TS616IDW(2002) データシートの表示（PDF） - STMicroelectronics

部品番号

コンポーネント説明

メーカー

TS616IDW
(Rev.:2002)

DUAL WIDE BAND OPERATIONAL AMPLIFIER WITH HIGH OUTPUT CURRENT

STMicroelectronics 

TS616

Component Calculation

Let us consider the equivalent circuit for a sin-

gle-ended configuration as shown in Figure 69.

Figure 69: Single ended equivalent circuit

+

Vi

_

½ R1

R2

R3

Rs1

Vo°

Vo

-1

½ RL

Vo

=

V

i





1

+

2--R--R---1--2--

+

RR-----23--

----------------1-----–----RR----------23--------------------

–

R---1--s---–-1---I-RR----o-----23--u-----t ,(

eq3)

By identification of both equations (eq2) and

(eq3), the synthesized impedance is, with

Rs1=Rs2=Rs:

Ro = 1-----–R-----RR-s---------23---- ,(eq4)

Figure 70: Equivalent schematic. Ro is the

synthesized impedance.

Ro

Iout

Let us consider the unloaded system. Assuming

the currents through R1, R2 and R3

are respectively:

2-R---V--1---i, (---V----i---R–-----V2----o----°---) and(---V----i--R-+----3-V-----o---)-

as Vo° equals Vo without load, the gain in this

case becomes :

G = V-----o----(--n----o----l-o----a----d----) = 1-----+-----2----R----R------1----2-------+-----RR----------23----

Vi

1 – RR-----23--

The gain, for the loaded system will be (eq1):

GL = V-----o----(--w-----i-V-t--h-i---l-o----a----d----) = 12-- 1-----+-----1-2---R----R--–----1----2-RR-----------+23-------RR----------23---- ,(eq1)

As shown in Figure70, this system is an ideal gen-

erator with a synthesized impedance as the inter-

nal impedance of the system. From this, the out-

put voltage becomes:

Vo = (ViG) – (RoIout),(eq2)

with Ro the synthesized impedance and Iout the

output current. On the other hand Vo can be ex-

pressed as:

Vi.Gi

1/2RL

Unlike the level Vo° required for a passive imped-

ance, Vo° will be smaller than 2Vo in our case. Let

us write Vo°=kVo with k the matching factor vary-

ing between 1 and 2. Assuming that the current

through R3 is negligible, (eq4) becomes the fol-

lowing :

Ro = R-----L-k---V-+----o-2--R--R---L--s---1--

After choosing the k factor, Rs will equal to

1/2RL(k-1).

A good impedance matching assumes:

R o = 12-- RL,(eq5)

(eq4) and (eq5) give :

RR-----23-- = 1 – 2--R--R---L--s- ,(eq6)

By fixing an arbitrary value of R2, (eq6) becomes :

R3 = 1-----–--R---2----2R----R-----L----s----

Finally, the values of R2 and R3 allow us to extract

R1 from (eq1), and it becomes:

25/27

Share Link: