Designing an Attenuator Pad for a Ham Transceiver

Robert Stampfli / KD8WK — July 2011

Recently, discussions in a ham forum I follow have revolved around the brief initial power overshoot condition that occurs in certain transceivers when they are keyed. When mated with sensitive linear amplifiers, these power peaks can result in the amp being briefly overdriven, with the potential for damage. Since such transceiver overshoots tend to be worse at lower power levels, and are sometimes all but non-existent at full power, some have suggested employing an attenuator pad between the transceiver and the power amplifier as a work-around, so as to mitigate the effects of the overshoot. And, in the limit, a pad can be devised such that the transceiver simply cannot put out enough power to overdrive and/or damage the amp. But, this begs the question: How does one design such a circuit? And that's what it is – a simple electrical engineering circuits problem. I've gone through the math for this, in one form or another, on several occasions in the past, but unfortunately never bothered to save the results. This time, I decided to solve the problem once and for all and record it in the form of a white paper for future reference.

So, what does an attenuator look like? Well, we can make a couple of observations by knowing a bit about its use. First, in this application, the source and intended load are purely resistive (each nominally 50Ω), so the attenuator components need only be resistive as well.

Most of us are already familiar with a somewhat similar device; it is our ubiquitous antenna tuner. We also know that most tuners are based on a Pi- or T-type design and consist of three components. That's because those are the two simplest configurations that can perform a full compliment of transformations. (It is possible to achieve some transformations with only two components, and there are some advantages to this, but it only works when the source and load are of unequal impedances.)

However, the antenna tuner has a slightly different function from an attenuation pad: The tuner is designed to transform impedances and internal resistive losses are actually a unwanted detriment. With an attenuator, the impedance transformation is one-to-one, and a reduction in power is the intent. Although different in functionality, the basic Pi and T architectures are equally applicable to pads. Either will work, and actually there are formulas for converting one to the other. In this white paper, I have arbitrarily chosen to model the T-type attenuator.

So, we start with a network of three resistors in the following configuration:

       +---- R1 ----+---- R3 ----+     
       |            |            |     
Transceiver         |           RL     
 RF Output          R2      (Amp input)
       |            |            |     

Actually, by applying some logic, we can deduce a few additional things about this circuit which will aid us in our analysis. Obviously, for ham radio applications, we need to design for a transceiver impedance of 50Ω. The amp's input impedance should also nominally be 50Ω. Thus

RL = 50 Ω

There is at least one other detail worth noting, and it's something which will make our task easier: Since both the source and load present a 50Ω impedance, the network must be electrically symmetric. In other words

R1 = R3

(This is true of the components' resistance only – the power ratings of the two resistors will not be the same.)

For simplicity, since both R1 and R3 have the same resistance, going forward we'll refer to that resistance as simply R.

Next, let's define a few terms: To make the analysis easier, we'll give a name to the combined resistance of R2, R3, and RL (i.e., the resistance seen from the right side of R1, through these components, to ground). We'll call this RT. We'll refer to the power entering the network (the power output from the transceiver) as Pi, and the power leaving the network (the power supplied to RL) as Po. Finally, we'll define the function "PD(·)" to denote power dissipation, nominally in watts, although the units really don't matter as long as Pi and Po are consistent. Thus, when we use the term Rx, we'll be referring to component Rx's electrical resistance, but when we use the term PD(Rx), we'll be referring to its power dissipation.

What we are trying to derive is a set of formulas for calculating the resistance (as well as a power rating) of R1, R2, and R3, given Pi and Po. So, let's get started.

From the transceiver's perspective, the resistance of the network it sees looks like the following: R1 + ( R2 || ( R3 + RL )). Of course, to achieve our design criteria, the total resistance of this conglomeration of resistances must be equal to 50Ω. Rearranging the terms a bit, substituting R for both R1 and R3, and substituting 50Ω for RL, the above circuit may be represented mathematically as

R2 (R + 50)   + R = 50 Ω
R2 + (R + 50)

Solving for R2

R2 =   502R2   Ω
2 R

Now, let's consider: Our total network can be modeled by two series resistances, R1 and RT, and together these must present an impedance of 50Ω to the source. Therefore, we can conclude

RT = 50 – R

In a series circuit, power divides in proportion to the resistances. Thus

PD(R1) = Pi   R


PD(RT) = Pi   50 – R

In a parallel network, power divides in inverse proportion to the resistances:

PD(R2) = PD(RT)   RT

After plugging the values previously calculated for PD(RT), RT, and R2, into the above equation and simplifying, we get

PD(R2) = Pi   2 R (50 – R)
50 (50 + R)

In a similar manner we can calculate the power dissipated in the R3+RL leg of the circuit as

PD(R3+RL) = Pi   (50 – R)2
50 (50 + R)

Using the above equation and again dividing the power between the two resistances in proportion to their resistance, we determine

PD(R3) = Pi   R (50 – R)2
50 (50 + R)2

And for PD(RL)

PD(RL) = Po = Pi   (50 – R)2
(50 + R)2

And finally, with a bit of algebra, we can solve the above equation for R

R = 50   1 – √Po/Pi   Ω
1 + √Po/Pi

Whew, that was a lot of work! But wait: We know Pi and Po, and from these we can now calculate R. And from R we can calculate everything else we need to design our network.

But first, let's check some boundary conditions to see if it all makes sense: Suppose we wanted to design a network which passes all power to the load. In this case Pi = Po, so Po/Pi = 1. Plugging this into the formula for R makes R = 0. This in turn causes R2 to become infinite. That certainly seems reasonable.

Next, suppose we wanted to design a network that absorbed all the power and delivered none of it to the load: In that case Po would be zero, which yields R = 50Ω and R2 = 0. Yes, that works out nicely, too.

OK, let's get serious: Suppose we wanted to design a 6 dB attenuator for a 100 watt transceiver, so that the power delivered to the load (the amplifier) is 25 watts. In that case Pi = 100 and Po = 25. Plugging these numbers into the equation for R, we get R = 16.66Ω and R2 = 66.66Ω. PD(R1) = 33.33 watts, PD(R2) = 33.33 watts, and PD(R3) = 8.33 watts. Hmm, that leaves exactly 25 watts for the load. Pretty cool!

So, there you have it. Select the values for Pi and Po you desire and you're off and running.

A few caveats: As I mentioned earlier, you can go through a similar set of calculations for a Pi network to come up with an analogous set of formulas. For some attenuations, the Pi configuration may yield more preferable component and power dissipation values. For others, the T will work out better.

Also, please note that at RF frequencies, the choice of resistor can be very important. Some resistors are also inductive, especially the higher wattage ones. These must be avoided in this environment. And at VHF and UHF frequencies and above, even the placement of the components can affect performance, so plan carefully.

Finally, be aware that, unless you construct the network within your amplifier, or make other arrangements, the attenuation will be in the circuit regardless of whether you are receiving or transmitting. In most applications, this won't make much of a difference, although it can limit weak signal reception.