Article

Matching a 2N5642: the AN721 input and output networks

Motorola's AN721 is one of the most copied application notes in RF, and section 4.2 is the reason. Here is that design worked through on a chart you can drag, with both the input and the output network.

The problem

Match a 2N5642 to deliver 20 W between a 50 ohm source and a 50 ohm load at 175 MHz, from a 28 V supply, with a maximum input Q of 10.

Two separate matching jobs: the input side, where the transistor's base looks like almost nothing, and the output side, where you have to work out what the collector wants to see before you can match to it.

The input: 1.94 + j1.1 ohms

That number comes off the data sheet, and it is the interesting part of this example. Under drive, the base of a power transistor looks like about two ohms. You are transforming 50 ohms down by a factor of 25, which is a hard thing to do gracefully.

Set the load to 1.94 + j1.1 and the frequency to 175 MHz. The dot sits jammed against the left edge of the chart, which is what a two ohm load looks like. VSWR is 25.8:1.

Draw a Q circle of 10 in the Overlays panel, because that is a design constraint here and not an afterthought. AN721 specifies a maximum input Q of 10, and the whole point of drawing it is that you can see when you are about to violate it.

Three components

Series inductor first. It climbs the constant resistance circle through the load. You stop when you reach the Q circle, not when it looks nice: 16.8 nH gets you to a loaded Q of 10.09, which is the budget, spent.

Shunt capacitor next. This is the one that does the actual transformation, swinging round the constant conductance circle until it meets the 50 ohm constant resistance circle. 38.12 pF.

Series capacitor last. Now the real part is right and only reactance is left over, so a series capacitor slides down the 50 ohm circle into the centre. 10.52 pF.

Look at that bandwidth: 13 MHz out of 175. That is a loaded Q of 10 for you, and it is exactly why AN721 sets a ceiling on it. Push the Q higher and the amplifier stops working at the edges of the band; let it go lower and you cannot make the transformation with three components.

The output: work out the load first

The output side has a step before the matching, and it is the step people skip. You are not matching the transistor's output impedance. You are presenting the collector with the resistance it needs to see in order to deliver 20 W without the collector voltage swinging into saturation.

Rc = (Vcc - Vce(sat))^2 / (2 x Pout)

With 28 V and 20 W out, that is 15.6 ohms. This is a design choice, not a measurement: you are choosing the load line.

Then the transistor's own output capacitance, 40 pF from the data sheet, sits in parallel with it. Convert that parallel combination to its series equivalent and you get 10.6 - j7.3 ohms, which is what the matching network actually sees.

QuickSmith will do that conversion for you: put in the parallel pair and use All values and equivalents in the readout, which prints the series and parallel forms of the same impedance side by side.

Then it is an ordinary two element match

Two components instead of three, because 10.6 ohms to 50 ohms is a ratio of five rather than twenty five, and the L-network's natural Q of about 2 is already inside budget. Nothing needed constraining.

What this example teaches that a simpler one does not

• The load is a design decision, not a given. On the output side you calculated the resistance from the power and supply you wanted. Change the power and the whole network changes.
• Device parasitics are part of the load. The 40 pF of collector capacitance is not something to compensate for later. It goes into the number you match to.
• Q is a specification. Drawing the Q circle first and stopping the first inductor when it touches is a different way of working from picking values and checking afterwards.
• Big transformation ratios need three components. Twenty five to one through two elements would have put the loaded Q near 5, which sounds better, but it cannot be done with the values available and it is not what the note calls for.

Following along

Both networks are in Help → Guided examples as tours that build them a component at a time, saying what each one does. The written versions are Example 6 and the output match.

Every number above is re-solved by the regression suite on each run, against what AN721 states, so the prose here and what the program computes cannot drift apart.