Steve Eddy
Member of the Trade: The Audio Guild
Aka: TempAccount555
- Joined
- Sep 28, 2003
- Posts
- 6,609
- Likes
- 554
Quote:
I'm not sure exactly what it is you're describing here, but it's certainly not a 100k ladder attenuator.
A ladder attenuator works just like a regular "series" attenuator except the series and shunt portions are split up into two switches, with one switch handling the series resistance and the other the shunt resistance.
The total of each pair of resistors equals the total resistance of the attenuator, which in the case of a 100k attenuator would be 100k. So at the midpoint of attenuation (-6dB), each resistor would be 50k.
The resistors in the ground leg deck at the full attenuation end would be approaching zero ohms, and would not be anywhere close to the attenuator's value.
Edit: Perhaps you mean the series deck at near full attenuation?
k
Originally Posted by jnewman /img/forum/go_quote.gif It makes perfect sense if you're looking at the ground leg deck of the attenuator at the full-attenuation end. A ladder attenuator with 2dB steps starting at -60dB would have 15 or so resistors on the ground leg side that are equal to the nominal attenuator impedance. The photo has 10 visible resistor markings. Example: I have one of the old 23-position Elna ladder attenuator switches. If I made a 100K attenuator, the ground leg resistor for -60dB, -55dB, -50dB, -45dB, and -40dB would all be 100K resistors. The signal legs would be, respectively, 100 ohm, 180 ohm, 340 ohm, 560 ohm, and 1000 ohm. Yes, ideally the -40dB step would actually be 1000 ohm / 99K, but there is no 99K resistor in typical product lines. The closest value is still 100K. Next closest is 97.6K. |
I'm not sure exactly what it is you're describing here, but it's certainly not a 100k ladder attenuator.
A ladder attenuator works just like a regular "series" attenuator except the series and shunt portions are split up into two switches, with one switch handling the series resistance and the other the shunt resistance.
The total of each pair of resistors equals the total resistance of the attenuator, which in the case of a 100k attenuator would be 100k. So at the midpoint of attenuation (-6dB), each resistor would be 50k.
The resistors in the ground leg deck at the full attenuation end would be approaching zero ohms, and would not be anywhere close to the attenuator's value.
Edit: Perhaps you mean the series deck at near full attenuation?
k