Applying SWR to Circuit Efficiency - RF Cafe Forums
Post subject: Applying SWR to Circuit Efficiency
Unread postPosted: Mon Mar 14, 2005 2:27 pm
Joined: Mon Mar 14, 2005 1:00 pm
Understanding and applying SWR measurements.
matching two circuits to a 50Ohm load at 915MHz, using a vector
network analyzer. My matching network is a shunt capacitor followed
by series inductor, a standard L-matching network. Both circuits
have the same topology but use a different diode. While matching
the circuits, I also record SWR and use this for efficiency calculations.
After the circuits are matched, I connect each to an RF signal
generator (50Ohm impedance) which is set to 915MHz, 10dBm and then
measure the output voltage of my circuit. Voltage is taken across
a load resistor.
One of my goals is to calculate the circuit
efficiency. I calculate efficiency as the output power of the circuit
divided by input power to the circuit. Input power is set by the
signal generator and output power is calculated by V^2 over R (my
load). I am also interested in output voltage (and hence output
power) alone. So I have two measures for how to rate my circuits,
efficiency and output power.
What I have seen is that a circuit
with lower output voltage but high SWR seems to be more efficient
than a circuit with higher output voltage but low SWR. When looking
at efficiency with SWR applied I am confused by the results. I apply
SWR to Pin as a reduction of this value. SWR tells me that the input
power to the circuit is the incident power (10dBm) minus the reflected
power. I am confident that this part of my calculation is correct.
Here is some data from my tests:
SWR, Pin, Pout,
Eff w/o SWR, Eff w/ SWR
3.00, 10dBm, 8.09dBm, 64.5%, 86.0%
SWR, Pin, Pout, Eff w/o SWR, Eff w/ SWR
10dBm, 8.78dBm, 75.5%, 77.2%
What does it mean that Circuit
1 can operate more efficiently (when applying the SWR) but have
a lower output voltage than the second circuit? Does it make sense
that I should be able to increase the efficiency of Circuit 2 while
maintaining the low SWR? If I could do this, then the output power/voltage
The efficiency calculations that do not
account for SWR make sense to me but the other calculations do not.
Circuit 1 may have a higher efficiency but the power of my source
still needs to be 10dBm for this to happen, it is not like I can
reduce the power with a better match (lower SWR) and have the same
postPosted: Mon Apr 04, 2005 1:44 am
are you measuring SWR
at Pin =10dBm or based on small signal measurement?
Post subject: Efficiency
Unread postPosted: Fri Apr 08, 2005 12:47 pm
What kind of circuit are you working with?
a diode: is this a frequency multiplier, a parametric amplifier,
or some other circuit?
Nonlinear circuits do not work the
same as linear circuits.