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lucycling Post subject: Applying SWR to Circuit Efficiency Unread
postPosted: Mon Mar 14, 2005 2:27 pm Offline Lieutenant
Joined: Mon Mar 14, 2005 1:00 pm Posts: 1 Location: Pittsburgh,
PA Understanding and applying SWR measurements. I am 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: Circuit 1 SWR, Pin, Pout, Eff
w/o SWR, Eff w/ SWR 3.00, 10dBm, 8.09dBm, 64.5%, 86.0% Circuit
2 SWR, Pin, Pout, Eff w/o SWR, Eff w/ SWR 1.35, 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 should increase.
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 efficiency. Any thoughts? Thank
you, Kevin Top Profile j_almira Post subject:
Unread postPosted: Mon Apr 04, 2005 1:44 am are you measuring
SWR at Pin =10dBm or based on small signal measurement? Regards,
Jean Top Guest Post subject: Efficiency
Unread postPosted: Fri Apr 08, 2005 12:47 pm Hello! What
kind of circuit are you working with? You mention a diode: is
this a frequency multiplier, a parametric amplifier, or some other circuit?
Nonlinear circuits do not work the same as linear circuits.
Good Luck! Posted
11/12/2012
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