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# Switch design

Revised July 8, 2011

Click here to go to our comparison of FET and PIN diode switches

Click here to go to our page on switch FETs and FET switches

Click here to go to an example of how to use Agilent's ADS to model two-state devices

Here you will learn how to design a solid-state switch.

### SPST switch (switch arm) design

What we define as a switch arm is essentially what would be used to create a SPST switch. We will later tell you how to combine switch arms into SPDT and multi-throw switches. The first thing to consider is whether to use series or shunt elements (or both...)

#### Shunt versus series elements

A shunt FET or PIN diode is a very effective means of shutting off a signal between two ports. But in the case of a SPDT switch, a short circuit on one arm would short out the signal on the other arm. So for a purely shunt structure, a quarter wave section is needed between the common node and the switch elements. The limitation of this structure is bandwidth. You will only get a useful switch over 60 percent bandwidth using shunt elements alone.

Shunt elements become the preferred approach at millimeterwaves. This is because you will never be able to create a truly open circuit with a series element, due to its physical geometry. This matters a great deal for series switches, but is not such a big deal for shunt switches.

Often, a combination of series and shunt elements are used.

New for January 2011! Here's an important consideration for designing with shunt switch elements that you need to know about!

### Some SPDT reflective switch examples

Below are five topologies for reflective SPDT switches, each is used to switch port 1 to port 2 or port 3. Note that FETs are shown, but we could just as well have used PIN diodes. The top two have a single switch element in each arm. These are the simplest switches, but don't offer as much isolation as the others. Expect no more than 30 dB isolation, less if you are above X-band. The quarterwave arms in the shunt switch are used so that the "short circuit" that the FET creates in the off arm behaves like an open circuit at the common node. Visit our page on quarterwave tricks for more info on circuits that use quarter-wavelength transmission lines. The series switch is smaller, and offers more bandwidth than the shunt switch. Both require two complimentary control voltages (for example, -5V/0V logic for MESFETs and some PHEMTs).

 Series SPDT switch Shunt SPDT switch

The next switch on the left is a series/shunt switch. This offers excellent performance in very small size. SPDT MMIC switches use this topology more than any other, 40 dB isolation is no problem.

On the right is an even higher isolation switch, because it adds a third element (another shunt FET). If you need 60 dB isolation, this is the way to go. However, it's time for another Microwaves101 Rule of Thumb!

Switch isolation is often limited by package isolation. If you design a 60 dB switch, you should think carefully about how to package it!

 Series/shunt SPDT switch Series/shunt/shunt SPDT switch

The final switch below (shunt/shunt) is also the biggest. But it can also offer the lowest insertion loss, which is the opposite of what you'd think. This is because the double-shunt structure gets as close to true short circuit as anything, which means that it appears to be a better open circuit at the common node.

 Shunt/shunt SPDT switch

### Non-symmetric SPDT switches

Something to consider... there is no reason that SPDT switches have to be symmetric. A clever way to design a switch with a single control voltage is to use a series element in one arm, and a shunt element in the other!

### The Rule of Six dB

Suppose you design a SPST switch with 40 dB isolation, and one dB insertion loss. Then you use it as a building block for a SPDT switch. Poof, you now have 46 dB isolation, and two dB loss. What is going on?

These two effects are separable, but let's start with the higher loss of the SPDT compared to the SPST switch. First of all, the off-arm of the SPDT must be designed to present an open circuit to the common node of the switch. If a perfect open were possible, the loss of the SPDT would be exactly the same as the loss of the SPST. Too bad you will never experience perfection. In practice, the reflection coefficient of the off-arm could be anywhere from -1 dB to -3 dB match if you did a good job, or as bad as -10 dB match if you didn't get it done. The effect of the off-arm reflection coefficient on insertion loss, given that its reflection angle is zero degrees (closer to an open than a short), is illustrated in the graph below. If your off-arm presents a -1 dB match, you only messed up the SPDT loss by 0.25 dB, and you are the man. As an extreme case, if the off-arm presents a perfect fifty-ohm match, it would eat up 3.5 dB of loss from the on-arm. This makes sense when you think about it: when both arms are matched to 50 ohms, you would expect a 50% voltage split (3 dB) between them. The transferred power is just a little bit worse, due to mismatch loss in a fifty-ohm system, since the combined impedance of the two 50-ohm arms in parallel is 25 ohms.

Increase in SPDT loss versus return loss of idled arm

The increased isolation going from a SPST to a SPDT switch is explained in the two figures below. The first figure describes a SPST switch in the isolation state, with -40 dB isolation. Here the off-arm is modeled by a 4950 ohm series resistor. This forms a 20:1 voltage divider to the isolated output load Z0, which translates to -40 dB drop in signal (=20*log(1/100).

When a second arm is placed at the common node to create a SPDT switch, a second voltage divider is formed between the source impedance Z0 and the on-arm load resistance Z0 as shown below. The net effect of this divider is a 2:1 voltage divider in series with the SPST 20:1 divider, which translates to an extra six dB. This six dB is constant to all switch designs: a 10 dB SPST switch would have 16 dB isolation in a SPDT configuration. Who says you never get anything for free in microwaves?

### Designing high-speed switches

Switching time estimates for FET switches can be simply calculated by looking at the R-C time constant of the bias choke resistor and the FET off capacitance. For example, if a 2K ohm resistor is used on a one millimeter FET, the time constant would be 0.3 nanoseconds (1000 times 0.32 picofarads). Such a design would be capable of settling within 1 nanosecond, which would be considered a "high-speed" switch.

Often the biggest limitation on switching speed is the driver circuit. As we said before, the 74HCT04 hex inverter is good for maybe 10 nanoseconds as a FET switch driver. The fastest switch driver circuits use discrete transistors. Impelliwho? can fix you up with a one-nanosecond PIN diode driver if you ask politely!

Mechanical switches, as well as MEMS switches, have inherently long switching times. Consult your switch vendor if you have questions on these.

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