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lucycling
Post subject: Applying SWR to Circuit Efficiency
Unread postPosted: Mon Mar 14, 2005 2:27 pm
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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


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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


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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!





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