Author: Jindrich Kolorenc, http://orcid.org/0000-0003-2627-8302

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Efficiency of Resistance-Loaded Follower

Figure 1: Sketch of the circuit. The power-supply voltage

is VC , the emitter resistor is R, and the load resistor is RL .

I consider a resistor-loaded emitter (or source) follower shown in figure 1. Without

any signal, the emitter sits at a voltage V which leads to a quiescent collector

current Iq D V =R. With a cosine signal VL .t/ D VL cos.2 t=T / applied across

the load RL , the instantaneous load and collector currents are

t

VL

IL .t/ D

cos 2

;

(1)

RL

T

t V

t

VL

L

IC .t/ D Iq C

cos 2

C

cos 2

:

(2)

RL

T

R

T

The average power flowing to the circuit from the power supply reads as

Z

1 T

VC V

PC D

dt VC IC .t/ D VC Iq D

T 0

R

since the cosine currents average to zero. The power delivered to the load is

Z

Z

t 1 V2

1 T VL2

1 T

L

PL D

dt VL .t/ IL .t/ D

dt

cos2 2

D

:

T 0

T 0

RL

T

2 RL

(3)

(4)

To maximize PL , I need to find the largest signal amplitude VL that is still not

clipped. The clipping at the negative swing starts when the collector current IC .t/,

equation (2), approaches zero for t D T =2,

0 D Iq

VL

RL

VL

R

)

VL.

/

D

Iq

V

V

D

D

:

1

1

R

1Cx

C

1C

R

RL

RL

(5)

To make this and the subsequent expressions shorter, I abbreviated R=RL as x. The

clipping at the positive swing happens when the emitter hits the supply rail,

VC D V C VL

)

1

VL.C/ D VC

V:

(6)

VL

VC

V(+)

L

V(−)

L

0

V

VC

0

Figure 2: Clipping at the positive (red) and

negative (blue) signal swing. The area without clipping is indicated with gray color, the

largest unclipped amplitude corresponds to

the crossing of the two lines.

Apparently, VL. / grows with increasing V whereas VL.C/ decreases. The optimal

setting is when neither of the two clipping conditions happens before the other,

that is, when the two lines (figure 2) intersect:

VC

V D

V

1Cx

)

V D VC

1Cx

:

2Cx

(7)

The largest unclipped amplitude then is

VL D

VC

:

2Cx

(8)

Inserting the optimal voltage across the emitter resistor V , equation (7), and the

maximal signal amplitude VL , equation (8), into formula for the circuit efficiency

D

PL

1 R VL2

D

PC

2 RL VC V

(9)

I get

1

1

1

x

:

(10)

2 1Cx 2Cx

For x 1 the efficiency grows as x=4 and for x 1 it decreases

as 1=.2x/. The maximum is somewhere in between, at the point where

d=dx D 0. This derivative can be written as

D

d

1

2 x2

D

(11)

dx

2 .1 C x/2 .2 C x/2

p

and it vanishes for x D 2. This means that the optimal value of the emitter

resistor comes out as

p

R D RL 2 1:41RL

(12)

2

0.9

1

2

4

8

5

0.8

7

V / VC

0.7

5

8

0.6

4

7

0.5

3

6

2

0.4

3

1

0.3

0

0

1

2

3

4

5

R / RL

efficiency (%)

6

Figure 3: Efficiency

as a function of the

emitter resistor R and

its bias voltage V .

The dashed line corresponds to the symmetrical clipping and

the white dot indicates the maximal efficiency 8:58%.

and its optimal biasing given by equation (7) is

p

V D VC 2=2 0:71VC :

(13)

p

Finally, the maximal efficiency is found by inserting x D 2 into equation (10),

which yields

2

1 p

D

2 1 8:58% :

(14)

2

To illustrate the dependence of the efficiency on the relevant circuit parameters, I plot equation (9) as a function of the emitter resistor R and its bias voltage V

in figure 3. The actual formula is

.C/ . / 2

i2 1

1 R h

1 R

V

V

VC

. /

.C/

D

min VL ; VL

D

min L ; L

2 RL

VC V

2 RL

VC

VC

V

1xh

y i2

D

min 1 y;

; (15)

2y

1Cx

where I introduced y D V =VC besides x D R=RL in the last step.

3

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