# attenuator resistor formulas ?

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#### fran1942

##### Junior Member level 3
Hello, I am trying to work out the resistor values for an 'O' attenuator.
I am using the attached formula to work out resistor 3's value.
My values to insert into this formula are:

Input impedance (Z1) = 50 Ohm
Load impedance (Z2) = 600 Ohm
a =1.995

However when I substitute these figures I am getting a negative number result.
Is this correct ? I mean if you get a negative result, do you just make it positive and use that for your resistor value ?

Thanks for any help.

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• formula.jpg
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#### Kral

I'm not familiar with the 'O' attenuator. Can you post a schematic?

fran1942

### fran1942

Points: 2

#### fran1942

##### Junior Member level 3
yes, here is it. Thanks for your help.

So, when I do the calculation for R3 as per my attached formula, I get a negative number. Obviously I cannot have a negative Ohm resistor, so how do I interpret this outcome ?
Thank you for any insight.

#### Attachments

• oatt.jpg
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#### enjunear

What you're showing is a standard PI pad attenuator (because it looks like the Greek letter, pi). An 'O' pad is also called a balanced PI pad, and consists of 4 resistors in a square. Those are generally installed on balanced transmission lines and can be analyzed like a standard (unbalanced) PI pad by halving the shunt resistance values and halving the voltages.

I haven't run across an unequally terminated PI pad attenuator before, but I'm sure they're out there. If you can suffer the insertion loss hit, you can keep going down this path. The simplest way to find an answer for me would be to simulate the circuit in something like ADS or MWO and setup an optimization simulation to minimize in return loss and insertion loss. If you can find the derivation of those equations in the textbook, the assumptions for the equations will often tell you what setup criterion you're violating (perhaps you need to swap Z1 and Z2?).

The more RF-efficient method would be to use a matching circuit comprised of L's and C's to convert from 50 ohms to 600 ohms. A two-part ("L") match can be found quickly for a single frequency/narrowband system. For wider bandwidths, you can scale up to multiple parts, but tuning the design becomes much more time intensive (but can also be done with an optimizer to perform many iterations very quickly).

fran1942

### fran1942

Points: 2

#### godfreyl

There's a few problems here. The picture you showed in post 3 is a PI attenuator, not an O attenuator. Your formula for R3 is correct for a PI attenuator or an O attenuator but your formula for R1 is only correct for a T attenuator.

Most importantly, for Z1 = 50 Ohms and Z2 = 600 Ohms, the minimum attenuation is about 16.63dB, so "a" must be greater than about 6.78.

Kral and fran1942

Points: 2

Points: 2