Power = (( Voltage * 1.414) + 0.15) ^2 / 50
and I normally use this whenever I want an accurate power measurement rather than rely on the readings of analogue meters.
What I don't know is whether this is accurate enough to measure power at levels as low as 1mW. So I asked the question on the Elecraft email reflector, rashly forgetting that the function of of the reflector is for users to flame each other and speculate on or redesign Elecraft products. Don, W3FPR, who was usually good for an answer to a technical question, has left the reflector after he was flamed for some imagined breach of netiquette. Frankly, the reflector is now worthless. I wish Elecraft would create an announcements-only mailing list so that one could stay informed of new developments and firmware updates without having to see all the ego clashes and endless questions about problems with USB serial adapters.When I measured, using the DL1 and my DVM, the output from the K3 transverter port with the level set to 1mW, I got a reading of 36mV. Plugging that into the formula, I get 0.8mW. But the 0.15 is, I presume, a "fudge factor" to compensate for the voltage drop in the detector diode, which gives me a result of 0.45mW even if zero voltage was detected. So I'm wondering if my 0.8mW is within the limits of error of my measurement method and that I should leave my transverter drive level as it is. I assembled my K3 myself and the transverter board was added later so I don't believe the level has ever been set at the factory.
TRy asking support?
ReplyDeleteLet me know what you find, I have the same issue.
73 de M0XDF, K3 #174
Hi Julian,
ReplyDeleteI assume the dummy load uses a diode and capacitor as a peak detector.
So by adding the voltage reading you get to the diode knee voltage (0.15 I guess) you should have the peak voltage across the 50 ohm resistor.
In your case if the knee is 0.15V and you have 0.036V reading. The combined voltage is 0.186V.
Thus the peak power is
V^2 / R = 0.035/50 = 0.7mW
Normally power measurements are stated in RMS so you need to multiply peak power by 0.707 giving 0.5mW
However, I am suspicious of the 0.15V as the 1N5711 has a forward voltage drop of 0.4V at 1mA and higher at higher currents.
A 1 milliwatt signal would produce 0.22V across 50 ohms so 0.4V would be too high to give meaningful results.
It may be useful to make a simple rf probe using a 1n34a germanium diode, a capacitor and a resistor. Then you can read the result on a dvm.
N5ESE has a simple probe schematic on his website.
The germanium diode is necessary for the low knee voltage they have.
If you have an oscilloscope that has the bandwidth, you could read the voltage across the dummy load. That would be the most accurate.
I hope this does not cloud the issue, but an rf probe is a very convenient item and costs almost nothing to make.
73s
I got a response from Don, W3FPR. It won't have been posted to the reflector, and it is interesting, so I will post it here:
ReplyDelete"Actually, the formula in the DL1 manual is overly complex and for low power levels it is wrong. Measure the diode voltage drop and substitute the actual value for your DL1 diode for the "0.33" in the formula P=(Vdvm+0.33)^2/25.
Bob Friess (the DL1 designer) has reviewed this formula derivation and has approved it for inclusion in an updated manual. I have included the derivation details below FYI. The 0.33 volts is what one individual measured with his diode, substitute your actual value.
---------------- DL1 Power Formula derivation -----------------------
The detector used in the DL1 is of the peak reading type (the diode is in series with the DVM.)
The zero to peak voltage across the lower 25 ohm resistor is the DVM reading *plus* the diode forward voltage. The zero to peak voltage across the entire dummy load is 2 times the voltage above.
Converting to RMS, you multiply by sqrt2/2 - so the RMS voltage is 2*(Vdvm+Vdiode)*sqrt2/2. The 2s cancel each other, so the voltage reduces to (Vdvm+Vdiode)*sqrt2. To calculate power, square the voltage and divide by the resistance.
So the power calculation becomes ((Vdvm+Vdiode)*sqrt2)^2/R. Since squaring the square root of 2 equals 2, this reduces to 2*(Vdvm+Vdiode)^2/R.
Since your measured diode forward voltage is 0.33 volts and the dummy load is 50 ohms and plugging those numbers in, we have the result of 2*(Vdvm+0.33)^2/50. Divide everything by 2 and you end up with (Vdvm+0.33)^2/25.
This formula is not an approximation, but the exact result produced by formula reduction."