Please, how to calculate output impedance for passive attenuator?

NewBuyer · « **on:** 17 Jun 2007, 08:47 am »

I am curious, how to (at least roughly) determine the input/output impedance values for passive attenuators.

For instance, take the well-known example of the Endler Stepped Attenuators, that attach directly to the input of an amp. These are shunt design attenuators with a single 4k resistor in-line with the incoming (positive) signal, and a varying resistor value on the outgoing (negative) signal that depends on which switch setting is chosen. The standard attenuation values range from -4db (loudest setting) to -46db, in eleven discrete steps.

How does one calculate the input impedance and output impedance values, for the 11 switch settings on such an attenuator?

Thanks to anybody that can please help with this question.

JohnR · « **Reply #1 on:** 17 Jun 2007, 10:16 am »

Since it's a shunt design attenuator, the minimum input impedance will be 4k.

The maximum output impedance will be whatever the total attenuator impedance is specified at, in parallel with (preamp output impedance plus 4k). So about 4k, then, most likely.

(I think I got that right.)

ashok · « **Reply #2 on:** 17 Jun 2007, 03:22 pm »

This picture might help.

VS, ZS represent the preamp or source. ZS is the output impedance of the preamp/source. R1 is fixed at 4K, and R2 varies from a minimum of 0 to a maximum of R2.

Minimum input impedance that the attenuator will present to the source is R1 = 4K (wiper of the pot connected to ground).

Maximum input impedance that the attenuator will present to the source is R1 in series with the parallel combination of R2 and the amplifier's input impedance. If R2 << input impedance of the amp, then max. input impedance = R1 + R2.

For the output impedance calculation, VS will be shorted. It is a voltage source. ZS will then connect to ground. So, looking into the attenuator from the amplifier terminals, the output impedance of the attenuator is (R1 + ZS) in parallel with R2. R1 and ZS are fixed. R2 varies from a minimum of 0 to a maximum of R2.

Minimum output impedance of attenuator = 0
Maximum output impedance of attenuator = (R1 + ZS) || R2. If ZS << R1 << R2, output impedance is effectively R1.

Ashok

NewBuyer · « **Reply #3 on:** 22 Jun 2007, 06:27 am »

Very helpful information, thank you.

Regarding the R2 value for the given example attenuator, please tell me, if I'm on the right track here:

The formula for decibel attenuation (I think) is:
db=20*log[(R1+R2)/R2], using a base-10 logarithm.

Since R1 equals 4k, then an 11-step attenuator ranging from 4db attenuation, to 44db attenuation, would require the following R2 values:

4db attenuation: R2=6.8389k (loudest setting)
8db attenuation: R2=2.6457k
...(etc)...
40db attenuation: R2=0.0404k
44db attenuation: R2=0.0254k (quietest setting)

Please, does this look correct?

1000a · « **Reply #4 on:** 22 Jun 2007, 07:29 am »

are you trying this route to compare to your TVCs?

Audiovista · « **Reply #5 on:** 22 Jun 2007, 11:37 am »

Quote from: NewBuyer on 22 Jun 2007, 06:27 am

Very helpful information, thank you.

Regarding the R2 value for the given example attenuator, please tell me, if I'm on the right track here:

The formula for decibel attenuation (I think) is:
db=20*log[(R1+R2)/R2], using a base-10 logarithm.

Since R1 equals 4k, then an 11-step attenuator ranging from 4db attenuation, to 44db attenuation, would require the following R2 values:

4db attenuation: R2=6.8389k (loudest setting)
8db attenuation: R2=2.6457k
...(etc)...
40db attenuation: R2=0.0404k
44db attenuation: R2=0.0254k (quietest setting)

Please, does this look correct?

Looks about right (I checked the first value only). I usually do attennuation as 20*log(Vout/Vin)=20*log [R2/(R1+R2)], which would give you "negative gain" in dB....

NewBuyer · « **Reply #6 on:** 23 Jun 2007, 07:52 am »

Thank you! So far so good. So we now have reference values for the varying resistor R2.

Still using the Endler 11-step 4k attenuators as an example.
Consider a load with 14k input impedance, and a source with 600 ohm (0.6k) output impedance.

Then the input impedance formula for the attenuator is:
R1 + 1/( 1/14 + 1/R2 ). Since R1 is fixed at 4k, this becomes 4 + 1/( 1/14 + 1/R2 )

The output impedance formula is:
1/( 1 /(R1 + 0.6) + 1/R2 ). Since R1 is fixed at 4k, this becomes 1/( 1/4.6 + 1/R2 )

So using the previously found R2 values, we now have the following attenuator impedances:

4db attenuation: input impedance 8.5945k, output impedance 2.7502k (loudest setting)
8db attenuation: input impedance 6.2252k, output impedance 1.6796k
...(etc)...
40db attenuation: input impedance 4.0403k, output impedance 0.0400k
44db attenuation: input impedance 4.0254k, output impedance 0.0253k (quietest setting)

Do I finally have this right please?

NewBuyer · « **Reply #7 on:** 24 Jun 2007, 02:35 am »

Please, can anybody confirm if I am on the right track? I have wanted to understand the impedances of these passive attenuators for so long, it would be exciting to know if I've finally "got it"...

randytsuch · « **Reply #8 on:** 24 Jun 2007, 05:39 am »

At one point, I found this page
http://www.mhsoft.nl/Mysystem/InputOutputImpedance.asp

it will calculate input and output impedance for you, but you need to make a few simple measurements to do it.

Randy

JohnR · « **Reply #9 on:** 24 Jun 2007, 11:00 am »

I didn't crunch this through a calculator, but your formulas and numbers look about right to me.

Those attenuators look like a good deal.