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BJT single stage amplifier designing

FET single stage amplifier designing

BJT multistage amplifier designing

Power amplifier designing

Solved problems

2 stage BJT amplifier design

Power amplifier design

[Assume other required data]

mod(Av) = (hfe * RL')/(hie + (ðh * RL') )

Where ðh = ((hie * hoe) - (hfe * hre)) [Get hfe, hre, hie,hoe from data sheet]

** ðh = 0.069**

**Hence**

**75 = (330 * Rc)/(4500 + (0.069 * Rc)**

**Hence**

**4.331 * Rc = 4500**

**Hence Rc = 1039.02 ohm**

**We use HSV for better gain**

**Hence Rc = 1.2 Kohm**

Ic peak = Vo peak / RL'
**Ic peak = 5.66/Rc = 4.714 mA**

Icq = Ic peak + Ic min

**Assume Ic min = 0**
**Hence Icq = 4.714 mA**

Selection of Re

Re = Vre/Icq
**Hence Re = 1/(4.714 * (10 ^ -3)) = 318 ohm**

Select lower std value of Re so that voltage drop across Re is less which increases the voltage swing of o/p

**Hence Re = 270ohm**

Select higher std val

**Hence Vcc = 18V**

s = (1 + hfe max)/(1 + ((hfe max * Re )/(Rb + Re))

**Hence 10 = 1 + 450)/(1 + ((450 * 270)/(Rb + 270))**

**Hence Rb = 2485.1 V**

**Vr2 = Vbe + Vre = 07 + (4.714 * 0.27) = 1.87 V**

**Vr1 = Vcc - Vr2 = 16.02 V**

**Assume Vbe = 0.6V**

**R1/R2 = Vr1/Vr2 .............(A)**
**[Get R1 in terms of R2 & substitute in Rb]**

**Hence R1 = 8.124 * R2**

**Rb = R1 parallel R2 = (R1 * R2)/(R1 + R2) = (8.124 * R2)/9.124 =
0.89 * R2**

**Hence 2.485.1 = 0.89 * R2**

**Hence R2 = 2790.99 ohm**
**Select lower standard value to make circuit indepent of beta**

**Hence R2 = 2.7 K ohm**

Substitute in (A) to find R1

**R1 = 8.124 * 2700 = 21934.8 K ohm**

Select higher standard value so that circuit draws minimum current
from supply

**Hence R1 = 22 K ohm**

FL = 25 Hz

**Ce = 1/(2*pi * FL * Xce) = 235.78 µ F**

__Hence Ce = 270 µ F | 50 V__

Xcb = ((Rb) parallel (hie))

**Cb = 1/(2 * pi * FL * Xcb) = 4.06 µ F**

**Hence Cb = 4.2 µ F | 50 V**

Xcc = Rc + RL [If RL[load resistance] is not specified thenassume amplifier is connected to a similar next stage. Hence RL = (Rb)parallel (hie)]

**Cc = 1/(2 * pi * FL * Xcc) = 5.3 µ F**

**Hence Cc = 5.6 µ F | 50 V**

[Draw the figure with designed values. Do all this in 30 minutes (1.8
min per mark)]

1: Design for device
parameter variations

2: Design for midpoint biasing

3: Design for Zero thermal drift

4: Graphical methord

[Do not write text included in square bracket ]

Vgs = Vp * (1 - sqrt(Id/Idss))

Vgs max = - 1.4 V, Vgs min = - 0.29 V,

__Vgsq = (Vgs max + Vgs min)/2 = -0.845 V__

Select higher std value

Hence __Rs = 620 ohm__

Gain of JFET amplifier, mod(Av) = gm * RL'

Hence RL' = 2719 ohm

Assume RL = infinity

RL' = (rd) parallel (Rd)

Hence Rd = 2.87 K ohm

Select higher standard value

Hence __Rd = 3.3 K ohm__

Select higher std value

Also

Vg = Vdd * R2/(R1 + R2)

Hence R1 = 0.47 * R2..................(A)

__Assume R2 = 1 M ohm__

Substitute in (A)

__Hence R1 = 470 Kohm__

Xcg = (R1) parallel (R2) = 319.72 ohm

Cg = 1/(2 * pi * FL * Xcg) = 0.0248 µ F

Select higher standard value
__Hence Cg = 0.027 µ F | 25 V__

Assume RL = Ri = (R1) parallel (R2)

Cd = 1/(2 * pi * FL * Xcd)

Cd = 0.0248 µ F

Select higher standard value

__Hence Cd = 0.027 µ F | 25 V__

Cs = 1/(2 * pi * FL * Xcs) = 39.17 µ F

Select HSV
__Hence Cs = 42 µ F | 25 V__

[Draw the figure with designed values]

We use only typical values for midpoint biasing

[Draw the figure]

Hence
__Vgsq = Vp typ * (1 - sqrt(Idq/Idss typ)) = 0.75 V__

Select HSV

__Hence Rs = 220 ohm__

Gain of JFET amplifier, mod(Av) = gm * RL'

RL is not given

Hence assume RL = infinity

RL' = (rd) parallel (Rd)

Rd = 3.03 Kohm

Select HSV
__Hence Rd = 3.3 ohm__

Select higher std value

__Hence Vdd = 20 V__

Cg = 1/(2 * pi * FL * Xcg) = 0.0159 µF

Select HSV

Hence__ Cg = 0.018 µF | 50 V__

If RL is not specified, assume RL = Ri = Rg

Hence Cd = 0.001 µF

Select HSV

Hence __Cd = 0.01 µF | 50 V__

Cg = 1/(2 * pi * FL * Xcs) = 76.89 µF

Select HSV

Hence __Cg = 81 µF | 50 V__

[Draw the circuit with designed values]

We use self bias circuit

[Draw the figure]

Hence mod(Vgsq) = 1.87 V

__Hence Vgsq = -1.87 V__

hence

Rs = mod(Vgsq/Idq) = 4.206 Kohm

Select nearest std value

Gain of JFET amplifier, mod(Av) = gm * RL'

RL is not given

assume RL = infinity

RL' = (rd) parallel (Rd)

Hence Rd = 8.33 Kohm

Select higher standard value
__Hence Rd = 9.1 K ohm__

Select higher std value

Xcg = (Rg)

Cg = 1/(2 * pi * FL * Xcg) = 0.0079 µF

Select higher std value

__Hence Cg = 0.01 µF | 25 V__

If RL is not specified, assume RL = Rg

Hence Cd = 0.0079 µF

Select higher std val
__Hence Cd = 0.01 µF | 25 V__

Cg = 1/(2 * pi * FL * Xcs) = 11.91 µF

Select HSV

__Hence Cg = 12 µF | 25 V__

[Draw the circuit with designed values]

[In graphical methord draw the graph of Ids against Vds [Values given in data sheet.You will be given the value/s of or range of Ids (2 values(max or min) or range of values for device parameter variation & single value(typ) for other methods)]

[Plot the required value/s of Vgs & find Vgsq
& continue with the usual method. The answers in the 2 methods will
differ a lot for the same problem. For device parameter variation use max
& min curve to calculate Vgs max & min resp. For other methods
use typ curve unless mentioned otherwise ]

Av = Av1 * Av2

(Av1/Av2) = (Ro1/Ro2)

Assume (Ro1/R02) = 0.5

Hence Av2 = 2* Av1

Hence Av = 2 * ((Av1) ^ 2)

Hence 1000 = 2 * ((Av1) ^ 2)

Hence Av1 = 31.62

&Av2 = 63.24

We use BC 147A

63.24 = (220 * Rc2)/2700

Hence Rc2 = 776.18

Take higher std value

__Hence Rc2 = 820 ohm__

Hence Vceq2 = 4.617 ohm

Icq2 = Ic2peak + Ic2min

Assume Ic2min = 0

Ic2 peak = (Vopeak/Rc2) = 3.449 mA

__Hence Icq2 = 3.449 mA__

Hence Re2 = Vre2/Icq2

Hence Re2 = 289 ohm

Take LSV

Hence Re2 = 270 ohm

Hence Vcc = 10.445 V

__Take Vcc = 12 V__

Assume s = 8

Hence 8 = (1 + hfe)/(1 + ((hfe * Re2)/(Rb2 + Re2)))

[Substitute hfe, Re2 & find Rb2]

Hence Rb2 = 1971.13 ohm [Do not standardise]

VR4 = Vbe + (Icq2 * Re2)

Hence VR4 = 1.63 V

VR3 = Vcc - VR4

Hence VR3 = 10.368 V

(VR3/VR4) = (R3/R4)

Hence (R3/R4) = 6.356

Now Rb2 = 1971.13 = (R3 * R4)/(R3 + R4) = (6.356 * R4)/7.356

Hence R4 = Rb/0.864 = 2281.22 ohm

Take lower standard value

__Hence R4 = 2.2 Kohm__

Hence R3 = 6.642 * R4 = 13917.2 ohm

Select higher standard value

__Hence R3 = 15 K ohm__

Av1 = 2000/Av2 = 29.93

| Av1 | = (hfe typ * RL1)/hie

Hence RL1 = (Rc1) parallel (Zin2)

Where Zin2 = (R3) parallel (R4) parallel (hie) = 1121.6 ohm

Hence Rc1 = 546.21 ohm

Taking higher std value

__Hence Rc1 = 560 ohm__

__Let Vceq1 = Vceq2 = 4.617 V____ Vrc1 = Vrc2 = 2.828 V____ Vre1 = Vre2 = 0.931 V____ Icq1 = Vrc1/Rc1 = 5.05 mA____ Re1 = Vre1/Icq1 = 184 ohm__

__R2 = R4 = 2.2 K ohm__

Ce1 = 1/(2*pi * FL * Xce1) = 106.1 µF

Select higher standard value

__Hence Ce1 = 120 µF__

Ce2 = 1/(2*pi * FL * Xce2) = 58 µF

Taking higher standard value

__Hence Ce2 = 62 µF__

Cb1 = 1/(2 * pi * FL * Xcb1) = 1.418 µF

Taking higher std value

__Hence Cb1 = 1.5 µF__

Xcb2 = Rc1 + ((Rb2) parallel (hie))

Cb2 = 1/(2 * pi * FL * Xcb) = 4.261 µF

Taking higher std values

__Hence Cb2 = 4.7 µF__

Xco = Rc + RL

Co = 1/(2 * pi * FL * Xco) = 0.819 µF

Taking higher std value

__HenceCo = 1 µF__

[Draw the figure with designed values. Do all this in about 36 - 40 minutes (1.8minper mark)]

Class A amplifier

Class B amplifier

Do not write text included in square bracket

Power transmitted to load, PL' = PL/nt = 5.555W
[nt = efficiency of transformer]

Assume nT = 90% or 0.9

Q = (Pq max)/(PL') = 2

Hence Pq max = 11.111 W

Select transistor with Pd > 2 * Pq max

Select ECN 149 with Pd max = 30 W

Selection of operating point

__Vceq = Vcc - Vre = 10.8 V__

Vce peak = Vceq - Vce sat = 9.8V

Ic peak = (2 * PL')/Vce peak = 1.134 A

Icq = Ic peak + Icmin

Assume Ic min = 0
__Hence Icq = 1.134 A__

Pre = sqr(Vre)/Re = 1.44 W
__Select Re = 1 ohm | 3 W__

__Ce = 1/(2 * pi * FL * RL) = 7957.74 µF__

__Since Ce is very high we leave Re unbypassed__

s = (1 + hfe max)/(1 + ((hfe max * Re )/(Rb + Re))

We have Rb = 9.89 ohm

Vr2 = Vbe + (Icq * Re) = 1.834 ohm

Vr1 = Vcc - Vr2 = 10.166 ohm

R1/R2 = Vr1/Vr2 = 5.543

Hence R1 = 5.543 * R2 .............(A)

Rb = R1 parallel R2 = (R1 * R2)/(R1 + R2) = (5.543 * R2)/6.543

Hence R2 = 11.67 ohm

Select lower standard value to make circuit indepent of beta

PR2 = ((VR2) ^ 2)/R2 = 0.336 W

__Hence R2 = 10 ohm | 0.75 W__

Substituting in (A) we get

R1 = 55.43 ohm

Select higher std value

Hence R1 = 56 ohm

PR1 = ((VR1) ^ 2)/R1 = 1.86 W

__Select R1 = 56 ohm | 3.75 W__

RL' = (sqr(N1/N2)) * RL

Hence (N1/N2) = 1.697

Select audio frequency transformer with turns ratio 1: 1.697

PL' FL = (Vce peak * Ic peak)/2 = 5.556 W

Pi dc = (Vcc * Icq) + (Vcc ^ 2)/(R1 + R2) = 15.79 W
**Hence n FL = 0.3518 or 35.18 %**

For a class A amp max power dissipation occurs under no signal condition

__Hence PD no signal = 30 W__

assume nT = 90% or 0.9

Q = (Pq max)/PL') = 1/5

PL = (I^2)/RL = 49/3 = 16.33W
PL' =PL/nT = 18.148 W

Hence Pqmax = 3.629 W

Select transistor with Pd > 2 * Pq max

Select transistor ECN 149 with Pdmax = 30W

(Vceo/2) <= Vcc <= Vceo

Select Vcc = 25 V

Vre = Vcc/10

Hence Vre = 2.5 V

__Vceq = Vcc - Vre = 25 - 2.5__

Hence Vceq = 22.5 V

Vce peak = Vceq - Vce sat = 21.5 V

Ic peak = (2 * PL')/Vce peak = 2.519 A

Assume Ic min = 0

Icq = Icpeak + Icmin = __2.519A__

Idc full wave = (2 * Idc peak)/pi = 0.967 A

Idc half wave = Idc peak/pi = 0.483 A

Pre = sqr(Vre)/Re = 1.225 W
__Select Re = 5.1 ohm |3W__

Ce = 1/(2 * pi * FL * RL)

Ce is too high hence leave Re unbypassed

s = (1 + hfe max)/(1 + ((hfe max * Re )/(Rb + Re))

Rb = 50.44 ohm [Show the calculations]

Vr2 = Vbe + (Idc half wave * Re) = 3.163 V

Vr1 = Vcc - Vr2 = 21.83 V

Assume Vbe = 0.6V

R1/R2 = Vr1/Vr2 = 6.903 .............(A)

Rb = R1 parallel R2 = (R1 * R2)/(R1 + R2)

Hence R2 = 57.747 ohm = 56 (LSV)

PR2 = (Vr1 ^ 2)/R1 = 0.178 W

Select lower standard value to make circuit indepent of beta
__Hence R2 = 56 ohm | 0.5 W__

Substitute in (A) to find R1 R1 = 386.56 ohm

PR1 = 1.22 W

Select higher standard value so that circuit draws minimum current from
supply
__Hence R1 = 390 ohm| 3 W__

RL' = (sqr(N1/N2)) * RL

Hence (N1/N2) = 2.172

Hence N1:N2 = 2.172 : 1

Power rating of primary > PL'

ie power rating of primary > 18.148 ohm

Select audio frequency transformer with turns ratio 1:(N1/N2)

Select centre tap transformer with turns ratio 2.172 : 2.172 : 1 &
power rating of 25 W

[Draw the circuit diagram with calculated values]