Constant impedance relay-resistor logarithmic attenuator

[Brian Guralnick]

Ya, I just realized that myself, how silly of me. It's the same for my circuit.

[Brian Guralnick]

It was mentioned on another link to this attenuator that it was patented. Whats the US patent #?

[Brian Guralnick]

Ok, I've adapted this pot to simulate a true single-channel pot. Now, any output load & any input load is usable. Also, a volume of 0 is now - (infinity) db. 127 will be -0.000000 db.

I've boxed the DPDT relays. The top half is wired to the n-o (normally open) side, and the bottom is on the n-c (normally closed).

I know it's double the resistors & double the switches, but, I want an authentic emulation of a real logarithmic pot.

Will it emulate correctly?

[x-pro]

Originally posted by Brian Guralnick
It was mentioned on another link to this attenuator that it was patented. Whats the US patent #?

It is NOT patented - only copyrighted . I did publish it in 2002 to prevent someone else from doing it as it is in a public domain now. The question was about your design not mine - I think it was something on patent pending on one of your diagrams?

x-pro.

[x-pro]

Originally posted by Kay
"this circuit is copyright of Creek Audio Limited, however it is already in a public domain for more than a year and half. "


Really?

I saw the circuit first at ELEKTOR 1991/7-8, page 23.

I post this circuit at July on

http://www.audiodiskussion.de/foren/....php?idx=37117

That is a different circuit, though a similar idea . I used only ONE switchover relay contact where that circuit uses two, as would be required for T or P -attenuators as well.

x-pro.

[x-pro]

Originally posted by Brian Guralnick


In the magazine schematic, your input impedance is listed with 10k input impedance. Does this mean I should drive the attenuator with a 0 ohm drive & expect the load of the attenuator to be 10kohm?

What's the level of affect on this attenuator as the output load changes? For example, if I make 2 driving 2 amps, and 1 amp has a load of 10.1k, and the other 9.9k, how messed up will the balance be?
I guess if the output load is unknown, this circuit should have an added output buffer. Sort of kills the ability to use this circuit externally as a passive device.

RF guys done this to match signal lengths & number of resistive elements as the relays are switched.

This is the 1 main beef I have about this circuit.

In my attenuator, the signal being fed through always appears to have 8 switches closed & another 8 open, it also appears that there are always 8 resistors in the signal path.

My final schematic makes this fairly easy to see:
http://pages.infinit.net/helloftp/attschemnew.png
Except now, I need to work out a log version.

You can drive it with any source impedance, however it will introduce an additional (but constant) attenuation. It needs to be loaded on 10 K, and if your load is higher impedance you may add some resistor in parallel to it to make it closer to 10K. Strictly speaking that somewhat lower load impedance, i.e. 5K would only introduce some noticable error for first 10-15 dB of attenuation. In practice it would be quite acceptable. I did use it as a passive device without much of a problem.

x-pro

[x-pro]

Originally posted by Brian Guralnick
Ok, I've adapted this pot to simulate a true single-channel pot. Now, any output load & any input load is usable. Also, a volume of 0 is now - (infinity) db. 127 will be -0.000000 db.

I've boxed the DPDT relays. The top half is wired to the n-o (normally open) side, and the bottom is on the n-c (normally closed).

I know it's double the resistors & double the switches, but, I want an authentic emulation of a real logarithmic pot.

Will it emulate correctly?

My circuit in a first message IS a correct emulation of a log pot. If you want a balanced version you wire two in parallel - one for "hot", one for "cold" input and take two outputs as well. Your idea on that diagram is not quite clear to me . However it would make sense (and I've done it in production) to wire the relays in "max" attenuation position when there is no power to the relays (opposite to what is shown on the diagram from the magazine).

x-pro

[Dzin]

There is a little problem with this circuit, the input impedance is really 10kOhms (if there is a 10k on the output too), but output impedance is changing with the actual relay-settings !!! So it's better to use pi or T-network, if you don't want to compute values of resistors and parallel or serial combinations, you can read something about it and use applet from my page, but it's only in Czech language. Have a nice day with a dictionary

http://dzin.hi-end.org/utlum.htm

Dzin

[Kay]

>I used only ONE switchover relay contact

Yeah, I see,
it will be better to minimize the relay contacts in signal path!

[Kay]

Hi Dzin,

>Czech language. Have a nice day with a dictionary


I like to read, but I don't like a dictionary


The T-switch is the the Elektor circuit, see link above.