Revised July 8,
here to go to our main page on microwave switches
here to go to our comparison of FET and PIN diode switches
here to go to our page on switch FETs and FET switches
here to go to an example of how to use Agilent's ADS to model
Here you will learn how to design
a solid-state 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...)
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!
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).
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
Switch isolation is often limited by package isolation. If you design
a 60 dB switch, you should think carefully about how to package
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.
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!
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?
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.