To establish a reference power source for my own lab
I have selected one particular 50 ohm dummy load to be my
It is a dummy load from Mini-Circuits.
The DC resistance is 50.35 ohms as measured with a Fluke 8842A
multimeter using four wires.
Figure 1. 50 ohm dummy loads from Mini-Circuits.|
While at R&S in Stockholm Oct 2009 I got a measurement of this particular dummy load measured through a particular BNC to N adaptor. (Now my reference BNC to N adaptor.) The result is shown in figure 2.
Figure 2. Calibrated Smith diagram of the reference
50 ohm dummy load.|
I do not have any reference calibration kit for my network analyzer. It is a HP 8712C and it has not been calibrated in 8 years. When I connect the reference dummy load to it I can read a resistive impedance of 49.90 ohms with a reactive component of +412 milliohms. Since the calibrated R&S analyzer shows 50.413 at 300 kHz with near zero reactance and an increase of only 0.1 ohms up to 100 MHz I can safely assume that the reference dummy load is 50.4 ohms resistive at 14.16 MHz. That means that a 50.0 ohm resistive load should indicate (49.50 +j0.41) on my HP8712C on 14.16 MHz.
The purpose of establishing a reference impedance is to do correct measurements of power levels. Once I knew the calibration I tested all dummy loads I could find and selected the best one which is now my reference dummy load for the future. The impedance is (49.9, j0.05), less than 0.25% error from the nominal impedance 50 ohm resistive.
Definition (for this article):
Power level: The power a particular unit would deliver to a load having exactly the nominal impedance (50 ohms.)
The impedance of the signal source may vary between 25 and 100 ohms (VSWR = 2) Therefore two signal sources having the same power level would not produce the same signal level when compared using a measurement tool that deviates from 50 ohms.
For simplicity I assume the appropriate length of lossles cable to be inserted for the impedance error to be resistive at the signal source to avoid imaginary numbers and to get the worst case. With a source impedance error of X ohms and a load impedance error of Y ohms the transferred power P is like this:
P(50+X)= U*U*(50+Y) / (50+Y+50+X)2
Here U is the electromotoric force of the signal source. To transfer 1 W to a 50 ohm load from VSWR=1 and VSWR=2 impedances the electromotoric force has to be as shown in table 1. The table also shows how much power is transferred to the load when the impedance error is 1% and 10% respectively.
Zs U P(50.5) P(55) (ohms) (V) (W) (W) 25 10.60660 0.99667 0.96680 50 14.14213 1.00000 1.00000 100 21.21320 1.00330 1.03018An error of 1% in the impedance of the load causes an uncertainty of 0.028 dB in the comparison between two signal sources with VSWR=2. A 10 times larger error causes a 10 times larger uncertainty. To properly compare signal sources that are not quite 50 ohms one has to compare the power that they deliver in a load that is accurately matched to 50 ohms.
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