Dummy Load
Posted: October 17th, 2013, 11:25 am
Here is a cheap to make 50 ohm load for working on radios also a very accurate way to calculate watts out to check your watt meters for accuracy.
Measuring Power Accurately
For power measurements place a 0.01uf disk ceramic of at least 250V rating be connected between the red and black binding posts. This will charge to the peak voltage applied to the 50-Ohm load, less the diode drop. You can then measure this voltage with your DVM.
Add 0.4V for the forward drop across the BAV21 for a total peak voltage reading of 100V (exmaple).
This is a peak voltage divide by the square root of two to get RMS voltage divide by 1.414.
100 divided by 1.414 equals 70.72 Vrms.
Calculate power take the RMS voltage, square it, and divide by the load impedance, which in our case is 50 Ohms.
70.72)^2 / 50 = 100W
output power, dependent on the accuracy of your DVM If your DVM is accurate, say within 1% on DC voltage measurements, you have nailed your rig's output power within 2%, or 2W
Consider a Bird Wattmeter. Their specified accuracy, when new & calibrated, is 5.0% of full-scale when the measurement is at 1/2 scale. So, for example, a 200W element used to measure the 100W output of your rig can fall within 10W, so your 100W rig might measure 90w to 110W and still be within the calibrated accuracy specification. That's 20W of ambiguity. When power is measured by looking at the peak voltage, accuracy is a function of your DVM accuracy, plus the distortion in your output signal, the total of which may be on the order of 2% - - That's a 98W to 102W measurement, substantially more accurate!
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Measuring Power Accurately
For power measurements place a 0.01uf disk ceramic of at least 250V rating be connected between the red and black binding posts. This will charge to the peak voltage applied to the 50-Ohm load, less the diode drop. You can then measure this voltage with your DVM.
Add 0.4V for the forward drop across the BAV21 for a total peak voltage reading of 100V (exmaple).
This is a peak voltage divide by the square root of two to get RMS voltage divide by 1.414.
100 divided by 1.414 equals 70.72 Vrms.
Calculate power take the RMS voltage, square it, and divide by the load impedance, which in our case is 50 Ohms.
70.72)^2 / 50 = 100W
output power, dependent on the accuracy of your DVM If your DVM is accurate, say within 1% on DC voltage measurements, you have nailed your rig's output power within 2%, or 2W
Consider a Bird Wattmeter. Their specified accuracy, when new & calibrated, is 5.0% of full-scale when the measurement is at 1/2 scale. So, for example, a 200W element used to measure the 100W output of your rig can fall within 10W, so your 100W rig might measure 90w to 110W and still be within the calibrated accuracy specification. That's 20W of ambiguity. When power is measured by looking at the peak voltage, accuracy is a function of your DVM accuracy, plus the distortion in your output signal, the total of which may be on the order of 2% - - That's a 98W to 102W measurement, substantially more accurate!
[Please login or register to view this link]