s) = [s ^ 2 +100 s +10000] / [(s +1) (s +10000)]
Draw the function A (s) = [s ^ 2 +100 s +10000] / [s ^ 2 +500 s +100]
BJT using the polarization characteristics
This is the characteristic input
This is the output characteristic
1) Draw the circuit biasing BJT with two generators (Vcc = 12V, Vbb = 1.5 V) and two resistance (Rb = 100K, Rc = 2K) and using the two features in the figure to find the resting point of entry (IBQ, Vbeq) and output (ICQ, Vceq)
Procedure: Plot the input load (pass for Vbb / Vbb and Rb) and output (pass for DC / DC and RC) are the points of intersection are the values.
2) Repeat the same exercise only doubling Rb
3) Repeat doubling Rc
4) Add Re = 0.33K
procedure: the only point where they pass through the lines (Vcc / (Rc + Re) and Vbb / (Rb + BRE)). The problem is to calculate the B and that you are looking for a mid-point in the output characteristics and calculating the ratio Icq / IBQ
5) Repeat with Rb = 10K, Re = 0K and say if the BJT is in saturation.
6) Repeat with Rc = 10K, Re = 0K and say if the BJT is in saturation.
7) Repeat with Rb = 10K, Re = 0.56K and say if the BJT is in saturation.
8) Repeat with Rc = 10K, Re = 0.22K and say if the BJT is in saturation.
BJT polarization without the use of characteristics and with the sole knowledge of hFE or beta using a circuit with two resistors.
It is suggested that you use a circuit with Vbb, Vcc, Rb and Rc and the value of beta = 200.
For the verification part of the Vbb, Vcc, Rc and Rb and the process the council is this:
assumed in the BJT region Ib active
calculation equation of the mesh input
calculating the beta Ic Vce
calculation equation of the mesh output
condition occurs if the active region
occurred hypothesize the saturation region
calculation Ib equation mesh input
Ic calculation equation of the mesh output
condition occurs saturation region
1) Vbb = 2V, Vcc = 15V, Rc and Rb = 47K = 1K
2) Vbb = 3V, Vcc = 15V, Rb = Rc = 2.2K and 47K
3) Vbb = 1V, Vcc = 15V, Rb = Rc = 2.2K and 47K
4) Vbb = 3V, Vcc = 15V, Rc = 4.7 K and Rb = 68K
for the project from Ic, Ib (Or beta), Vbe, Vce, Vbb, Vcc and this procedure is recommended:
calculation of the equation Rb mesh input
calculation of the equation Rc Mesh
found out the commercial value is passed to verification of the rest point by calculating
Ib Vbe, Vce and Ic
1) Vbb = 2V, Vcc = 15V, Ib = 15ua, Ic = 2mA, Vce = 6V, Vbe = 0.7V
2) Vbb = 3V, Vcc = 12V, Ib = 25uA, IC = 3mA, Vce = 8V, Vbe = 0.7V
3) Vbb = 2V, Vcc = 10V, Ib = 45ua, Ic = 3mA, VCE = 2V, Vbe = 0.7V
4) Vbb = 2V, Vcc = 10V, Ib = 25uA, IC = 2mA, Vce = 4V, Vbe = 0.7V
polarization without the use of the BJT features and with only the knowledge of hFE or beta using a circuit with 3 resistors.
polrizzazione In the circuit with three resistors of the procedure is similar
only changing the loop equations and the project must be added the possibility of project
Re Ic is about 10% of Vcc.
Check or analysis
1) Vbb = 2V, Vcc = 15V, 1K = Rc, Re and Rb = 47K = 0.33K
2) Vbb = 3V, Vcc = 15V, Rc = 2.2K, D = 0.27 K and Rb = 47K
3) Vbb = 1V, Vcc = 15V, Rc = 2.2K, D = 0.68K and Rb = 47K
4) Vbb = 3V, Vcc = 15V, Rc = 4.7K, D = 0.56 K and Rb = 68K
Project:
1) Vbb = 2V, Vcc = 15V, Ib = 15ua, Ic = 2mA, Vce = 6V, Vbe = 0.7V
2) Vbb = 3V, Vcc = 12V, Ib = 25uA, IC = 3mA, Vce = 8V, Vbe = 0.7V
3) Vbb = 2V, Vcc = 10V, Ib = 45ua, Ic = 3mA, VCE = 2V, Vbe = 0.7V
4) Vbb = 2V, Vcc = 10V, Ib = 25uA, IC = 2mA, Vce = 4V, Vbe = 0.7V
BJT polarization without the use of characteristics and with the sole knowledge of hFE or beta using a circuit with 4 resistors.
In polrizzazione circuit with four resistors
a similar process is only changing the loop equations and the project must be added
the idea of \u200b\u200bthe value of the divider input.
Verification or analysis
1) Vcc = 15V, Rc = 1K, Re = 0.33K, R1 = 47K and R2 = 5.6K
2) Vcc = 15V, Rc = 2.2K, D = 0.27K, R1 = 33K and R2 = 6.8K
3) Vcc = 18V, Rc = 2.2K, D = 0.68K, R1 = 33K and R2 = 8.2K
4) Vcc = 18V, Rc = 4.7K, D = 0.56K, R1 = 33K and R2 = 6.8K
Project:
1) Vcc = 15V, Ib = 15ua, Ic = 2mA, Vce = 6V, Vbe = 0.7V
2) Vcc = 12V, Ib = 25uA, IC = 3mA, Vce = 8V, Vbe = 0.7V
3) Vcc = 10V, Ib = 45ua, Ic = 3mA, VCE = 2V, Vbe = 0.7V
4) Vcc = 10V, Ib = 25uA, IC = 2mA, Vce = 4V, Vbe = 0.7V
EC Amplifier
1) Draw a complete CE amplifier stage, calculate Ai, r'i, Av, Ri, Avs alpha and knowing that Rs = 1K, R1 = 47K, R2 = 5.6K, D = 0.39K, Rc = 2.2K, hie = 1K, hfe = 220, RL = 5k
2) Draw a complete CE amplifier stage, calculate Ai, r'i, Av, Ri, Avs alpha and knowing that Rs = 1K, R1 = 33K, R2 = 5.6K, D = 0.39K, Rc = 2.2K, hie = 1K, hfe = 220, RL = 5k
3) Draw a complete CE amplifier stage, calculate Ai, r'i, Av, Ri, Avs alpha and knowing that Rs = 1K , R1 = 47K, R2 = 6.8K, D = 0.39K, Rc = 2.2K, 2K = hie, hfe = 180, RL = 10k
4) Draw a complete CE amplifier stage, calculate Ai, r'i, Av , Ri, Avs alpha and knowing that Rs = 1K R1 = 47K, R2 = 5.6K, D = 0.68K, Rc = 4.7K, 2K = hie, and hfe = 200 RL = 5k
5) Draw a complete CE amplifier stage, calculate Ai, r'i, Av, Ri, Avs alpha and knowing that Rs = 1K, R1 = 33K, R2 = 5.6K, D = 0.56K, Rc = 3.9K, 1.5K = hie, hfe = 220, RL = 5k
Multistage
To solve the exercises you need:
1) Draw the complete multistage
2) To analyze the static part of each stage (draft or check)
3) Draw the dynamic circuit of each stage
4) Determine the transfer functions starting from 'last stage
Mult1:
EC + CC
You know the rest point CE (Ic = 3mA, VCE = 5V), while in DC should be determined with R3 = R4 = 33K = 1K and RE2. We know that Rs = 1K RL = 5.6K, hie = 1K, hfe = 220 (identical in both stages)
Mult2:
EC + DC
You know the rest point of EC (Ic = 3mA, VCE = 5V ) as necessary to define, DC = 33K R3 = R4 = 1K and RE2. We know that Rs = 1K RL = 5.6K, hie = 1K, hfe = 220 (identical in both stages) and the output is the collector.
Mult3:
EC + DC
You know the rest point of EC (Ic = 3mA, VCE = 5V), while in DC should be determined with R3 = R4 = 33K = 1K and RE2. We know that Rs = 1K RL = 5.6K, hie = 1K, hfe = 220 (identical in both stages) and the output is nell'emettitore.
Op
1) Designing a reversing OA to amplify -5
2) Draw a noninverting OA to amplify 3
3) Design an adder inverting -5 to amplify two signals of amplitude 0.1 V and 0.15 V entry to
4) Draw a sottratore that makes a difference of two signals of amplitude 1 V and 0.55 V at places
5) Draw a non-inverting adder 2 to amplify the two signals of amplitude 0.1 V and 0.15 V entry to
6) Draw an open loop comparator with OA and the output obtained by placing two input signals sinusiodali with the same frequency but phase and amplitude of 2V and 5V
7) Draw the circuit of a Schmitt trigger with OA and the output obtained by placing two input signals sinusiodali with the same phase and frequency but amplitude 2V and 5V
Operational and filters
In circuits with OA:
- z1 connects the generator or the mass with the inverting terminal
- z2 connects the inverting output terminal
- z3 connects the generator or the mass with the non-inverting terminal
- z4 connecting the mass with the non-inverting terminal
- Given an inverting amplifier with z1 and z2 = 10R = R +1 / (Cs) Av calculate and draw its graph with Bode knowing that R = 1K and C = 10nF
- Given an inverting amplifier with z1 = R +1 / (Cs) and z2 = 10R +1 / (Cs) Av calculate and draw its graph with Bode knowing that R = 1K and C = 10nF
- Given an inverting amplifier with z1 and z2 = R + Ls = 10R +1 / (Cs) Av calculate and draw its graph with Bode knowing that R = 1K, L = 10mH and C = 10nF
- Given an inverting amplifier with z1 and z2 = R = 9R / / (R +1 / (Cs)) Av calculate and draw its graph with Bode knowing that R = 1K and C = 10nF
- Given a non-inverting amplifier with R = z1, z2 = 10R, = R z3 and z4 = 1 / (Cs) Av calculate and draw its graph with Bode knowing that R = 1K and C = 10nF
- Given a non-inverting amplifier with R = z1, z2 = 10R, z3 = 1 / (Cs) and z4 = R Av calculate and draw its graph with Bode knowing that R = 1K and C = 10nF
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