Download transformer coupled amplifiers

Transformer Coupling
1
Introduction



Figure shown the transformer coupling to a load resistance
such as a Loudspeaker in a radio.
With a transformer – the small load resistance can be
stepped up to a higher
impedance level


This improve the voltage gain – also since no signal power is
wasted in a collector resistor, all the ac power is delivered to
the final load resistor
Very popular at audio frequencies (20 Hz - 20 kHz)
2
AMPLIFIER
VCC
RB1
Rgen
RC
CC
CC
Vin(ac)
RB2
RL
RE
Vout
AMPLIFIER OBJECTIVE
PROVIDE V, I , POWER to the LOAD most efficiently
3
AMPLIFIER POWER LOSS
VCC
RB1
Ptransistor  VCEQ  I CQ

VBE  I B
Rgen
Vin(ac)

I CQ  RC
PRB1

I RB1  RB1
PRB 2

I RB 2  RB 2
2
CC
CC
(ignore )
PRC
RC
RB2
RL
RE
Vout
if RC not load
2
2
4
PRE

I E  RE
2
TRANSFORMERS V & I transformation
Nprim : Nsec
I & V ONLY ac
Iprim

primary turns
sec ondary turns

N prim
N sec
V prim
Vsec
V prim  I prim
ideally
Pin
Isec
Vsec
Vprim
primary turns
sec ondary turns
n :1

n
1

I sec
I prim
 Vsec  I sec


Pout
5
TRANSFORMERS Z & R transformation
Zprim
Nprim : Nsec
Iprim
I & V ONLY ac
Vprim
V prim

Rin, prim
 n 2  Rsec  n 2  RL
I prim

n  Vsec
 I sec 


 n 
Z in, prim
V
 n 2 sec
I sec
n :1
Isec
Vsec
Zsec
 n 2  Z @ sec
6
TRANSFORMER COUPLED AMPLIFIER
VCC
n :1
RL
Vou
t
n :1
VCC
Vin(ac)
RB1
Rgen
Vin(ac)
RL
Vout
CC
RB2
7
TRANSFORMER COUPLED AMPLIFIER
n :1
VCC
VCC
RB1
RB1
RprimDC
Rgen
RB2
RL
Vout
CC
Vin(ac)
RB2
DC EQUIVALENT CIRCUIT
IDEALLY Rprim,DC = 0
8
TRANSFORMER COUPLED AMPLIFIER
VCC
RB1
Rgen
Rsec,ac
RB1
CC
Vin(ac)
n :1
VCC
Rgen
RB2
Vin(ac)
RL
Vout
CC
RB2
ac EQUIVALENT CIRCUIT
Rprim,ac = n2 RL
9
R prim  RE
DC & AC LOAD LINES
20
19
slope 
18
17
1
R prim
16
15
14
13
mA
12
‘ideal’ DC load line
11
10
Ic
slope 
TRANSFORMER COUPLED AMPLIFIER
1
DC LOAD LINE
if RE in ciruit
9
8
7
Q
6
5
‘Q’ point
determined by
BIAS CIRCUIT
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Vce VOLTS
VCEQ = VCC
10
TRANSFORMER COUPLED AMPLIFIER
ac LOAD LINE
DC & AC LOAD LINES
20
OPERATION
19
18
maximum output voltage
17
16
VCC
R prim , ac
max swing
15
14
‘ideal’ DC load line
13
mA
12
11
I CQ

VCC
Rac  RDC
I CQ

VCC
2
(k  RL )  ( RDC )
Ic
10
VCEQ  VCC  ( RDC  I CQ )
9
8
7
ac load line
Q
6
5
4
slope 
3
2
‘Q’ point
determined by
BIAS CIRCUIT
1
R prim , ac
1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
VCEQ = VCC
Vce VOLTS
2VCC
11
TRANSFORMER COUPLED AMPLIFIERS
Advantages:
electrical isolation: SAFETY
 NO XC voltage drop : if @ input
higher efficiency : no RC power loss
zero or 180 degree phase shift : DOT NOTATION
impedance matching : better power transfer
voltage / current transformation : can be used as a ‘gain’ stage
Disadvantages:
 COST
 WEIGHT
 frequency response
 SIZE : reduces as frequency increases
12
EXAMPLE:
k :1
VCC
20 V
RL
RB1
5
Ri
Rgen
k :1
680
8
CC
4 :1
4 :1
Vin(ac)
Vout
RB2
100
RE
10
  50,VBE (ON )  0.7V , R priDC  5
Determine the amplifier
i) Q-point
ii) Operating Q-point for the max. o/p
iii) i/p impedance, Ri
iv) o/p load power
13
SOLUTION:
i)
Q-point
VBB  2.564V
RBB  87.18
I BQ 
VBB  VBE
2.564  0.7

 3.12mA
RBB  (   1) RE 87.18  (51)(10)
I CQ  156.1mA
VCEQ  VCC  I C RC  (   1) I B RE  20  (3.12m)(5)  (51)(3.12m)(10)  18.39V
ii) Operating Q-point for the max. o/p
VCC
max swing
I CQ 
Rac  RDC
I CQ

VCC
20

 139.86 mA
2
2
(k  RL )  ( RDC )
(4  8)  (5  10)
VCEQ  VCC  ( RDC  I CQ )  20  (15  0.1399) 
11.48 V
14
iii) i/p impedance, Ri
R
Rin i
Rgen
Rinb
k :1
k :1
CC
Ri
RL
Vin(ac)
RB2
RB1
Vout
RE
Ri

k 2  ( RB1 RB 2 Rinb )
Ri

4 2  (680 // 100 //( r  (1   ) RE ))
Ri
 16  (680 // 100 //( 8.33  (51)(10)  16 x(680 // 100 // 518.33)  74.63
15
iii) o/p load power,
I CQ  RLac
2
Pout 
Pout
2
RLac  k 2 RL
(0.131) 2  4 2  8

2
Pout 
0.922 W
16