Electronic Design

# Program Designs Tee Attenuator To Minimize Stress

Symmetrical Tee attenuators are designed on an image-impedance basis with RO ohms as the matching resistance. The figure(a) shows the attenuator for RA = RB, power input PJ-1, and power output PJ in watts, where J is a counter variable. Definitions of the terms are:

```RA = RB = RO(1 - √PJ/PJ - 1)
RC = (RO2 - RA2)/2RA
PA = PJ - 1(RA/RO)
PB = PJ(RB/RO)
PC = PDSP - (PA + PB)```

where PDSP is the power dissipation.

Note that single-attenuator designs have the most power-dissipation stress. Cascaded attenuators evenly distribute the necessary dissipation \\[PDSP = (PIN - POUT)/N\\] to reduce stress (Fig. b). Those terms are defined as:

```PDSP = (PIN  - POUT)/N
PJ = PIN  - JPDSP
AJ = 10 log(PJ - 1/PJ) dB```

The program will design one to 10 attenuators in cascade. It prompts for the number of attenuators needed, image impedance RO, power input PIN, and power output POUT. Then for each attenuator, it displays the computed resistor values (RA, RB, RC), resistor dissipations (PA, PB, PC), and attenuation A in decibels.

Consider a single-attenuator design for RO = 100 Ω, PIN = 10 W, and POUT = 1 W. Using the program produces the following results:

```RA = RB = 68.377 Ω, RC = 38.935 Ω
PA = 6.838 W, PB = 0.684 W
PC = 1.479 W```

But a three-network attenuator design with the same requirements ends in:

 Network 1: RA = RB = 16.334 Ω RC = 297.943 Ω PA = 1.633 W PB = 1.143 W PC = 0.223 W A1 = 1.549 dB Network 2: RA = RB = 24.407 Ω RC = 192.655 Ω PA = 1.708 W PB = 0.976 W PC = 0.315 W A2 = 2.430 dB Network 3: RA = RB = 50.000 Ω Rc = 75.000 Ω PA = 2.000 W PB = 0.500 W PC = 0.500 W A3 = 6.021 dB 