Amplifier Design(Sem 3 &4 Mum university)

BJT single stage amplifier designing
FET single stage amplifier designing
BJT multistage amplifier designing
Power amplifier designing
Solved problems

   These pages are meant  mainly for engineering students studying in second year( Mumbai university only). If you are from some other university then confirm with your professors whether these formulae/methods are valid for your university.

Designing of single stage BJT audio frequency amplifier(CEamp)

  [Do not write text included in square bracket]

Step 1: Circuit diagram

  [We use voltage divider circuit as it provides Q point independent of temperature & beta]
  [Draw the figure]


Step 2: Selection of Rc

RL' = Rc parallel RL    [RL = load resistance connected between Vo & ground. Assume zero if not given]

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

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

[if min voltage gain is specified, use hfe min.  If some specific voltagegain is specified, use hfe typ]

Calculate RL' & Rc
Select higher std value for Rc to increase voltage gain [if min voltage gain is specified or nothing is specified. If max voltage gain                                                                                   is specified use lower std value. If some specific voltage gain is                                                                                                 specified, use nearest std val]

Step 3: Selection of operating point (Vceq, Icq)

[If Vcc is given,]

Vceq = Vcc/2

[If Vcc not given]

Vceq = 1.5(Vo peak + Vce sat)

Ic peak = Vo peak / RL'

Icq = Ic peak + Ic min
[Unless specified] Assume Ic min = 0 or 0.005 mA

Step 4: Selection of Re

[If Vcc not given]
Assume Vre = 1

[If Vcc given]
Vre = 10% of Vcc

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

Step 5: Selection of Vcc

[This step is necessary if Vcc is not given]
Vcc = Vceq + Icq(Rc + Re)

Assume higher std val [typically 9,12,15,18]

Step 6: Selection of R1 & R2

[If stability factor not given] assume s = 8

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

Find Rb[Do not standardise]

Vr2 = Vbe + Vre
Vr1 = Vcc - Vr2

Assume Vbe = 0.6V [for Si, 0.3 for Ge if not specified]

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

Rb = R1 parallel R2 = (R1 * R2)/(R1 + R2)
Find R2
Select lower standard value to make circuit indepent of beta

Substitute in (A) to find R1
Select higher standard value so that circuit draws minimum current from supply

Step 7: Selection of coupling capacitors

Select higher standard value for all capacitors

Selection of Ce:

Xce = Re/10
Ce = 1/(2*pi * FL * Xce)     [FL = lower cutoff frequency.Assume FL = 20 Hz (For all capacitors)if not specified]

Selection of  Cb:

[Rb = R1 parallel R2]

[If Rs[Source resistance] is not specified assume Rs = 0]

Xcb = Rs +((Rb) parallel (hie))

Cb = 1/(2 * pi * FL * Xcb)

Selection of Cc:

[Rb = R1 parallel R2]

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)

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

To find input impedence[If required]:

[Rb = R1 parallel R2]

Zin = (Rb) parallel (hie)

To find output impedence[If required]:

Zo = RL'

Standard values