LAST MODIFIED:

An attenuator circuit allows a known source of power to be reduced by a predetermined factor usually expressed as decibels. A powerful advantage of an attenuator is since it is made from non-inductive resistors, the attenuator is able to change a source or load, which might be reactive, into one which is precisely known and resistive. This power reduction is achieved by the attenuator without introducing distortion.

Shown in figure 1 below is the most common attenuator circuit known as the "pi attenuator network". Included in this figure are the formulae for calculating the required resistances R1 and R2. An attenuator may be used in either audio or radio signal circuits.

Figure 1. - pi attenuator network schematic and formula for equal impedances

Note the above formulae are for **equal source and load impedances** of whatever value.

The factor K is called the ratio of current, voltage, or power corresponding to a given value of attenuation "A" expressed in decibels. It is the more difficult calculation which proceeds as follows:

"K" is the number 10 raised to the power of the value of attenuation "A" in dB, divided by 20.

K = 10 ^{("A" / 20)}

As just one practical example, let's look at 3 dB attenuation and calculate our "K" factor.

K = 10 ^{(3 dB / 20)} = 10 ^{(0.15)} = 1.4125

On my calculator I entered 10 and pressed Y^{X} then entered .15

That's really the hardest part of designing a pi attenuator network.

Perhaps the most common use of an attenuator is in 50 ohm radio circuits. To this end I have included a small table in figure 2 below which depicts common power reduction values of 3 db, 6 dB, 10 dB and 20 dB. I suggest you review the topic decibels if you have difficulty understanding the expression dB and the impact of the numerical values.

Figure 2. - Resistor values for a 50 ohm pi attenuator network - equal source and load

Under the columns of "actual" I have included resistors from the E24 series of resistor values. For most practical purposes these values will suffice.

If, as I suggest you reviewed the topic decibels you would know that decibel values **add** together.

Assume we have a 50 ohm amateur transmitter with an output power of five watts. We want to reduce that power down to 250 milli-watts. This means we must **reduce the power** by 4.75 watts.

This power reduction MUST be dissipated in our pi network attenuator. This is a very important point to keep in the back of your mind, I will tell you later as to why.

By simple calculation we know that this is 4.75 / 5.0 = .95 or a 95% reduction in power.

Looked at in another way all we want is 5% of the original power or 1 /20th. On my calculator 5% is expressed as .05 and log .05 = 1.301 which from our tutorial on decibels we know to multiply by 10 because we are dealing in power levels. This means we're looking for a **power reduction** of 13 dB.

Using the formula in figure 1 we will now construct a 13 dB "pi network attenuator" designed for equal source and load impedances of 50 ohms.

We start by determining our "K" factor for 13 dB.

K = 10 ^{(13 dB / 20)} = 10 ^{(0.65)} = 4.4668

Determine R1 where (K + 1) = 4.4668 + 1 = 5.4668 and (K - 1) = 4.4668 - 1 = 3.4668 and (K + 1) / (K - 1) = 5.4668 / 3.4668 = 1.5769

As you can see these mathematics are "real" hard. We multiply this answer 1.5769 by our impedance "Z" which is, 1.5769 X 50 = 78.84 ohms for R1. This is depicted in figure 2 below.

Figure 3. - A practical example of a pi attenuator at work

Now we calculate R2 and again our "K" value for a 13 dB attenuator remains at 4.4668

First we take the square of "K" which of course is 19.95, subtract 1 = 18.95, then divide by "K". This becomes 18.95 / 4.4668 = 4.24

This figure is then multiplied by our impedance "Z" which has been divided by two. In our case we get 4.24 X (50 / 2) = 106 ohms. Now could I make anything simpler to follow?

I'm mighty glad you asked me that question. Overlooked it "meself" once until I noticed smoke arising from my relatively expensive step attenuator box.

Yes..... don't ever forget your power calculations. All are based on ohms law which of course you should know. You do know it don't you? Thought so.

Hokay, if we started with 5W of power into 50 ohms what was the initial RMS voltage at point 1 in our figure above?

We know that P = E ^{2} / R so 5 = E ^{2} / 50

From simple high school algebra we are looking for the square root of 50 ohms X 5W or the square root of 250 which is 15.81V RMS.

Now it tends to get a bit messy. I'll try the relatively simple way. If we start with the final RMS voltage at point 2 in our figure above and doing the exact same calculations we know we are looking for the square root of 50 X 0.25 which equals 3.54V RMS at point 2.

If we started out with 15.81V and ended up with 3.54V then we have lost 15.81 - 3.54 = 12.27V across a 106 ohm resistor. Ohms law again gives us [(12.27 X 12.27) / 106] = 1.42W

Using the same principles, at point 1 with 15.81V across R1 of 78.84 ohms we get, yet again by ohms law, [(15.81 X 15.81) / 78.84] = 3.17W

Also using the same principles, at point 2 with 3.54V across the right hand R1 of 78.84 ohms we get, [(3.54 X 3.54) / 78.84] = 0.16W

Adding these wattages together we get 3.17 + 1.42 + 0.16 = 4.75W

The critical point here is, these **non-inductive** resistors MUST be able to safely dissipate these amounts of power.

Because we have odd resistor values I find it more convenient to parallel higher wattage (non-inductive) resistors together. Look out for 1 and 2 watt types. For example the left hand R1 might comprise 5 X 390 ohm ohm 1W resistors in parallel. This is the equivalent of a 78 ohm 5W resistor.

The 106 ohm resistor could be 2 X 220 ohm in parallel making a 110 ohm 2W resistor. In the real world, depending on the tolerance type you buy, say 5% 2 X 220 ohms could equal anything between 104.5 and 115.5 ohms.

If you parallel odd resistor values e.g. 220 ohm and 200 ohm to get a nominal 104.76 ohm resistor (and you can) be aware the power dissipation will obviously be different across each resistor. Nothing wrong with the idea, just be wary.

Wirewound types are **NOT** acceptable for RF because they are inductive, they have reactances. In audio applications I wouldn't worry, but I could be wrong.

This attenuator network is somewhat different obviously. Notice the different but somewhat similar calculations. I've never used one of these and beyond that I can offer no other comment.

Figure 4. - "T" network attenuator

To me the calculations are no more difficult or easier than the pi network attenuator.

1. Complete your calculations of resistor values in a methodical manner and double check them.

2. Do take power dissipation into account. Allow a good safety margin.

3. Use non-inductive resistors for RF applications.

4. If sufficient demand exists I will present a paper on designing attenuators for unequal terminations.

5. We discussed the "pi" and "T" type attenuators here. Among others, there are the "H", "L", "O" and "U", bridge types, balanced and unbalanced types, ladder types and even attenuators using potentiometers (audio). Again if sufficient demand exists.........

decibels

ohms law

resistors

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Updated 12th November, 2000