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1 -<br />

P a e /rom an ¿njineer teb00%<br />

No. 36-Wien<br />

By J. F. SODARO<br />

Bridge Design<br />

Fig. 1: Conventional Wien bridge circuit<br />

THE Wien Bridge network shown<br />

in Fig. 1 is used as a frequency<br />

selective circuit in oscillators and<br />

for audio frequency measurements.<br />

When used with an oscillator,<br />

this circuit is connected<br />

to a suitable amplifier with regen-<br />

erative feedback.<br />

For the mea-<br />

surement of frequency, the unknown<br />

frequency is applied across<br />

upper and lower terminals and a<br />

null detector across right and left -<br />

side terminals. In the design<br />

of this network it is customary to<br />

make RI= R_= R and C1 =C._=<br />

C and R;5 /Ra = 2. For these conditions<br />

the bridge is balanced at<br />

the frequency given by<br />

f =<br />

2,RC<br />

(I)<br />

As an oscillator selective network,<br />

R4 is an incandescent lamp<br />

which controls amplitude by increasing<br />

resistance with current<br />

and thus approaching the bridge<br />

balance value. In this case the<br />

bridge is never quite balanced since<br />

an exact balance would cut -off regenerative<br />

coupling and stop oscillations.<br />

The 'nomograph shown in Fig. 2<br />

may be used to evaluate Eq. (1) .<br />

To use this nomograph locate the<br />

required frequency on an f scale<br />

and pivot a straight -edge about this<br />

point to select combinations of<br />

values on the R and C scales that<br />

give the desired frequency. If the<br />

resistance value is fixed, place the<br />

straight -edge on this R scale value<br />

and on the variable condenser maximum<br />

value on C. Read the low<br />

frequency limit where the straightedge<br />

crosses f. Now pivot the<br />

straight -edge on R to minimum condenser<br />

value on C. Read the high<br />

frequency limit where the straightedge<br />

crosses f. The R, C, and f<br />

scales have split stems. Use left -<br />

side scales to obtain magnitude and<br />

approximate answers. Use right-<br />

ZOOM<br />

10M-<br />

1 M-<br />

10K -<br />

R<br />

10<br />

=9<br />

-= 8<br />

100 -<br />

10 -<br />

7<br />

FZ__ -<br />

t<br />

0.001 -<br />

o 0.01 =<br />

z O U -<br />

W 0.1 =<br />

N m -<br />

w<br />

I<br />

0_<br />

=<br />

= J 10 =<br />

=- 5 r<br />

- U<br />

_ 100 =<br />

=4<br />

- 3<br />

-2<br />

L1000 =<br />

IO Kc<br />

100 Kc =<br />

I<br />

MC =<br />

IOMc=<br />

f<br />

side scales to obtain additional significant<br />

figures.<br />

As an example assume that R is<br />

60,000 ohms and C is 0.005 µf. Find<br />

the balance frequency if the ratio<br />

of R.4 to RI is two -to -one. Using<br />

left -side scale stems connect 60 k<br />

on R to 0.005 µ on C with a straight<br />

line. Estimate 500 cps where this<br />

line crosses f. Next use right -side<br />

scale stems and connect 6 on R and<br />

5 on C with a line. Read 5.3 where<br />

the line crosses f. Thus, the balance<br />

frequency is 530 cps.<br />

- 1.5<br />

- 2<br />

3<br />

4<br />

5<br />

6<br />

- 7<br />

8<br />

- 9<br />

ç-1- - -<br />

C- 2<br />

= 3<br />

4<br />

6<br />

7<br />

r<br />

I-<br />

C<br />

IAJ --10<br />

.I,U -<br />

U -<br />

-Q----<br />

D-<br />

U<br />

.OI,IJ -<br />

0.001 AI 1000 ALU -<br />

I00AJAJ -<br />

10 ALU -<br />

=8<br />

- 7<br />

= 6<br />

5<br />

= 4<br />

-3<br />

-2<br />

J. F. SODARO, Registered Engineer, 2924<br />

Selby Ave., Los Angeles 64, Calif.<br />

TeleTech & ELECTRONIC INDUSTRIES June 1956<br />

- I<br />

1.5<br />

IAJAJ -<br />

- I<br />

93

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