0
\$\begingroup\$

I wish to design a balanced pi type attenuator to give attenuation of 20dB and characteristic resistance of 500 ohms.

I have found out the necessary resistances required to design the attenuator. However upon simulation on ISIS proteus, the results are not what I expected. For input of 40V, the output is 7.92V instead of 4V. Where am I making a mistake?

circuit I designed

\$\endgroup\$
1

2 Answers 2

1
\$\begingroup\$

Try adding the 500 ohm source resistance and 500 ohm load resistance... If you remove the three attenuator resistors, you'll get 6 dB attenuation from V1-to-load resistor R3.
If you insert the three attenuator PI-resistors, you'll get 26 dB attenuation from V1-to-load resistor R3.

Normally, V1 is not accessible as a bare voltage source...its source resistance of 500 ohms (R4) is part of the source. You would measure attenuator input across R1 (below). Then the attenuator output across R3 (below) would be -20 dB.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
3
  • \$\begingroup\$ Allright, I will try this \$\endgroup\$
    – Sams
    Commented Apr 19, 2020 at 18:01
  • \$\begingroup\$ Iam getting approx. 2V from your approach for 40 V input \$\endgroup\$
    – Sams
    Commented Apr 19, 2020 at 18:05
  • \$\begingroup\$ Yes, that's -26 dB from V1-to-R3. Try measuring attenuation from top-of-R1 to R3. Should be -20 dB. \$\endgroup\$
    – glen_geek
    Commented Apr 19, 2020 at 18:15
0
\$\begingroup\$

Where am I making a mistake?

You forgot to add a 500 ohm termination resistor across the load side. Doing this makes the net resistance seen after the 2475 ohm resistor into 275 ohms and this provides 20 db attenuation. Your source also needs to drive via 500 ohms and this means that without an explicit attenuator in place, there is an implicit attenuation of 2:1.

I wish to design a balanced pi type attenuator to give attenuation of 20 dB and characteristic resistance of 500 ohms.

Also, your circuit isn't a balanced attenuator because you have one side earthed. Maybe you mean balanced in respect of both input and output being the same? Wiki quote from the above link: -

In telecommunications and professional audio, a balanced line or balanced signal pair is a transmission line consisting of two conductors of the same type, each of which have equal impedances along their lengths and equal impedances to ground and to other circuits.

Wiki also tells us about unbalanced transmission lines: -

In electrical engineering, an unbalanced line is a transmission line, often coaxial cable, whose conductors have unequal impedances with respect to ground; as opposed to a balanced line. Micro strip and single-wire lines are also unbalanced lines.

\$\endgroup\$
6
  • \$\begingroup\$ I don't get it. Isn't the necessary condition for balanced attenuators is that souce impedance = load impedence \$\endgroup\$
    – Sams
    Commented Apr 19, 2020 at 18:00
  • \$\begingroup\$ It all depends on what you decide "balanced" means. Normally "balanced" means a balanced transmission line consisting of two conductors that have the same impedance balance to earth. You've interpreted "balanced" as meaning the impedances are the same on both sides and this is less well-known. \$\endgroup\$
    – Andy aka
    Commented Apr 19, 2020 at 18:02
  • \$\begingroup\$ hey thanks man, your solution worked. However why doesn't the second definition of balanced wrong.As it is not giving results, why is this definition still in use? \$\endgroup\$
    – Sams
    Commented Apr 19, 2020 at 18:10
  • \$\begingroup\$ I can only quote what wiki says and back this up from decades of being an EE that I've never heard your version of the word balanced. \$\endgroup\$
    – Andy aka
    Commented Apr 19, 2020 at 18:13
  • \$\begingroup\$ Thanks for your help .Looks like I have missed a detail \$\endgroup\$
    – Sams
    Commented Apr 19, 2020 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.